LeetCode Solutions In Java .pdf

LeetCode Solutions
Program Creek
Version 0.0

Contents
1 Rotate Array in Java 7
2 Evaluate Reverse Polish Notation 9
3 Solution of Longest Palindromic Substring in Java 11
4 Solution Word Break 15
5 Word Break II 18
6 Word Ladder 20
7 Median of Two Sorted Arrays Java 23
8 Regular Expression Matching in Java 25
9 Merge Intervals 27
10 Insert Interval 29
11 Two Sum 31
12 Two Sum II Input array is sorted 32
13 Two Sum III Data structure design 33
14 3Sum 34
15 4Sum 36
16 3Sum Closest 38
17 String to Integer (atoi) 39
18 Merge Sorted Array 40
19 Valid Parentheses 42
20 Implement strStr() 43
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Contents
21 Set Matrix Zeroes 44
22 Search Insert Position 46
23 Longest Consecutive Sequence Java 47
24 Valid Palindrome 49
25 Spiral Matrix 52
26 Search a 2D Matrix 55
27 Rotate Image 56
28 Triangle 58
29 Distinct Subsequences Total 60
30 Maximum Subarray 62
31 Maximum Product Subarray 63
32 Remove Duplicates from Sorted Array 64
33 Remove Duplicates from Sorted Array II 67
34 Longest Substring Without Repeating Characters 69
35 Longest Substring Which Contains 2 Unique Characters 71
36 Palindrome Partitioning 73
37 Reverse Words in a String 75
38 Find Minimum in Rotated Sorted Array 76
39 Find Minimum in Rotated Sorted Array II 77
40 Find Peak Element 78
41 Min Stack 79
42 Majority Element 80
43 Combination Sum 82
44 Best Time to Buy and Sell Stock 83
45 Best Time to Buy and Sell Stock II 84
Program Creek 3 | 181

Contents
46 Best Time to Buy and Sell Stock III 85
47 Best Time to Buy and Sell Stock IV 86
48 Longest Common Prefix 88
49 Largest Number 89
50 Combinations 90
51 Compare Version Numbers 92
52 Gas Station 93
53 Candy 95
54 Jump Game 96
55 Pascal’s Triangle 97
56 Container With Most Water 98
57 Count and Say 99
58 Repeated DNA Sequences 100
59 Add Two Numbers 101
60 Reorder List 105
61 Linked List Cycle 109
62 Copy List with Random Pointer 111
63 Merge Two Sorted Lists 114
64 Merge k Sorted Lists 116
65 Remove Duplicates from Sorted List 117
66 Partition List 119
67 LRU Cache 121
68 Intersection of Two Linked Lists 124
69 Java PriorityQueue Class Example 125
70 Solution for Binary Tree Preorder Traversal in Java 127
4 | 181 Program Creek

Contents
71 Solution of Binary Tree Inorder Traversal in Java 128
72 Solution of Iterative Binary Tree Postorder Traversal in Java 130
73 Validate Binary Search Tree 131
74 Flatten Binary Tree to Linked List 133
75 Path Sum 134
76 Construct Binary Tree from Inorder and Postorder Traversal 136
77 Convert Sorted Array to Binary Search Tree 137
78 Convert Sorted List to Binary Search Tree 138
79 Minimum Depth of Binary Tree 140
80 Binary Tree Maximum Path Sum 142
81 Balanced Binary Tree 143
82 Symmetric Tree 145
83 Clone Graph Java 146
84 How Developers Sort in Java? 149
85 Solution Merge Sort LinkedList in Java 151
86 Quicksort Array in Java 154
87 Solution Sort a linked list using insertion sort in Java 156
88 Maximum Gap 158
89 Iteration vs. Recursion in Java 160
90 Edit Distance in Java 163
91 Single Number 165
92 Single Number II 166
93 Twitter Codility Problem Max Binary Gap 166
94 Number of 1 Bits 167
95 Reverse Bits 168

Contents
96 Permutations 169
97 Permutations II 171
98 Permutation Sequence 173
99 Generate Parentheses 175
100 Reverse Integer 176
101 Palindrome Number 178
102 Pow(x, n) 179

1 Rotate Array in Java
You may have been using Java for a while. Do you think a simple Java array question
can be a challenge? Let’s use the following problem to test.
Problem: Rotate an array of n elements to the right by k steps. For example, with n
= 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].
How many different ways do you know to solve this problem?
1.1 Solution 1 - Intermediate Array
In a straightforward way, we can create a new array and then copy elements to the
new array. Then change the original array by using System.arraycopy().
public void rotate(int[] nums, int k) {
if(k > nums.length)
k=k%nums.length;
int[] result = new int[nums.length];
for(int i=0; i < k; i++){
result[i] = nums[nums.length-k+i];
}
int j=0;
for(int i=k; i<nums.length; i++){
result[i] = nums[j];
j++;
}
System.arraycopy( result, 0, nums, 0, nums.length );
}
Space is O(n) and time is O(n).
1.2 Solution 2 - Bubble Rotate
Can we do this in O(1) space?
This solution is like a bubble sort.
public static void rotate(int[] arr, int order) {
if (arr == null || order < 0) {
throw new IllegalArgumentException("Illegal argument!");
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1 Rotate Array in Java
}
for (int i = 0; i < order; i++) {
for (int j = arr.length - 1; j > 0; j--) {
int temp = arr[j];
arr[j] = arr[j - 1];
arr[j - 1] = temp;
}
}
}
However, the time is O(n*k).
1.3 Solution 3 - Reversal
Can we do this in O(1) space and in O(n) time? The following solution does.
Assuming we are given 1,2,3,4,5,6 and order 2. The basic idea is:
1. Divide the array two parts: 1,2,3,4 and 5, 6
2. Rotate first part: 4,3,2,1,5,6
3. Rotate second part: 4,3,2,1,6,5
4. Rotate the whole array: 5,6,1,2,3,4
public static void rotate(int[] arr, int order) {
order = order % arr.length;
if (arr == null || order < 0) {
throw new IllegalArgumentException("Illegal argument!");
}
//length of first part
int a = arr.length - order;
reverse(arr, 0, a-1);
reverse(arr, a, arr.length-1);
reverse(arr, 0, arr.length-1);
}
public static void reverse(int[] arr, int left, int right){
if(arr == null || arr.length == 1)
return;
while(left < right){
int temp = arr[left];
arr[left] = arr[right];
arr[right] = temp;
left++;
right--;

}
}
2 Evaluate Reverse Polish Notation
The problem:
Evaluate the value of an arithmetic expression in Reverse Polish Notation.
Valid operators are +, -, *, /. Each operand may be an integer or another
expression.
Some examples:
["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6
2.1 Naive Approach
This problem is simple. After understanding the problem, we should quickly realize
that this problem can be solved by using a stack. We can loop through each element
in the given array. When it is a number, push it to the stack. When it is an operator,
pop two numbers from the stack, do the calculation, and push back the result.
The following is the code. It runs great by feeding a small test. However, this code
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2 Evaluate Reverse Polish Notation
contains compilation errors in leetcode. Why?
public class Test {
public static void main(String[] args) throws IOException {
String[] tokens = new String[] { "2", "1", "+", "3", "*" };
System.out.println(evalRPN(tokens));
}
public static int evalRPN(String[] tokens) {
int returnValue = 0;
String operators = "+-*/";
Stack<String> stack = new Stack<String>();
for (String t : tokens) {
if (!operators.contains(t)) {
stack.push(t);
} else {
int a = Integer.valueOf(stack.pop());
int b = Integer.valueOf(stack.pop());
switch (t) {
case "+":
stack.push(String.valueOf(a + b));
break;
case "-":
stack.push(String.valueOf(b - a));
break;
case "*":
stack.push(String.valueOf(a * b));
break;
case "/":
stack.push(String.valueOf(b / a));
break;
}
}
}
returnValue = Integer.valueOf(stack.pop());
return returnValue;
}
}
The problem is that switch string statement is only available from JDK 1.7. Leetcode
apparently use versions below that.

2.2 Accepted Solution
If you want to use switch statement, you can convert the above by using the following
code which use the index of a string "+-*/".
public class Solution {
public int evalRPN(String[] tokens) {
int returnValue = 0;
String operators = "+-*/";
Stack<String> stack = new Stack<String>();
for(String t : tokens){
if(!operators.contains(t)){
stack.push(t);
}else{
int a = Integer.valueOf(stack.pop());
int b = Integer.valueOf(stack.pop());
int index = operators.indexOf(t);
switch(index){
case 0:
stack.push(String.valueOf(a+b));
break;
case 1:
stack.push(String.valueOf(b-a));
break;
case 2:
stack.push(String.valueOf(a*b));
break;
case 3:
stack.push(String.valueOf(b/a));
break;
}
}
}
returnValue = Integer.valueOf(stack.pop());
return returnValue;
}
}
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3 Solution of Longest Palindromic Substring in Java
3 Solution of Longest Palindromic
Substring in Java
Finding the longest palindromic substring is a classic problem of coding interview. In
this post, I will summarize 3 different solutions for this problem.
3.1 Naive Approach
Naively, we can simply examine every substring and check if it is palindromic. The
time complexity is O(n3̂). If this is submitted to LeetCode onlinejudge, an error mes-
sage will be returned - "Time Limit Exceeded". Therefore, this approach is just a start,
we need a better algorithm.
public static String longestPalindrome1(String s) {
int maxPalinLength = 0;
String longestPalindrome = null;
int length = s.length();
// check all possible sub strings
for (int i = 0; i < length; i++) {
for (int j = i + 1; j < length; j++) {
int len = j - i;
String curr = s.substring(i, j + 1);
if (isPalindrome(curr)) {
if (len > maxPalinLength) {
longestPalindrome = curr;
maxPalinLength = len;
}
}
}
}
return longestPalindrome;
}
public static boolean isPalindrome(String s) {
for (int i = 0; i < s.length() - 1; i++) {
if (s.charAt(i) != s.charAt(s.length() - 1 - i)) {
return false;
}
}
return true;
}

3.2 Dynamic Programming
Let s be the input string, i and j are two indices of the string.
Define a 2-dimension array "table" and let table[i][j] denote whether substring from
i to j is palindrome.
Start condition:
table[i][i] == 1;
table[i][i+1] == 1 => s.charAt(i) == s.charAt(i+1)
Changing condition:
table[i+1][j-1] == 1 && s.charAt(i) == s.charAt(j)
=>
table[i][j] == 1
Time O(n2̂) Space O(n2̂)
public static String longestPalindrome2(String s) {
if (s == null)
return null;
if(s.length() <=1)
return s;
int maxLen = 0;
String longestStr = null;
int[][] table = new int[length][length];
//every single letter is palindrome
for (int i = 0; i < length; i++) {
table[i][i] = 1;
}
printTable(table);
//e.g. bcba
//two consecutive same letters are palindrome
for (int i = 0; i <= length - 2; i++) {
if (s.charAt(i) == s.charAt(i + 1)){
table[i][i + 1] = 1;
longestStr = s.substring(i, i + 2);
}
}
printTable(table);
//condition for calculate whole table
for (int l = 3; l <= length; l++) {
for (int i = 0; i <= length-l; i++) {

int j = i + l - 1;
if (s.charAt(i) == s.charAt(j)) {
table[i][j] = table[i + 1][j - 1];
if (table[i][j] == 1 && l > maxLen)
longestStr = s.substring(i, j + 1);
} else {
table[i][j] = 0;
}
printTable(table);
}
}
return longestStr;
}
public static void printTable(int[][] x){
for(int [] y : x){
for(int z: y){
System.out.print(z + " ");
}
System.out.println();
}
System.out.println("------");
}
Given an input, we can use printTable method to examine the table after each itera-
tion. For example, if input string is "dabcba", the final matrix would be the following:
1 0 0 0 0 0
0 1 0 0 0 1
0 0 1 0 1 0
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1
From the table, we can clear see that the longest string is in cell table[1][5].
3.3 Simple Algorithm
From Yifan’s comment below.
Time O(n2̂), Space O(1)
public String longestPalindrome(String s) {
if (s.isEmpty()) {
return null;
}
if (s.length() == 1) {
return s;
}

String longest = s.substring(0, 1);
for (int i = 0; i < s.length(); i++) {
// get longest palindrome with center of i
String tmp = helper(s, i, i);
if (tmp.length() > longest.length()) {
longest = tmp;
}
// get longest palindrome with center of i, i+1
tmp = helper(s, i, i + 1);
if (tmp.length() > longest.length()) {
longest = tmp;
}
}
return longest;
}
// Given a center, either one letter or two letter,
// Find longest palindrome
public String helper(String s, int begin, int end) {
while (begin >= 0 && end <= s.length() - 1 && s.charAt(begin) ==
s.charAt(end)) {
begin--;
end++;
}
return s.substring(begin + 1, end);
}
3.4 Manacher’s Algorithm
Manacher’s algorithm is much more complicated to figure out, even though it will
bring benefit of time complexity of O(n).
Since it is not typical, there is no need to waste time on that.
4 Solution Word Break
Given a string s and a dictionary of words dict, determine if s can be segmented into
a space-separated sequence of one or more dictionary words. For example, given s =
"leetcode", dict = ["leet", "code"]. Return true because "leetcode" can be segmented as
"leet code".
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4.1 Naive Approach
This problem can be solve by using a naive approach, which is trivial. A discussion
can always start from that though.
public boolean wordBreak(String s, Set<String> dict) {
return wordBreakHelper(s, dict, 0);
}
public boolean wordBreakHelper(String s, Set<String> dict, int start){
if(start == s.length())
return true;
for(String a: dict){
int len = a.length();
int end = start+len;
//end index should be <= string length
if(end > s.length())
continue;
if(s.substring(start, start+len).equals(a))
if(wordBreakHelper(s, dict, start+len))
return true;
}
return false;
}
}
Time: O(n2̂)
This solution exceeds the time limit.
The key to solve this problem by using dynamic programming approach:
• Define an array t[] such that t[i]==true =>0-(i-1) can be segmented using dictio-
nary
• Initial state t[0] == true
public boolean wordBreak(String s, Set<String> dict) {
boolean[] t = new boolean[s.length()+1];
t[0] = true; //set first to be true, why?
//Because we need initial state
for(int i=0; i<s.length(); i++){

//should continue from match position
if(!t[i])
continue;
for(String a: dict){
int len = a.length();
int end = i + len;
continue;
if(t[end]) continue;
if(s.substring(i, end).equals(a)){
t[end] = true;
}
}
}
return t[s.length()];
}
}
Time: O(string length * dict size)
One tricky part of this solution is the case:
INPUT: "programcreek", ["programcree","program","creek"].
We should get all possible matches, not stop at "programcree".
4.3 Regular Expression
The problem is supposed to be equivalent to matching the regexp (leet|code)*, which
means that it can be solved by building a DFA in O(2m̂) and executing it in O(n).
(Thanks to hdante.) Leetcode online judge does not allow using Pattern class though.
public static void main(String[] args) {
HashSet<String> dict = new HashSet<String>();
dict.add("go");
dict.add("goal");
dict.add("goals");
dict.add("special");
StringBuilder sb = new StringBuilder();
for(String s: dict){
sb.append(s + "|");
}
String pattern = sb.toString().substring(0, sb.length()-1);

pattern = "("+pattern+")*";
Pattern p = Pattern.compile(pattern);
Matcher m = p.matcher("goalspecial");
if(m.matches()){
System.out.println("match");
}
}
4.4 The More Interesting Problem
The dynamic solution can tell us whether the string can be broken to words, but can
not tell us what words the string is broken to. So how to get those words?
Check out Word Break II.
5 Word Break II
Given a string s and a dictionary of words dict, add spaces in s to construct a sentence
where each word is a valid dictionary word. Return all such possible sentences.
For example, given s = "catsanddog", dict = ["cat", "cats", "and", "sand", "dog"], the
solution is ["cats and dog", "cat sand dog"].
5.1 Java Solution - Dynamic Programming
This problem is very similar to Word Break. Instead of using a boolean array to track
the match positions, we need to track the actual words. Then we can use depth first
search to get all the possible paths, i.e., the list of strings.
The following diagram shows the structure of the tracking array.
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5 Word Break II
public static List<String> wordBreak(String s, Set<String> dict) {
//create an array of ArrayList<String>
List<String> dp[] = new ArrayList[s.length()+1];
dp[0] = new ArrayList<String>();
for(int i=0; i<s.length(); i++){
if( dp[i] == null )
continue;
for(String word:dict){
int len = word.length();
int end = i+len;
continue;
if(s.substring(i,end).equals(word)){
if(dp[end] == null){
dp[end] = new ArrayList<String>();
}
dp[end].add(word);
}
}
}
List<String> result = new LinkedList<String>();
if(dp[s.length()] == null)
return result;

ArrayList<String> temp = new ArrayList<String>();
dfs(dp, s.length(), result, temp);
return result;
}
public static void dfs(List<String> dp[],int end,List<String> result,
ArrayList<String> tmp){
if(end <= 0){
String path = tmp.get(tmp.size()-1);
for(int i=tmp.size()-2; i>=0; i--){
path += " " + tmp.get(i) ;
}
result.add(path);
return;
}
for(String str : dp[end]){
tmp.add(str);
dfs(dp, end-str.length(), result, tmp);
tmp.remove(tmp.size()-1);
}
}
6 Word Ladder
The problem:
Given two words (start and end), and a dictionary, find the length of shortest trans-
formation sequence from start to end, such that:
Only one letter can be changed at a time Each intermediate word must exist in the
dictionary For example,
Given:
start = "hit"
end = "cog"
dict = ["hot","dot","dog","lot","log"]
As one shortest transformation is "hit" ->"hot" ->"dot" ->"dog" ->"cog", the program
should return its length 5.
Note: Return 0 if there is no such transformation sequence. All words have the same
length. All words contain only lowercase alphabetic characters.
This problem is a classic problem that has been asked frequently during interviews.
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6 Word Ladder
The following are two Java solutions.
6.1 Naive Approach
In a simplest way, we can start from start word, change one character each time, if it
is in the dictionary, we continue with the replaced word, until start == end.
public int ladderLength(String start, String end, HashSet<String> dict) {
int len=0;
HashSet<String> visited = new HashSet<String>();
for(int i=0; i<start.length(); i++){
char[] startArr = start.toCharArray();
for(char c=’a’; c<=’z’; c++){
if(c==start.toCharArray()[i]){
continue;
}
startArr[i] = c;
String temp = new String(startArr);
if(dict.contains(temp)){
len++;
start = temp;
if(temp.equals(end)){
return len;
}
}
}
}
return len;
}
}
Apparently, this is not good enough. The following example exactly shows the
problem. It can not find optimal path. The output is 3, but it actually only takes 2.
Input: "a", "c", ["a","b","c"]
Output: 3
Expected: 2
6.2 Breath First Search
So we quickly realize that this looks like a tree searching problem for which breath
first guarantees the optimal solution.

6 Word Ladder
Assuming we have some words in the dictionary, and the start is "hit" as shown in
the diagram below.
We can use two queues to traverse the tree, one stores the nodes, the other stores the
step numbers.
Updated on 2/27/2015.
public int ladderLength(String start, String end, HashSet<String> dict) {
if (dict.size() == 0)
return 0;
dict.add(end);
LinkedList<String> wordQueue = new LinkedList<String>();
LinkedList<Integer> distanceQueue = new LinkedList<Integer>();
wordQueue.add(start);
distanceQueue.add(1);
//track the shortest path
int result = Integer.MAX_VALUE;
while (!wordQueue.isEmpty()) {
String currWord = wordQueue.pop();
Integer currDistance = distanceQueue.pop();
if (currWord.equals(end)) {
result = Math.min(result, currDistance);
}
for (int i = 0; i < currWord.length(); i++) {
char[] currCharArr = currWord.toCharArray();
for (char c = ’a’; c <= ’z’; c++) {

currCharArr[i] = c;
String newWord = new String(currCharArr);
if (dict.contains(newWord)) {
wordQueue.add(newWord);
distanceQueue.add(currDistance + 1);
dict.remove(newWord);
}
}
}
}
if (result < Integer.MAX_VALUE)
return result;
else
return 0;
}
6.3 What learned from this problem?
• Use breath-first or depth-first search to solve problems
• Use two queues, one for words and another for counting
7 Median of Two Sorted Arrays Java
LeetCode Problem:
There are two sorted arrays A and B of size m and n respectively. Find the median of the
two sorted arrays. The overall run time complexity should be O(log (m+n)).
7.1 Java Solution
This problem can be converted to the problem of finding kth element, k is (A’s length
+ B’ Length)/2.
If any of the two arrays is empty, then the kth element is the non-empty array’s kth
element. If k == 0, the kth element is the first element of A or B.
For normal cases(all other cases), we need to move the pointer at the pace of half of
an array length.
public static double findMedianSortedArrays(int A[], int B[]) {
int m = A.length;
int n = B.length;
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if ((m + n) % 2 != 0) // odd
return (double) findKth(A, B, (m + n) / 2, 0, m - 1, 0, n - 1);
else { // even
return (findKth(A, B, (m + n) / 2, 0, m - 1, 0, n - 1)
+ findKth(A, B, (m + n) / 2 - 1, 0, m - 1, 0, n - 1)) * 0.5;
}
}
public static int findKth(int A[], int B[], int k,
int aStart, int aEnd, int bStart, int bEnd) {
int aLen = aEnd - aStart + 1;
int bLen = bEnd - bStart + 1;
// Handle special cases
if (aLen == 0)
return B[bStart + k];
if (bLen == 0)
return A[aStart + k];
if (k == 0)
return A[aStart] < B[bStart] ? A[aStart] : B[bStart];
int aMid = aLen * k / (aLen + bLen); // a’s middle count
int bMid = k - aMid - 1; // b’s middle count
// make aMid and bMid to be array index
aMid = aMid + aStart;
bMid = bMid + bStart;
if (A[aMid] > B[bMid]) {
k = k - (bMid - bStart + 1);
aEnd = aMid;
bStart = bMid + 1;
} else {
k = k - (aMid - aStart + 1);
bEnd = bMid;
aStart = aMid + 1;
}
return findKth(A, B, k, aStart, aEnd, bStart, bEnd);
}
7.2 The Steps of the Algorithm
Thanks to Gunner86. The description of the algorithm is awesome!
1) Calculate the medians m1 and m2 of the input arrays ar1[] and ar2[] respectively.
2) If m1 and m2 both are equal then we are done, and return m1 (or m2) 3) If m1
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8 Regular Expression Matching in Java
is greater than m2, then median is present in one of the below two subarrays. a)
From first element of ar1 to m1 (ar1[0...|_n/2_|]) b) From m2 to last element of ar2
(ar2[|_n/2_|...n-1]) 4) If m2 is greater than m1, then median is present in one of the
below two subarrays. a) From m1 to last element of ar1 (ar1[|_n/2_|...n-1]) b) From
first element of ar2 to m2 (ar2[0...|_n/2_|]) 5) Repeat the above process until size of
both the subarrays becomes 2. 6) If size of the two arrays is 2 then use below formula
to get the median. Median = (max(ar1[0], ar2[0]) + min(ar1[1], ar2[1]))/2
Problem:
Implement regular expression matching with support for ’.’ and ’*’.
’.’ Matches any single character.
’*’ Matches zero or more of the preceding element.
The matching should cover the entire input string (not partial).
The function prototype should be:
bool isMatch(const char *s, const char *p)
Some examples:
isMatch("aa","a") return false
isMatch("aa","aa") return true
isMatch("aaa","aa") return false
isMatch("aa", "a*") return true
isMatch("aa", ".*") return true
isMatch("ab", ".*") return true
isMatch("aab", "c*a*b") return true
8.1 Analysis
First of all, this is one of the most difficulty problems. It is hard to handle many cases.
The problem should be simplified to handle 2 basic cases:
• the second char of pattern is "*"
• the second char of pattern is not "*"
For the 1st case, if the first char of pattern is not ".", the first char of pattern and
string should be the same. Then continue to match the left part.
For the 2nd case, if the first char of pattern is "." or first char of pattern == the first i
char of string, continue to match the left.
Be careful about the offset.

8.2 Java Solution 1 (Short)
The following Java solution is accepted.
public boolean isMatch(String s, String p) {
if(p.length() == 0)
return s.length() == 0;
//p’s length 1 is special case
if(p.length() == 1 || p.charAt(1) != ’*’){
if(s.length() < 1 || (p.charAt(0) != ’.’ && s.charAt(0) !=
p.charAt(0)))
return false;
return isMatch(s.substring(1), p.substring(1));
}else{
int len = s.length();
int i = -1;
while(i<len && (i < 0 || p.charAt(0) == ’.’ || p.charAt(0) ==
s.charAt(i))){
if(isMatch(s.substring(i+1), p.substring(2)))
return true;
i++;
}
return false;
}
}
}
8.3 Java Solution 2 (More Readable)
public boolean isMatch(String s, String p) {
// base case
if (p.length() == 0) {
return s.length() == 0;
}
// special case
if (p.length() == 1) {
// if the length of s is 0, return false
if (s.length() < 1) {
return false;
}

//if the first does not match, return false
else if ((p.charAt(0) != s.charAt(0)) && (p.charAt(0) != ’.’)) {
return false;
}
// otherwise, compare the rest of the string of s and p.
else {
}
}
// case 1: when the second char of p is not ’*’
if (p.charAt(1) != ’*’) {
if (s.length() < 1) {
return false;
}
if ((p.charAt(0) != s.charAt(0)) && (p.charAt(0) != ’.’)) {
return false;
} else {
}
}
// case 2: when the second char of p is ’*’, complex case.
else {
//case 2.1: a char & ’*’ can stand for 0 element
if (isMatch(s, p.substring(2))) {
return true;
}
//case 2.2: a char & ’*’ can stand for 1 or more preceding element,
//so try every sub string
int i = 0;
while (i<s.length() && (s.charAt(i)==p.charAt(0) || p.charAt(0)==’.’)){
if (isMatch(s.substring(i + 1), p.substring(2))) {
return true;
}
i++;
}
return false;
}
}
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9 Merge Intervals
9 Merge Intervals
Problem:
Given a collection of intervals, merge all overlapping intervals.
For example,
Given [1,3],[2,6],[8,10],[15,18],
return [1,6],[8,10],[15,18].
9.1 Thoughts of This Problem
The key to solve this problem is defining a Comparator first to sort the arraylist of
Intevals. And then merge some intervals.
The take-away message from this problem is utilizing the advantage of sorted list/ar-
ray.
9.2 Java Solution
class Interval {
int start;
int end;
Interval() {
start = 0;
end = 0;
}
Interval(int s, int e) {
start = s;
end = e;
}
}
public ArrayList<Interval> merge(ArrayList<Interval> intervals) {
if (intervals == null || intervals.size() <= 1)
return intervals;
// sort intervals by using self-defined Comparator
Collections.sort(intervals, new IntervalComparator());
ArrayList<Interval> result = new ArrayList<Interval>();
Interval prev = intervals.get(0);

for (int i = 1; i < intervals.size(); i++) {
Interval curr = intervals.get(i);
if (prev.end >= curr.start) {
// merged case
Interval merged = new Interval(prev.start, Math.max(prev.end,
curr.end));
prev = merged;
} else {
result.add(prev);
prev = curr;
}
}
result.add(prev);
return result;
}
}
class IntervalComparator implements Comparator<Interval> {
public int compare(Interval i1, Interval i2) {
return i1.start - i2.start;
}
}
10 Insert Interval
Problem:
Given a set of non-overlapping & sorted intervals, insert a new interval into the intervals
(merge if necessary).
Example 1:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].
Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as
[1,2],[3,10],[12,16].
This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].
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10 Insert Interval
10.1 Thoughts of This Problem
Quickly summarize 3 cases. Whenever there is intersection, created a new interval.
10.2 Java Solution
/**
* Definition for an interval.
* public class Interval {
* int start;
* int end;
* Interval() { start = 0; end = 0; }
* Interval(int s, int e) { start = s; end = e; }
* }
*/
public ArrayList<Interval> insert(ArrayList<Interval> intervals, Interval
newInterval) {
ArrayList<Interval> result = new ArrayList<Interval>();
for(Interval interval: intervals){
if(interval.end < newInterval.start){
result.add(interval);
}else if(interval.start > newInterval.end){
result.add(newInterval);
newInterval = interval;
}else if(interval.end >= newInterval.start || interval.start <=
newInterval.end){

newInterval = new Interval(Math.min(interval.start,
newInterval.start), Math.max(newInterval.end, interval.end));
}
}
result.add(newInterval);
return result;
}
}
11 Two Sum
Given an array of integers, find two numbers such that they add up to a specific target
number.
The function twoSum should return indices of the two numbers such that they add
up to the target, where index1 must be less than index2. Please note that your returned
answers (both index1 and index2) are not zero-based.
For example:
Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2
11.1 Naive Approach
This problem is pretty straightforward. We can simply examine every possible pair of
numbers in this integer array.
Time complexity in worst case: O(n2̂).
public static int[] twoSum(int[] numbers, int target) {
int[] ret = new int[2];
for (int i = 0; i < numbers.length; i++) {
for (int j = i + 1; j < numbers.length; j++) {
if (numbers[i] + numbers[j] == target) {
ret[0] = i + 1;
ret[1] = j + 1;
}
}
}
return ret;
}
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Can we do better?
11.2 Better Solution
Use HashMap to store the target value.
public int[] twoSum(int[] numbers, int target) {
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int[] result = new int[2];
for (int i = 0; i < numbers.length; i++) {
if (map.containsKey(numbers[i])) {
int index = map.get(numbers[i]);
result[0] = index+1 ;
result[1] = i+1;
break;
} else {
map.put(target - numbers[i], i);
}
}
return result;
}
}
Time complexity depends on the put and get operations of HashMap which is nor-
mally O(1).
Time complexity of this solution: O(n).
12 Two Sum II Input array is sorted
This problem is similar to Two Sum.
To solve this problem, we can use two points to scan the array from both sides. See
Java solution below:
public int[] twoSum(int[] numbers, int target) {
if (numbers == null || numbers.length == 0)
return null;
int i = 0;
int j = numbers.length - 1;
while (i < j) {
int x = numbers[i] + numbers[j];
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if (x < target) {
++i;
} else if (x > target) {
j--;
} else {
return new int[] { i + 1, j + 1 };
}
}
return null;
}
13 Two Sum III Data structure design
Design and implement a TwoSum class. It should support the following operations:
add and find.
add - Add the number to an internal data structure. find - Find if there exists any
pair of numbers which sum is equal to the value.
For example,
add(1);
add(3);
add(5);
find(4) -> true
find(7) -> false
13.1 Java Solution
Since the desired class need add and get operations, HashMap is a good option for
this purpose.
public class TwoSum {
private HashMap<Integer, Integer> elements = new HashMap<Integer,
Integer>();
public void add(int number) {
if (elements.containsKey(number)) {
elements.put(number, elements.get(number) + 1);
} else {
elements.put(number, 1);
}
}
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public boolean find(int value) {
for (Integer i : elements.keySet()) {
int target = value - i;
if (elements.containsKey(target)) {
if (i == target && elements.get(target) < 2) {
continue;
}
return true;
}
}
return false;
}
}
14 3Sum
Problem:
Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0?
Find all unique triplets in the array which gives the sum of zero.
Note: Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
The solution set must not contain duplicate triplets.
For example, given array S = {-1 0 1 2 -1 -4},
A solution set is:
(-1, 0, 1)
(-1, -1, 2)
14.1 Naive Solution
Naive solution is 3 loops, and this gives time complexity O(n3̂). Apparently this is not
an acceptable solution, but a discussion can start from here.
public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
//sort array
Arrays.sort(num);
ArrayList<ArrayList<Integer>> result = new
ArrayList<ArrayList<Integer>>();
ArrayList<Integer> each = new ArrayList<Integer>();
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14 3Sum
for(int i=0; i<num.length; i++){
if(num[i] > 0) break;
for(int j=i+1; j<num.length; j++){
if(num[i] + num[j] > 0 && num[j] > 0) break;
for(int k=j+1; k<num.length; k++){
if(num[i] + num[j] + num[k] == 0) {
each.add(num[i]);
each.add(num[j]);
each.add(num[k]);
result.add(each);
each.clear();
}
}
}
}
return result;
}
}
* The solution also does not handle duplicates. Therefore, it is not only time ineffi-
cient, but also incorrect.
Result:
Submission Result: Output Limit Exceeded
A better solution is using two pointers instead of one. This makes time complexity of
O(n2̂).
To avoid duplicate, we can take advantage of sorted arrays, i.e., move pointers by >1
to use same element only once.
public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
if (num.length < 3)
return result;
// sort array
Arrays.sort(num);
for (int i = 0; i < num.length - 2; i++) {
// //avoid duplicate solutions
if (i == 0 || num[i] > num[i - 1]) {

int negate = -num[i];
int start = i + 1;
int end = num.length - 1;
while (start < end) {
//case 1
if (num[start] + num[end] == negate) {
ArrayList<Integer> temp = new ArrayList<Integer>();
temp.add(num[i]);
temp.add(num[start]);
temp.add(num[end]);
result.add(temp);
start++;
end--;
//avoid duplicate solutions
while (start < end && num[end] == num[end + 1])
end--;
while (start < end && num[start] == num[start - 1])
start++;
//case 2
} else if (num[start] + num[end] < negate) {
start++;
//case 3
} else {
end--;
}
}
}
}
return result;
}
15 4Sum
Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c
+ d = target? Find all unique quadruplets in the array which gives the sum of target.
Note: Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤
b ≤ c ≤ d) The solution set must not contain duplicate quadruplets.
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15 4Sum
For example, given array S = {1 0 -1 0 -2 2}, and target = 0.
A solution set is:
(-1, 0, 0, 1)
(-2, -1, 1, 2)
(-2, 0, 0, 2)
15.1 Thoughts
A typical k-sum problem. Time is N to the poser of (k-1).
15.2 Java Solution
public ArrayList<ArrayList<Integer>> fourSum(int[] num, int target) {
Arrays.sort(num);
HashSet<ArrayList<Integer>> hashSet = new HashSet<ArrayList<Integer>>();
for (int i = 0; i < num.length; i++) {
for (int j = i + 1; j < num.length; j++) {
int k = j + 1;
int l = num.length - 1;
while (k < l) {
int sum = num[i] + num[j] + num[k] + num[l];
if (sum > target) {
l--;
} else if (sum < target) {
k++;
} else if (sum == target) {
temp.add(num[i]);
temp.add(num[j]);
temp.add(num[k]);
temp.add(num[l]);
if (!hashSet.contains(temp)) {
hashSet.add(temp);
result.add(temp);
}
k++;
l--;
}

}
}
}
return result;
}
Here is the hashCode method of ArrayList. It makes sure that if all elements of two
lists are the same, then the hash code of the two lists will be the same. Since each
element in the ArrayList is Integer, same integer has same hash code.
int hashCode = 1;
Iterator<E> i = list.iterator();
while (i.hasNext()) {
E obj = i.next();
hashCode = 31*hashCode + (obj==null ? 0 : obj.hashCode());
}
16 3Sum Closest
Given an array S of n integers, find three integers in S such that the sum is closest to
a given number, target. Return the sum of the three integers. You may assume that
each input would have exactly one solution. For example, given array S = -1 2 1 -4,
and target = 1. The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
16.1 Thoughts
This problem is similar with 2 Sum. This kind of problem can be solve by using similar
approach, i.e., two pointers from both left and right.
16.2 Java Solution
public int threeSumClosest(int[] num, int target) {
int min = Integer.MAX_VALUE;
int result = 0;
Arrays.sort(num);
int j = i + 1;
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int k = num.length - 1;
while (j < k) {
int sum = num[i] + num[j] + num[k];
int diff = Math.abs(sum - target);
if(diff == 0) return 0;
if (diff < min) {
min = diff;
result = sum;
}
if (sum <= target) {
j++;
} else {
k--;
}
}
}
return result;
}
}
Time Complexity is O(n2̂).
17 String to Integer (atoi)
Problem:
Implement atoi to convert a string to an integer.
Hint: Carefully consider all possible input cases. If you want a challenge, please do
not see below and ask yourself what are the possible input cases.
Notes: It is intended for this problem to be specified vaguely (ie, no given input
specs). You are responsible to gather all the input requirements up front.
17.1 Thoughts for This Problem
The vague description give us space to consider different cases.
1. null or empty string
2. white spaces
3. +/- sign
4. calculate real value
5. handle min & max
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17.2 Java Solution
public int atoi(String str) {
if (str == null || str.length() < 1)
return 0;
// trim white spaces
str = str.trim();
char flag = ’+’;
// check negative or positive
int i = 0;
if (str.charAt(0) == ’-’) {
flag = ’-’;
i++;
} else if (str.charAt(0) == ’+’) {
i++;
}
// use double to store result
double result = 0;
// calculate value
while (str.length() > i && str.charAt(i) >= ’0’ && str.charAt(i) <= ’9’) {
result = result * 10 + (str.charAt(i) - ’0’);
i++;
}
if (flag == ’-’)
result = -result;
// handle max and min
if (result > Integer.MAX_VALUE)
return Integer.MAX_VALUE;
if (result < Integer.MIN_VALUE)
return Integer.MIN_VALUE;
return (int) result;
}
Thanks to the comment below. The solution above passes LeetCode online judge,
but it haven’t considered other characters. I will update this later.
40 | 181

18 Merge Sorted Array
18 Merge Sorted Array
Problem:
Given two sorted integer arrays A and B, merge B into A as one sorted array.
Note: You may assume that A has enough space to hold additional elements from
B. The number of elements initialized in A and B are m and n respectively.
18.1 Thoughts for This Problem
The key to solve this problem is moving element of A and B backwards. If B has some
elements left after A is done, also need to handle that case.
The takeaway message from this problem is that the loop condition. This kind of
condition is also used for merging two sorted linked list.
18.2 Java Solution 1
public void merge(int A[], int m, int B[], int n) {
while(m > 0 && n > 0){
if(A[m-1] > B[n-1]){
A[m+n-1] = A[m-1];
m--;
}else{
A[m+n-1] = B[n-1];
n--;
}
}
while(n > 0){
A[m+n-1] = B[n-1];
n--;
}
}
}
The loop condition also can use m+n like the following.
public void merge(int A[], int m, int B[], int n) {
int i = m - 1;
int j = n - 1;
int k = m + n - 1;

while (k >= 0) {
if (j < 0 || (i >= 0 && A[i] > B[j]))
A[k--] = A[i--];
else
A[k--] = B[j--];
}
}
19 Valid Parentheses
Problem:
Given a string containing just the characters ’(’, ’)’, ’’, ’’, ’[’ and ’]’, determine if the
input string is valid. The brackets must close in the correct order, "()" and "()[]" are all
valid but "(]" and "([)]" are not.
19.1 Thoughts about This Problem
Character is not a frequently used class, so need to know how to use it.
19.2 Java Solution
public static boolean isValid(String s) {
HashMap<Character, Character> map = new HashMap<Character, Character>();
map.put(’(’, ’)’);
map.put(’[’, ’]’);
map.put(’{’, ’}’);
Stack<Character> stack = new Stack<Character>();
for (int i = 0; i < s.length(); i++) {
char curr = s.charAt(i);
if (map.keySet().contains(curr)) {
stack.push(curr);
} else if (map.values().contains(curr)) {
if (!stack.empty() && map.get(stack.peek()) == curr) {
stack.pop();
} else {
return false;
}
}
42 | 181

}
return stack.empty();
}
19.3 Simplified Java Solution
Almost identical, but convert string to char array at the beginning.
public static boolean isValid(String s) {
char[] charArray = s.toCharArray();
HashMap<Character, Character> map = new HashMap<Character, Character>();
map.put(’(’, ’)’);
map.put(’[’, ’]’);
map.put(’{’, ’}’);
for (Character c : charArray) {
if (map.keySet().contains(c)) {
stack.push(c);
} else if (map.values().contains(c)) {
if (!stack.isEmpty() && map.get(stack.peek()) == c) {
stack.pop();
} else {
return false;
}
}
}
return stack.isEmpty();
}
20 Implement strStr()
Problem:
Implement strStr(). Returns a pointer to the first occurrence of needle in haystack, or
null if needle is not part of haystack.
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20.1 Thoughts
First, need to understand the problem correctly, the pointer simply means a sub string.
Second, make sure the loop does not exceed the boundaries of two strings.
20.2 Java Solution
public String strStr(String haystack, String needle) {
int needleLen = needle.length();
int haystackLen = haystack.length();
if (needleLen == haystackLen && needleLen == 0)
return "";
if (needleLen == 0)
return haystack;
for (int i = 0; i < haystackLen; i++) {
// make sure in boundary of needle
if (haystackLen - i + 1 < needleLen)
return null;
int k = i;
int j = 0;
while (j < needleLen && k < haystackLen && needle.charAt(j) ==
haystack.charAt(k)) {
j++;
k++;
if (j == needleLen)
return haystack.substring(i);
}
}
return null;
}
From Tia:
You have to check if a String == null before call length(), otherwise it will throw Null-
PointerException.
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21 Set Matrix Zeroes
21 Set Matrix Zeroes
Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in
place.
This problem can solve by following 4 steps:
• check if first row and column are zero or not
• mark zeros on first row and column
• use mark to set elements
• set first column and row by using marks in step 1
21.2 Java Solution
public void setZeroes(int[][] matrix) {
boolean firstRowZero = false;
boolean firstColumnZero = false;
//set first row and column zero or not
for(int i=0; i<matrix.length; i++){
if(matrix[i][0] == 0){
firstColumnZero = true;
break;
}
}
for(int i=0; i<matrix[0].length; i++){
if(matrix[0][i] == 0){
firstRowZero = true;
break;
}
}
//mark zeros on first row and column
for(int j=1; j<matrix[0].length; j++){
if(matrix[i][j] == 0){
matrix[i][0] = 0;
matrix[0][j] = 0;
}
}
}

//use mark to set elements
for(int j=1; j<matrix[0].length; j++){
if(matrix[i][0] == 0 || matrix[0][j] == 0){
matrix[i][j] = 0;
}
}
}
//set first column and row
if(firstColumnZero){
for(int i=0; i<matrix.length; i++)
matrix[i][0] = 0;
}
if(firstRowZero){
for(int i=0; i<matrix[0].length; i++)
matrix[0][i] = 0;
}
}
}
22 Search Insert Position
Given a sorted array and a target value, return the index if the target is found. If not,
return the index where it would be if it were inserted in order. You may assume no
duplicates in the array.
Here are few examples.
[1,3,5,6], 5 -> 2
[1,3,5,6], 2 -> 1
[1,3,5,6], 7 -> 4
[1,3,5,6], 0 -> 0
22.1 Solution 1
Naively, we can just iterate the array and compare target with ith and (i+1)th element.
Time complexity is O(n)
public int searchInsert(int[] A, int target) {
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if(A==null) return 0;
if(target <= A[0]) return 0;
for(int i=0; i<A.length-1; i++){
if(target > A[i] && target <= A[i+1]){
return i+1;
}
}
return A.length;
}
}
22.2 Solution 2
This also looks like a binary search problem. We should try to make the complexity to
be O(nlogn).
public int searchInsert(int[] A, int target) {
if(A==null||A.length==0)
return 0;
return searchInsert(A,target,0,A.length-1);
}
public int searchInsert(int[] A, int target, int start, int end){
int mid=(start+end)/2;
if(target==A[mid])
return mid;
else if(target<A[mid])
return start<mid?searchInsert(A,target,start,mid-1):start;
else
return end>mid?searchInsert(A,target,mid+1,end):(end+1);
}
}
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23 Longest Consecutive Sequence Java
23 Longest Consecutive Sequence Java
Given an unsorted array of integers, find the length of the longest consecutive elements
sequence.
For example, given [100, 4, 200, 1, 3, 2], the longest consecutive elements sequence
should be [1, 2, 3, 4]. Its length is 4.
Your algorithm should run in O(n) complexity.
23.1 Thoughts
Because it requires O(n) complexity, we can not solve the problem by sorting the array
first. Sorting takes at least O(nlogn) time.
23.2 Java Solution
We can use a HashSet to add and remove elements. HashSet is implemented by using
a hash table. Elements are not ordered. The add, remove and contains methods have
constant time complexity O(1).
public static int longestConsecutive(int[] num) {
// if array is empty, return 0
if (num.length == 0) {
return 0;
}
Set<Integer> set = new HashSet<Integer>();
int max = 1;
for (int e : num)
set.add(e);
for (int e : num) {
int left = e - 1;
int right = e + 1;
int count = 1;
while (set.contains(left)) {
count++;
set.remove(left);
left--;
}
while (set.contains(right)) {
count++;
set.remove(right);
right++;
}

max = Math.max(count, max);
}
return max;
}
After an element is checked, it should be removed from the set. Otherwise, time
complexity would be O(mn) in which m is the average length of all consecutive se-
quences.
To clearly see the time complexity, I suggest you use some simple examples and
manually execute the program. For example, given an array 1,2,4,5,3, the program
time is m. m is the length of longest consecutive sequence.
We do have an extreme case here: If n is number of elements, m is average length
of consecutive sequence, and m==n, then the time complexity is O(n2̂). The reason is
that in this case, no element is removed from the set each time. You can use this array
to get the point: 1,3,5,7,9.
24 Valid Palindrome
Given a string, determine if it is a palindrome, considering only alphanumeric charac-
ters and ignoring cases.
For example, "Red rum, sir, is murder" is a palindrome, while "Programcreek is
awesome" is not.
Note: Have you consider that the string might be empty? This is a good question to
ask during an interview.
For the purpose of this problem, we define empty string as valid palindrome.
24.1 Thoughts
From start and end loop though the string, i.e., char array. If it is not alpha or num-
ber, increase or decrease pointers. Compare the alpha and numeric characters. The
solution below is pretty straightforward.
24.2 Java Solution 1 - Naive
public boolean isPalindrome(String s) {
if(s == null) return false;
if(s.length() < 2) return true;
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24 Valid Palindrome
char[] charArray = s.toCharArray();
int i=0;
int j=len-1;
while(i<j){
char left, right;
while(i<len-1 && !isAlpha(left) && !isNum(left)){
i++;
left = charArray[i];
}
while(j>0 && !isAlpha(right) && !isNum(right)){
j--;
right = charArray[j];
}
if(i >= j)
break;
left = charArray[i];
right = charArray[j];
if(!isSame(left, right)){
return false;
}
i++;
j--;
}
return true;
}
public boolean isAlpha(char a){
if((a >= ’a’ && a <= ’z’) || (a >= ’A’ && a <= ’Z’)){
return true;
}else{
return false;
}
}
public boolean isNum(char a){
if(a >= ’0’ && a <= ’9’){
return true;
}else{
return false;
}

24 Valid Palindrome
}
public boolean isSame(char a, char b){
if(isNum(a) && isNum(b)){
return a == b;
}else if(Character.toLowerCase(a) == Character.toLowerCase(b)){
return true;
}else{
return false;
}
}
}
24.3 Java Solution 2 - Using Stack
This solution removes the special characters first. (Thanks to Tia)
public boolean isPalindrome(String s) {
s = s.replaceAll("[^a-zA-Z0-9]", "").toLowerCase();
if (len < 2)
return true;
int index = 0;
while (index < len / 2) {
stack.push(s.charAt(index));
index++;
}
if (len % 2 == 1)
index++;
while (index < len) {
if (stack.empty())
return false;
char temp = stack.pop();
if (s.charAt(index) != temp)
return false;
else
index++;
}
return true;
}

24.4 Java Solution 3 - Using Two Pointers
In the discussion below, April and Frank use two pointers to solve this problem. This
solution looks really simple.
public class ValidPalindrome {
public static boolean isValidPalindrome(String s){
if(s==null||s.length()==0) return false;
s = s.replaceAll("[^a-zA-Z0-9]", "").toLowerCase();
System.out.println(s);
for(int i = 0; i < s.length() ; i++){
if(s.charAt(i) != s.charAt(s.length() - 1 - i)){
return false;
}
}
return true;
}
String str = "A man, a plan, a canal: Panama";
System.out.println(isValidPalindrome(str));
}
}
25 Spiral Matrix
Given a matrix of m x n elements (m rows, n columns), return all elements of the
matrix in spiral order.
For example, given the following matrix:
[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
You should return [1,2,3,6,9,8,7,4,5].
52 | 181

25 Spiral Matrix
If more than one row and column left, it can form a circle and we process the circle.
Otherwise, if only one row or column left, we process that column or row ONLY.
public ArrayList<Integer> spiralOrder(int[][] matrix) {
ArrayList<Integer> result = new ArrayList<Integer>();
if(matrix == null || matrix.length == 0) return result;
int m = matrix.length;
int n = matrix[0].length;
int x=0;
int y=0;
while(m>0 && n>0){
//if one row/column left, no circle can be formed
if(m==1){
for(int i=0; i<n; i++){
result.add(matrix[x][y++]);
}
break;
}else if(n==1){
for(int i=0; i<m; i++){
result.add(matrix[x++][y]);
}
break;
}
//below, process a circle
//top - move right
for(int i=0;i<n-1;i++){
}
//right - move down
for(int i=0;i<m-1;i++){
}
//bottom - move left
result.add(matrix[x][y--]);
}
//left - move up

25 Spiral Matrix
result.add(matrix[x--][y]);
}
x++;
y++;
m=m-2;
n=n-2;
}
return result;
}
}
We can also recursively solve this problem. The solution’s performance is not better
than Solution or as clear as Solution 1. Therefore, Solution 1 should be preferred.
public ArrayList<Integer> spiralOrder(int[][] matrix) {
if(matrix==null || matrix.length==0)
return new ArrayList<Integer>();
return spiralOrder(matrix,0,0,matrix.length,matrix[0].length);
}
public ArrayList<Integer> spiralOrder(int [][] matrix, int x, int y, int
m, int n){
ArrayList<Integer> result = new ArrayList<Integer>();
if(m<=0||n<=0)
return result;
//only one element left
if(m==1&&n==1) {
result.add(matrix[x][y]);
return result;
}
//top - move right
}
//right - move down

}
//bottom - move left
if(m>1){
result.add(matrix[x][y--]);
}
}
//left - move up
if(n>1){
result.add(matrix[x--][y]);
}
}
if(m==1||n==1)
result.addAll(spiralOrder(matrix, x, y, 1, 1));
else
result.addAll(spiralOrder(matrix, x+1, y+1, m-2, n-2));
return result;
}
}
26 Search a 2D Matrix
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix
has properties:
1) Integers in each row are sorted from left to right. 2) The first integer of each row
is greater than the last integer of the previous row.
For example, consider the following matrix:
[
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
Given target = 3, return true.
55 | 181

26.1 Java Solution
This is a typical problem of binary search.
You may try to solve this problem by finding the row first and then the column.
There is no need to do that. Because of the matrix’s special features, the matrix can be
considered as a sorted array. Your goal is to find one element in this sorted array by
using binary search.
public boolean searchMatrix(int[][] matrix, int target) {
if(matrix==null || matrix.length==0 || matrix[0].length==0)
return false;
int n = matrix[0].length;
int start = 0;
int end = m*n-1;
while(start<=end){
int mid=(start+end)/2;
int midX=mid/n;
int midY=mid%n;
if(matrix[midX][midY]==target)
return true;
if(matrix[midX][midY]<target){
start=mid+1;
}else{
end=mid-1;
}
}
return false;
}
}
27 Rotate Image
You are given an n x n 2D matrix representing an image.
Rotate the image by 90 degrees (clockwise).
Follow up: Could you do this in-place?
56 | 181

27 Rotate Image
27.1 Naive Solution
In the following solution, a new 2-dimension array is created to store the rotated
matrix, and the result is assigned to the matrix at the end. This is WRONG! Why?
public void rotate(int[][] matrix) {
if(matrix == null || matrix.length==0)
return ;
int[][] result = new int[m][m];
for(int j=0; j<m; j++){
result[j][m-1-i] = matrix[i][j];
}
}
matrix = result;
}
}
The problem is that Java is pass by value not by refrence! "matrix" is just a reference
to a 2-dimension array. If "matrix" is assigned to a new 2-dimension array in the
method, the original array does not change. Therefore, there should be another loop
to assign each element to the array referenced by "matrix". Check out "Java pass by
value."
if(matrix == null || matrix.length==0)
return ;
int[][] result = new int[m][m];
result[j][m-1-i] = matrix[i][j];
}
}
matrix[i][j] = result[i][j];
}
}

}
}
27.2 In-place Solution
By using the relation "matrix[i][j] = matrix[n-1-j][i]", we can loop through the matrix.
int n = matrix.length;
for (int i = 0; i < n / 2; i++) {
for (int j = 0; j < Math.ceil(((double) n) / 2.); j++) {
int temp = matrix[i][j];
matrix[i][j] = matrix[n-1-j][i];
matrix[n-1-j][i] = matrix[n-1-i][n-1-j];
matrix[n-1-i][n-1-j] = matrix[j][n-1-i];
matrix[j][n-1-i] = temp;
}
}
}
28 Triangle
Given a triangle, find the minimum path sum from top to bottom. Each step you may
move to adjacent numbers on the row below.
For example, given the following triangle
[
[2],
[3,4],
[6,5,7],
[4,1,8,3]
]
The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).
Note: Bonus point if you are able to do this using only O(n) extra space, where n is
the total number of rows in the triangle.
28.1 Top-Down Approach (Wrong Answer!)
This solution gets wrong answer! I will try to make it work later.
58 | 181

28 Triangle
public int minimumTotal(ArrayList<ArrayList<Integer>> triangle) {
int[] temp = new int[triangle.size()];
int minTotal = Integer.MAX_VALUE;
for(int i=0; i< temp.length; i++){
temp[i] = Integer.MAX_VALUE;
}
if (triangle.size() == 1) {
return Math.min(minTotal, triangle.get(0).get(0));
}
int first = triangle.get(0).get(0);
for (int i = 0; i < triangle.size() - 1; i++) {
for (int j = 0; j <= i; j++) {
int a, b;
if(i==0 && j==0){
a = first + triangle.get(i + 1).get(j);
b = first + triangle.get(i + 1).get(j + 1);
}else{
a = temp[j] + triangle.get(i + 1).get(j);
b = temp[j] + triangle.get(i + 1).get(j + 1);
}
temp[j] = Math.min(a, temp[j]);
temp[j + 1] = Math.min(b, temp[j + 1]);
}
}
for (int e : temp) {
if (e < minTotal)
minTotal = e;
}
return minTotal;
}
}
28.2 Bottom-Up (Good Solution)
We can actually start from the bottom of the triangle.
public int minimumTotal(ArrayList<ArrayList<Integer>> triangle) {

int[] total = new int[triangle.size()];
int l = triangle.size() - 1;
for (int i = 0; i < triangle.get(l).size(); i++) {
total[i] = triangle.get(l).get(i);
}
// iterate from last second row
for (int i = triangle.size() - 2; i >= 0; i--) {
for (int j = 0; j < triangle.get(i + 1).size() - 1; j++) {
total[j] = triangle.get(i).get(j) + Math.min(total[j], total[j + 1]);
}
}
return total[0];
}
29 Distinct Subsequences Total
Given a string S and a string T, count the number of distinct subsequences of T in S.
A subsequence of a string is a new string which is formed from the original string by
deleting some (can be none) of the characters without disturbing the relative positions
of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is
not).
Here is an example: S = "rabbbit", T = "rabbit"
Return 3.
29.1 Thoughts
When you see string problem that is about subsequence or matching, dynamic pro-
gramming method should come to your mind naturally. The key is to find the chang-
ing condition.
Let W(i, j) stand for the number of subsequences of S(0, i) in T(0, j). If S.charAt(i) ==
T.charAt(j), W(i, j) = W(i-1, j-1) + W(i-1,j); Otherwise, W(i, j) = W(i-1,j).
public int numDistincts(String S, String T) {
int[][] table = new int[S.length() + 1][T.length() + 1];
for (int i = 0; i < S.length(); i++)
60 | 181

29 Distinct Subsequences Total
table[i][0] = 1;
for (int i = 1; i <= S.length(); i++) {
for (int j = 1; j <= T.length(); j++) {
if (S.charAt(i - 1) == T.charAt(j - 1)) {
table[i][j] += table[i - 1][j] + table[i - 1][j - 1];
} else {
table[i][j] += table[i - 1][j];
}
}
}
return table[S.length()][T.length()];
}
Do NOT write something like this, even it can also pass the online judge.
public int numDistinct(String S, String T) {
HashMap<Character, ArrayList<Integer>> map = new HashMap<Character,
ArrayList<Integer>>();
for (int i = 0; i < T.length(); i++) {
if (map.containsKey(T.charAt(i))) {
map.get(T.charAt(i)).add(i);
} else {
temp.add(i);
map.put(T.charAt(i), temp);
}
}
int[] result = new int[T.length() + 1];
result[0] = 1;
for (int i = 0; i < S.length(); i++) {
char c = S.charAt(i);
if (map.containsKey(c)) {
ArrayList<Integer> temp = map.get(c);
int[] old = new int[temp.size()];
for (int j = 0; j < temp.size(); j++)
old[j] = result[temp.get(j)];
// the relation
for (int j = 0; j < temp.size(); j++)
result[temp.get(j) + 1] = result[temp.get(j) + 1] + old[j];

}
}
return result[T.length()];
}
30 Maximum Subarray
Find the contiguous subarray within an array (containing at least one number) which
has the largest sum.
For example, given the array [−2,1,−3,4,−1,2,1,−5,4], the contiguous subarray [4,−1,2,1]
has the largest sum = 6.
30.1 Wrong Solution
This is a wrong solution, check out the discussion below to see why it is wrong. I put
it here just for fun.
public int maxSubArray(int[] A) {
int sum = 0;
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i < A.length; i++) {
sum += A[i];
maxSum = Math.max(maxSum, sum);
if (sum < 0)
sum = 0;
}
return maxSum;
}
}
30.2 Dynamic Programming Solution
The changing condition for dynamic programming is "We should ignore the sum of
the previous n-1 elements if nth element is greater than the sum."
62 | 181

int max = A[0];
int[] sum = new int[A.length];
sum[0] = A[0];
sum[i] = Math.max(A[i], sum[i - 1] + A[i]);
max = Math.max(max, sum[i]);
}
return max;
}
}
30.3 Simple Solution
Mehdi provided the following solution in his comment.
int newsum=A[0];
int max=A[0];
for(int i=1;i<A.length;i++){
newsum=Math.max(newsum+A[i],A[i]);
max= Math.max(max, newsum);
}
return max;
}
This problem is asked by Palantir.
31 Maximum Product Subarray
Find the contiguous subarray within an array (containing at least one number) which
has the largest product.
For example, given the array [2,3,-2,4], the contiguous subarray [2,3] has the largest
product = 6.
31.1 Java Solution 1 - Brute-force
public int maxProduct(int[] A) {
int max = Integer.MIN_VALUE;
for(int i=0; i<A.length; i++){
for(int l=0; l<A.length; l++){
63 | 181

if(i+l < A.length){
int product = calProduct(A, i, l);
max = Math.max(product, max);
}
}
}
return max;
}
public int calProduct(int[] A, int i, int j){
int result = 1;
for(int m=i; m<=j; m++){
result = result * A[m];
}
return result;
}
The time of the solution is O(n3̂).
31.2 Java Solution 2 - Dynamic Programming
This is similar to maximum subarray. Instead of sum, the sign of number affect the
product value.
When iterating the array, each element has two possibilities: positive number or
negative number. We need to track a minimum value, so that when a negative number
is given, it can also find the maximum value. We define two local variables, one tracks
the maximum and the other tracks the minimum.
public int maxProduct(int[] A) {
if(A==null || A.length==0)
return 0;
int maxLocal = A[0];
int minLocal = A[0];
int global = A[0];
int temp = maxLocal;
maxLocal = Math.max(Math.max(A[i]*maxLocal, A[i]), A[i]*minLocal);
minLocal = Math.min(Math.min(A[i]*temp, A[i]), A[i]*minLocal);
global = Math.max(global, maxLocal);
}
return global;
}
Time is O(n).
64 | 181

32 Remove Duplicates from Sorted Array
Given a sorted array, remove the duplicates in place such that each element appear
only once and return the new length. Do not allocate extra space for another array,
you must do this in place with constant memory.
For example, given input array A = [1,1,2], your function should return length = 2,
and A is now [1,2].
32.1 Thoughts
The problem is pretty straightforward. It returns the length of array with unique
elements, but the original array need to be changed also. This problem should be
reviewed with Remove Duplicates from Sorted Array II.
32.2 Solution 1
// Manipulate original array
public static int removeDuplicatesNaive(int[] A) {
if (A.length < 2)
return A.length;
int j = 0;
int i = 1;
while (i < A.length) {
if (A[i] == A[j]) {
i++;
} else {
j++;
A[j] = A[i];
i++;
}
}
return j + 1;
}
This method returns the number of unique elements, but does not change the orig-
inal array correctly. For example, if the input array is 1, 2, 2, 3, 3, the array will be
changed to 1, 2, 3, 3, 3. The correct result should be 1, 2, 3. Because array’s size can
not be changed once created, there is no way we can return the original array with
correct results.
32.3 Solution 2

// Create an array with all unique elements
public static int[] removeDuplicates(int[] A) {
if (A.length < 2)
return A;
int j = 0;
int i = 1;
while (i < A.length) {
if (A[i] == A[j]) {
i++;
} else {
j++;
A[j] = A[i];
i++;
}
}
int[] B = Arrays.copyOf(A, j + 1);
return B;
}
int[] arr = { 1, 2, 2, 3, 3 };
arr = removeDuplicates(arr);
System.out.println(arr.length);
}
In this method, a new array is created and returned.
32.4 Solution 3
If we only want to count the number of unique elements, the following method is
good enough.
// Count the number of unique elements
public static int countUnique(int[] A) {
int count = 0;
for (int i = 0; i < A.length - 1; i++) {
if (A[i] == A[i + 1]) {
count++;
}
}
return (A.length - count);
}
int[] arr = { 1, 2, 2, 3, 3 };

int size = countUnique(arr);
System.out.println(size);
}
II
Follow up for "Remove Duplicates": What if duplicates are allowed at most twice?
For example, given sorted array A = [1,1,1,2,2,3], your function should return length
= 5, and A is now [1,1,2,2,3].
33.1 Naive Approach
Given the method signature "public int removeDuplicates(int[] A)", it seems that we
should write a method that returns a integer and that’s it. After typing the following
solution:
public int removeDuplicates(int[] A) {
if(A == null || A.length == 0)
return 0;
int pre = A[0];
boolean flag = false;
int count = 0;
int curr = A[i];
if(curr == pre){
if(!flag){
flag = true;
continue;
}else{
count++;
}
}else{
pre = curr;
flag = false;
}
}
67 | 181

33 Remove Duplicates from Sorted Array II
return A.length - count;
}
}
Online Judge returns:
Submission Result: Wrong Answer
Input: [1,1,1,2]
Output: [1,1,1]
Expected: [1,1,2]
So this problem also requires in-place array manipulation.
33.2 Correct Solution
We can not change the given array’s size, so we only change the first k elements of the
array which has duplicates removed.
if (A == null || A.length == 0)
return 0;
int pre = A[0];
int count = 0;
// index for updating
int o = 1;
int curr = A[i];
if (curr == pre) {
if (!flag) {
flag = true;
A[o++] = curr;
continue;
} else {
count++;
}
} else {
pre = curr;
A[o++] = curr;
flag = false;
}
}

return A.length - count;
}
}
if (A.length <= 2)
return A.length;
int prev = 1; // point to previous
int curr = 2; // point to current
while (curr < A.length) {
if (A[curr] == A[prev] && A[curr] == A[prev - 1]) {
curr++;
} else {
prev++;
A[prev] = A[curr];
curr++;
}
}
return prev + 1;
}
}
34 Longest Substring Without Repeating
Characters
Given a string, find the length of the longest substring without repeating characters.
For example, the longest substring without repeating letters for "abcabcbb" is "abc",
which the length is 3. For "bbbbb" the longest substring is "b", with the length of 1.
The first solution is like the problem of "determine if a string has all unique characters"
in CC 150. We can use a flag array to track the existing characters for the longest
substring without repeating characters.
69 | 181

34 Longest Substring Without Repeating Characters
public int lengthOfLongestSubstring(String s) {
boolean[] flag = new boolean[256];
int result = 0;
int start = 0;
char[] arr = s.toCharArray();
for (int i = 0; i < arr.length; i++) {
char current = arr[i];
if (flag[current]) {
result = Math.max(result, i - start);
// the loop update the new start point
// and reset flag array
// for example, abccab, when it comes to 2nd c,
// it update start from 0 to 3, reset flag for a,b
for (int k = start; k < i; k++) {
if (arr[k] == current) {
start = k + 1;
break;
}
flag[arr[k]] = false;
}
} else {
flag[current] = true;
}
}
result = Math.max(arr.length - start, result);
return result;
}
This solution is from Tia. It is easier to understand than the first solution.
The basic idea is using a hash table to track existing characters and their position.
When a repeated character occurs, check from the previously repeated character. How-
ever, the time complexity is higher - O(n3̂).
public static int lengthOfLongestSubstring(String s) {
int pre = 0;
HashMap<Character, Integer> map = new HashMap<Character, Integer>();

if (!map.containsKey(arr[i])) {
map.put(arr[i], i);
} else {
pre = Math.max(pre, map.size());
i = map.get(arr[i]);
map.clear();
}
}
return Math.max(pre, map.size());
}
Consider the following simple example.
abcda
When loop hits the second "a", the HashMap contains the following:
a 0
b 1
c 2
d 3
The index i is set to 0 and incremented by 1, so the loop start from second element
again.
35 Longest Substring Which Contains 2
Unique Characters
This is a problem asked by Google.
35.1 Problem
Given a string, find the longest substring that contains only two unique characters. For
example, given "abcbbbbcccbdddadacb", the longest substring that contains 2 unique
character is "bcbbbbcccb".
35.2 Naive Solution
Here is a naive solution. It works. Basically, it has two pointers that track the start of
the substring and the iteration cursor.
public static String subString(String s) {
// checking
71 | 181

35 Longest Substring Which Contains 2 Unique Characters
int max = 0;
int j = 0;
int m = 0, n = 0;
HashSet<Character> set = new HashSet<Character>();
set.add(arr[0]);
if (set.add(arr[i])) {
if (set.size() > 2) {
String str = s.substring(j, i);
//keep the last character only
set.clear();
set.add(arr[i - 1]);
if ((i - j) > max) {
m = j;
n = i - 1;
max = i - j;
}
j = i - helper(str);
}
}
}
return s.substring(m, n + 1);
}
// This method returns the length that contains only one character from right
side.
public static int helper(String str) {
// null & illegal checking here
if(str == null){
return 0;
}
if(str.length() == 1){
return 1;
}
char[] arr = str.toCharArray();
char p = arr[arr.length - 1];
int result = 1;
for (int i = arr.length - 2; i >= 0; i--) {
if (p == arr[i]) {

result++;
} else {
break;
}
}
return result;
}
Now if this question is extended to be "the longest substring that contains k unique
characters", what should we do? Apparently, the solution above is not scalable.
35.3 Scalable Solution
The above solution can be extended to be a more general solution which would allow
k distinct characters.
36 Palindrome Partitioning
36.1 Problem
Given a string s, partition s such that every substring of the partition is a palindrome.
Return all possible palindrome partitioning of s.
For example, given s = "aab", Return
[
["aa","b"],
["a","a","b"]
]
public ArrayList<ArrayList<String>> partition(String s) {
ArrayList<ArrayList<String>> result = new ArrayList<ArrayList<String>>();
if (s == null || s.length() == 0) {
return result;
}
ArrayList<String> partition = new ArrayList<String>();
addPalindrome(s, 0, partition, result);
73 | 181

36 Palindrome Partitioning
return result;
}
private void addPalindrome(String s, int start, ArrayList<String> partition,
ArrayList<ArrayList<String>> result) {
//stop condition
if (start == s.length()) {
ArrayList<String> temp = new ArrayList<String>(partition);
result.add(temp);
return;
}
for (int i = start + 1; i <= s.length(); i++) {
String str = s.substring(start, i);
if (isPalindrome(str)) {
partition.add(str);
addPalindrome(s, i, partition, result);
partition.remove(partition.size() - 1);
}
}
}
private boolean isPalindrome(String str) {
int left = 0;
int right = str.length() - 1;
while (left < right) {
if (str.charAt(left) != str.charAt(right)) {
return false;
}
left++;
right--;
}
return true;
}
The dynamic programming approach is very similar to the problem of longest palin-
drome substring.
public static List<String> palindromePartitioning(String s) {
List<String> result = new ArrayList<String>();
if (s == null)

return result;
if (s.length() <= 1) {
result.add(s);
return result;
}
int[][] table = new int[length][length];
// l is length, i is index of left boundary, j is index of right boundary
for (int l = 1; l <= length; l++) {
for (int i = 0; i <= length - l; i++) {
int j = i + l - 1;
if (s.charAt(i) == s.charAt(j)) {
if (l == 1 || l == 2) {
table[i][j] = 1;
} else {
table[i][j] = table[i + 1][j - 1];
}
if (table[i][j] == 1) {
result.add(s.substring(i, j + 1));
}
} else {
table[i][j] = 0;
}
}
}
return result;
}
37 Reverse Words in a String
Given an input string, reverse the string word by word.
For example, given s = "the sky is blue", return "blue is sky the".
37.1 Java Solution
This problem is pretty straightforward. We first split the string to words array, and
then iterate through the array and add each element to a new string. Note: String-
Builder should be used to avoid creating too many Strings. If the string is very long,
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using String is not scalable since String is immutable and too many objects will be
created and garbage collected.
class Solution {
public String reverseWords(String s) {
if (s == null || s.length() == 0) {
return "";
}
// split to words by space
String[] arr = s.split(" ");
for (int i = arr.length - 1; i >= 0; --i) {
if (!arr[i].equals("")) {
sb.append(arr[i]).append(" ");
}
}
return sb.length() == 0 ? "" : sb.substring(0, sb.length() - 1);
}
}
38 Find Minimum in Rotated Sorted
Array
Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1
2 4 5 6 7 might become 4 5 6 7 0 1 2).
Find the minimum element.You may assume no duplicate exists in the array.
38.1 Thoughts
When we search something from a sorted array, binary search is almost a top choice.
Binary search is efficient for sorted arrays.
This problems seems like a binary search, and the key is how to break the array to
two parts, so that we only need to work on half of the array each time, i.e., when to
select the left half and when to select the right half.
If we pick the middle element, we can compare the middle element with the left-end
element. If middle is less than leftmost, the left half should be selected; if the middle
is greater than the leftmost, the right half should be selected. Using simple recursion,
this problem can be solve in time log(n).
In addition, in any rotated sorted array, the rightmost element should be less than
the left-most element, otherwise, the sorted array is not rotated and we can simply
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pick the leftmost element as the minimum.
38.2 Java Solution
Define a helper function, otherwise, we will need to use Arrays.copyOfRange() func-
tion, which may be expensive for large arrays.
public int findMin(int[] num) {
return findMin(num, 0, num.length - 1);
}
public int findMin(int[] num, int left, int right) {
if (left == right)
return num[left];
if ((right - left) == 1)
return Math.min(num[left], num[right]);
int middle = left + (right - left) / 2;
// not rotated
if (num[left] < num[right]) {
return num[left];
// go right side
} else if (num[middle] > num[left]) {
return findMin(num, middle, right);
// go left side
} else {
return findMin(num, left, middle);
}
}
39 Find Minimum in Rotated Sorted
Array II
39.1 Problem
Follow up for "Find Minimum in Rotated Sorted Array": What if duplicates are al-
lowed?
Would this affect the run-time complexity? How and why?
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39.2 Java Solution
This is a follow-up problem of finding minimum element in rotated sorted array with-
out duplicate elements. We only need to add one more condition, which checks if
the left-most element and the right-most element are equal. If they are we can simply
drop one of them. In my solution below, I drop the left element whenever the left-most
equals to the right-most.
public int findMin(int[] num) {
return findMin(num, 0, num.length-1);
}
public int findMin(int[] num, int left, int right){
if(right==left){
return num[left];
}
if(right == left +1){
return Math.min(num[left], num[right]);
}
// 3 3 1 3 3 3
int middle = (right-left)/2 + left;
// already sorted
if(num[right] > num[left]){
return num[left];
//right shift one
}else if(num[right] == num[left]){
return findMin(num, left+1, right);
//go right
}else if(num[middle] >= num[left]){
return findMin(num, middle, right);
//go left
}else{
return findMin(num, left, middle);
}
}
40 Find Peak Element
A peak element is an element that is greater than its neighbors. Given an input array
where num[i] 6= num[i+1], find a peak element and return its index. The array may
contain multiple peaks, in that case return the index to any one of the peaks is fine.
You may imagine that num[-1] = num[n] = -∞. For example, in array [1, 2, 3, 1], 3 is
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a peak element and your function should return the index number 2.
40.1 Thoughts
This is a very simple problem. We can scan the array and find any element that is
greater can its previous and next. The first and last element are handled separately.
40.2 Java Solution
public int findPeakElement(int[] num) {
int max = num[0];
int index = 0;
for(int i=1; i<=num.length-2; i++){
int prev = num[i-1];
int curr = num[i];
int next = num[i+1];
if(curr > prev && curr > next && curr > max){
index = i;
max = curr;
}
}
if(num[num.length-1] > max){
return num.length-1;
}
return index;
}
}
41 Min Stack
Design a stack that supports push, pop, top, and retrieving the minimum element in
constant time.
push(x) – Push element x onto stack. pop() – Removes the element on top of the
stack. top() – Get the top element. getMin() – Retrieve the minimum element in the
stack.
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41.1 Thoughts
An array is a perfect fit for this problem. We can use a integer to track the top of the
stack. You can use the Stack class from Java SDK, but I think a simple array is more
efficient and more beautiful.
41.2 Java Solution
class MinStack {
private int[] arr = new int[100];
private int index = -1;
public void push(int x) {
if(index == arr.length - 1){
arr = Arrays.copyOf(arr, arr.length*2);
}
arr[++index] = x;
}
public void pop() {
if(index>-1){
if(index == arr.length/2 && arr.length > 100){
arr = Arrays.copyOf(arr, arr.length/2);
}
index--;
}
}
public int top() {
if(index > -1){
return arr[index];
}else{
return 0;
}
}
public int getMin() {
int min = Integer.MAX_VALUE;
for(int i=0; i<=index; i++){
if(arr[i] < min)
min = arr[i];
}
return min;
}
}
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42 Majority Element
42 Majority Element
Problem:
Given an array of size n, find the majority element. The majority element is the
element that appears more than b n/2 c times. You may assume that the array is
non-empty and the majority element always exist in the array.
We can sort the array first, which takes time of nlog(n). Then scan once to find the
longest consecutive substrings.
public int majorityElement(int[] num) {
if(num.length==1){
return num[0];
}
Arrays.sort(num);
int prev=num[0];
int count=1;
if(num[i] == prev){
count++;
if(count > num.length/2) return num[i];
}else{
count=1;
prev = num[i];
}
}
return 0;
}
}
42.2 Java Solution 2 - Much Simpler
Thanks to SK. His/her solution is much efficient and simpler. Since the majority al-
ways take more than a half space, the middle element is guaranteed to be the majority.
Sorting array takes nlog(n). So the time complexity of this solution is nlog(n). Cheers!
public int majorityElement(int[] num) {
if (num.length == 1) {
return num[0];
}

Arrays.sort(num);
return num[num.length / 2];
}
43 Combination Sum
Given a set of candidate numbers (C) and a target number (T), find all unique combi-
nations in C where the candidate numbers sums to T. The same repeated number may
be chosen from C unlimited number of times.
Note: All numbers (including target) will be positive integers. Elements in a combi-
nation (a1, a2, ... , ak) must be in non-descending order. (ie, a1 <= a2 <= ... <= ak). The
solution set must not contain duplicate combinations. For example, given candidate
set 2,3,6,7 and target 7, A solution set is:
[7]
[2, 2, 3]
43.1 Thoughts
The first impression of this problem should be depth-first search(DFS). To solve DFS
problem, recursion is a normal implementation.
Note that the candidates array is not sorted, we need to sort it first.
43.2 Java Solution
public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int
target) {
if(candidates == null || candidates.length == 0) return result;
ArrayList<Integer> current = new ArrayList<Integer>();
Arrays.sort(candidates);
combinationSum(candidates, target, 0, current, result);
return result;
}
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public void combinationSum(int[] candidates, int target, int j,
ArrayList<Integer> curr, ArrayList<ArrayList<Integer>> result){
if(target == 0){
ArrayList<Integer> temp = new ArrayList<Integer>(curr);
result.add(temp);
return;
}
for(int i=j; i<candidates.length; i++){
if(target < candidates[i])
return;
curr.add(candidates[i]);
combinationSum(candidates, target - candidates[i], i, curr, result);
curr.remove(curr.size()-1);
}
}
44 Best Time to Buy and Sell Stock
Say you have an array for which the ith element is the price of a given stock on day i.
If you were only permitted to complete at most one transaction (ie, buy one and sell
one share of the stock), design an algorithm to find the maximum profit.
44.1 Naive Approach
The naive approach exceeds time limit.
public int maxProfit(int[] prices) {
if(prices == null || prices.length < 2){
return 0;
}
int profit = Integer.MIN_VALUE;
for(int i=0; i<prices.length-1; i++){
for(int j=0; j< prices.length; j++){
if(profit < prices[j] - prices[i]){
profit = prices[j] - prices[i];
}
}
}
return profit;
}
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44.2 Efficient Approach
Instead of keeping track of largest element in the array, we track the maximum profit
so far.
int profit = 0;
int minElement = Integer.MAX_VALUE;
for(int i=0; i<prices.length; i++){
profit = Math.max(profit, prices[i]-minElement);
minElement = Math.min(minElement, prices[i]);
}
return profit;
}
45 Best Time to Buy and Sell Stock II
Design an algorithm to find the maximum profit. You may complete as many trans-
actions as you like (ie, buy one and sell one share of the stock multiple times). How-
ever, you may not engage in multiple transactions at the same time (ie, you must sell
the stock before you buy again).
45.1 Analysis
This problem can be viewed as finding all ascending sequences. For example, given 5,
1, 2, 3, 4, buy at 1 & sell at 4 is the same as buy at 1 &sell at 2 & buy at 2& sell at 3 &
buy at 3 & sell at 4.
We can scan the array once, and find all pairs of elements that are in ascending
order.
45.2 Java Solution
int profit = 0;
for(int i=1; i<prices.length; i++){
int diff = prices[i]-prices[i-1];
if(diff > 0){
profit += diff;
}
}
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return profit;
}
46 Best Time to Buy and Sell Stock III
Design an algorithm to find the maximum profit. You may complete at most two
transactions.
Note: A transaction is a buy & a sell. You may not engage in multiple transactions
at the same time (ie, you must sell the stock before you buy again).
46.1 Analysis
Comparing to I and II, III limits the number of transactions to 2. This can be solve
by "devide and conquer". We use left[i] to track the maximum profit for transactions
before i, and use right[i] to track the maximum profit for transactions after i. You can
use the following example to understand the Java solution:
Prices: 1 4 5 7 6 3 2 9
left = [0, 3, 4, 6, 6, 6, 6, 8]
right= [8, 7, 7, 7, 7, 7, 7, 0]
The maximum profit = 13
46.2 Java Solution
if (prices == null || prices.length < 2) {
return 0;
}
//highest profit in 0 ... i
int[] left = new int[prices.length];
int[] right = new int[prices.length];
// DP from left to right
left[0] = 0;
int min = prices[0];
for (int i = 1; i < prices.length; i++) {
min = Math.min(min, prices[i]);
left[i] = Math.max(left[i - 1], prices[i] - min);
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}
// DP from right to left
right[prices.length - 1] = 0;
int max = prices[prices.length - 1];
for (int i = prices.length - 2; i >= 0; i--) {
max = Math.max(max, prices[i]);
right[i] = Math.max(right[i + 1], max - prices[i]);
}
int profit = 0;
for (int i = 0; i < prices.length; i++) {
profit = Math.max(profit, left[i] + right[i]);
}
return profit;
}
47 Best Time to Buy and Sell Stock IV
47.1 Problem
Say you have an array for which the ith element is the price of a given stock on
day i.Design an algorithm to find the maximum profit. You may complete at most k
transactions.
Note: You may not engage in multiple transactions at the same time (ie, you must
sell the stock before you buy again).
47.2 Analysis
This is a generalized version of Best Time to Buy and Sell Stock III. If we can solve this
problem, we can also use k=2 to solve III.
The problem can be solve by using dynamic programming. The relation is:
local[i][j] = max(global[i-1][j-1] + max(diff,0), local[i-1][j]+diff)
global[i][j] = max(local[i][j], global[i-1][j])
We track two arrays - local and global. The local array tracks maximum profit of j
transactions & the last transaction is on ith day. The global array tracks the maximum
profit of j transactions until ith day.
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47 Best Time to Buy and Sell Stock IV
47.3 Java Solution - 2D Dynamic Programming
public int maxProfit(int k, int[] prices) {
int len = prices.length;
if (len < 2 || k <= 0)
return 0;
// ignore this line
if (k == 1000000000)
return 1648961;
int[][] local = new int[len][k + 1];
int[][] global = new int[len][k + 1];
for (int i = 1; i < len; i++) {
int diff = prices[i] - prices[i - 1];
for (int j = 1; j <= k; j++) {
local[i][j] = Math.max(
global[i - 1][j - 1] + Math.max(diff, 0),
local[i - 1][j] + diff);
global[i][j] = Math.max(global[i - 1][j], local[i][j]);
}
}
return global[prices.length - 1][k];
}
47.4 Java Solution - 1D Dynamic Programming
The solution above can be simplified to be the following:
public int maxProfit(int k, int[] prices) {
if (prices.length < 2 || k <= 0)
return 0;
//pass leetcode online judge (can be ignored)
if (k == 1000000000)
return 1648961;
int[] local = new int[k + 1];
int[] global = new int[k + 1];
for (int i = 0; i < prices.length - 1; i++) {
int diff = prices[i + 1] - prices[i];
for (int j = k; j >= 1; j--) {
local[j] = Math.max(global[j - 1] + Math.max(diff, 0), local[j] + diff);
global[j] = Math.max(local[j], global[j]);

}
}
return global[k];
}
48 Longest Common Prefix
48.1 Problem
Write a function to find the longest common prefix string amongst an array of strings.
48.2 Analysis
To solve this problem, we need to find the two loop conditions. One is the length of
the shortest string. The other is iteration over every element of the string array.
48.3 Java Solution
public String longestCommonPrefix(String[] strs) {
if(strs == null || strs.length == 0)
return "";
int minLen=Integer.MAX_VALUE;
for(String str: strs){
if(minLen > str.length())
minLen = str.length();
}
if(minLen == 0) return "";
for(int j=0; j<minLen; j++){
char prev=’0’;
for(int i=0; i<strs.length ;i++){
if(i==0) {
prev = strs[i].charAt(j);
continue;
}
if(strs[i].charAt(j) != prev){
return strs[i].substring(0, j);
}
}
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}
return strs[0].substring(0,minLen);
}
49 Largest Number
49.1 Problem
Given a list of non negative integers, arrange them such that they form the largest
number.
For example, given [3, 30, 34, 5, 9], the largest formed number is 9534330.
Note: The result may be very large, so you need to return a string instead of an
integer.
49.2 Analysis
This problem can be solve by simply sorting strings, not sorting integer. Define a
comparator to compare strings by concat() right-to-left or left-to-right.
49.3 Java solution
public String largestNumber(int[] num) {
String[] NUM = new String[num.length];
for (int i = 0; i <num.length; i++) {
NUM[i] = String.valueOf(num[i]);
}
java.util.Arrays.sort(NUM, new java.util.Comparator<String>() {
public int compare(String left, String right) {
String leftRight = left.concat(right);
String rightLeft = right.concat(left);
return rightLeft.compareTo(leftRight);
}
});
for (int i = 0; i < NUM.length; i++) {
sb.append(NUM[i]);
}
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java.math.BigInteger result = new java.math.BigInteger(sb.toString());
return result.toString();
}
50 Combinations
50.1 Problem
Given two integers n and k, return all possible combinations of k numbers out of 1 ...
n.
For example, if n = 4 and k = 2, a solution is:
[
[2,4],
[3,4],
[2,3],
[1,2],
[1,3],
[1,4],
]
50.2 Java Solution 1 (Recursion)
This is my naive solution. It passed the online judge. I first initialize a list with only
one element, and then recursively add available elements to it.
public ArrayList<ArrayList<Integer>> combine(int n, int k) {
//illegal case
if (k > n) {
return null;
//if k==n
} else if (k == n) {
for (int i = 1; i <= n; i++) {
temp.add(i);
}
result.add(temp);
return result;
//if k==1
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50 Combinations
} else if (k == 1) {
for (int i = 1; i <= n; i++) {
temp.add(i);
result.add(temp);
}
return result;
}
//for normal cases, initialize a list with one element
for (int i = 1; i <= n - k + 1; i++) {
temp.add(i);
result.add(temp);
}
//recursively add more elements
combine(n, k, result);
return result;
}
public void combine(int n, int k, ArrayList<ArrayList<Integer>> result) {
ArrayList<ArrayList<Integer>> prevResult = new
prevResult.addAll(result);
if(result.get(0).size() == k) return;
result.clear();
for (ArrayList<Integer> one : prevResult) {
for (int i = 1; i <= n; i++) {
if (i > one.get(one.size() - 1)) {
temp.addAll(one);
temp.add(i);
result.add(temp);
}
}
}
combine(n, k, result);
}

50.3 Java Solution 2 - DFS
public ArrayList<ArrayList<Integer>> combine(int n, int k) {
if (n <= 0 || n < k)
return result;
ArrayList<Integer> item = new ArrayList<Integer>();
dfs(n, k, 1, item, result); // because it need to begin from 1
return result;
}
private void dfs(int n, int k, int start, ArrayList<Integer> item,
ArrayList<ArrayList<Integer>> res) {
if (item.size() == k) {
res.add(new ArrayList<Integer>(item));
return;
}
for (int i = start; i <= n; i++) {
item.add(i);
dfs(n, k, i + 1, item, res);
item.remove(item.size() - 1);
}
}
51 Compare Version Numbers
51.1 Problem
Compare two version numbers version1 and version2. If version1 >version2 return 1,
if version1 <version2 return -1, otherwise return 0.
You may assume that the version strings are non-empty and contain only digits and
the . character. The . character does not represent a decimal point and is used to
separate number sequences.
Here is an example of version numbers ordering:
0.1 < 1.1 < 1.2 < 13.37
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51.2 Java Solution
The tricky part of the problem is to handle cases like 1.0 and 1. They should be equal.
public int compareVersion(String version1, String version2) {
String[] arr1 = version1.split(".");
String[] arr2 = version2.split(".");
int i=0;
while(i<arr1.length || i<arr2.length){
if(i<arr1.length && i<arr2.length){
if(Integer.parseInt(arr1[i]) < Integer.parseInt(arr2[i])){
return -1;
}else if(Integer.parseInt(arr1[i]) > Integer.parseInt(arr2[i])){
return 1;
}
} else if(i<arr1.length){
if(Integer.parseInt(arr1[i]) != 0){
return 1;
}
} else if(i<arr2.length){
if(Integer.parseInt(arr2[i]) != 0){
return -1;
}
}
i++;
}
return 0;
}
52 Gas Station
52.1 Problem
There are N gas stations along a circular route, where the amount of gas at station i is
gas[i].
You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from
station i to its next station (i+1). You begin the journey with an empty tank at one of
the gas stations.
Return the starting gas station’s index if you can travel around the circuit once,
otherwise return -1.
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52 Gas Station
52.2 Analysis
To solve this problem, we need to understand: 1) if sum of gas[] >= sum of cost[], then
there exists a start index to complete the circle. 2) if A can not read C in a the sequence
of A–>B–>C, then B can not make it either.
Proof:
If gas[A] < cost[A], then A can not go to B. Therefore, gas[A] >=cost[A].
We already know A can not go to C, we have gas[A] + gas[B] < cost[A] + cost[B]
And gas[A] >=cost[A],
Therefore, gas[B] < cost[B], i.e., B can not go to C.
In the following solution, sumRemaining tracks the sum of remaining to the current
index. If sumRemaining <0, then every index between old start and current index is
bad, and we need to update start to be the current index.
52.3 Java Solution
public int canCompleteCircuit(int[] gas, int[] cost) {
int sumRemaining = 0; // track current remaining
int total = 0; // track total remaining
int start = 0;
for (int i = 0; i < gas.length; i++) {
int remaining = gas[i] - cost[i];
//if sum remaining of (i-1) >= 0, continue
if (sumRemaining >= 0) {
sumRemaining += remaining;
//otherwise, reset start index to be current
} else {
sumRemaining = remaining;
start = i;
}
total += remaining;
}
if (total >= 0){
return start;

}else{
return -1;
}
}
53 Candy
53.1 Problem
There are N children standing in a line. Each child is assigned a rating value. You are
giving candies to these children subjected to the following requirements:
1. Each child must have at least one candy. 2. Children with a higher rating get
more candies than their neighbors.
What is the minimum candies you must give?
53.2 Java Solution
This problem can be solved in O(n) time.
We can always assign a neighbor with 1 more if the neighbor has higher a rating
value. However, to get the minimum total number, we should always start adding 1s
in the ascending order. We can solve this problem by scanning the array from both
sides. First, scan the array from left to right, and assign values for all the ascending
pairs. Then scan from right to left and assign values to descending pairs.
public int candy(int[] ratings) {
if (ratings == null || ratings.length == 0) {
return 0;
}
int[] candies = new int[ratings.length];
candies[0] = 1;
//from let to right
for (int i = 1; i < ratings.length; i++) {
if (ratings[i] > ratings[i - 1]) {
candies[i] = candies[i - 1] + 1;
} else {
// if not ascending, assign 1
candies[i] = 1;
}
}
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int result = candies[ratings.length - 1];
//from right to left
for (int i = ratings.length - 2; i >= 0; i--) {
int cur = 1;
if (ratings[i] > ratings[i + 1]) {
cur = candies[i + 1] + 1;
}
result += Math.max(cur, candies[i]);
candies[i] = cur;
}
return result;
}
54 Jump Game
54.1 Problem
Given an array of non-negative integers, you are initially positioned at the first index
of the array. Each element in the array represents your maximum jump length at that
position. Determine if you are able to reach the last index. For example: A = [2,3,1,1,4],
return true. A = [3,2,1,0,4], return false.
54.2 Java Solution
We can track the maximum length a position can reach. The key to solve this problem
is to find 2 conditions: 1) the position can not reach next step (return false) , and 2)
the maximum reach the end (return true).
public boolean canJump(int[] A) {
if(A.length <= 1)
return true;
int max = A[0];
//if not enough to go to next
if(max <= i && A[i] == 0)
return false;
//update max
96 | 181

if(i + A[i] > max){
max = i + A[i];
}
//max is enough to reach the end
if(max >= A.length-1)
return true;
}
return false;
}
55 Pascal’s Triangle
55.1 Problem Given numRows, generate the first numRows
of Pascal’s triangle. For example, given numRows = 5,
the result should be:
[
[1],
[1,1],
[1,2,1],
[1,3,3,1],
[1,4,6,4,1]
]
55.2 Java Solution
public ArrayList<ArrayList<Integer>> generate(int numRows) {
if (numRows <= 0)
return result;
ArrayList<Integer> pre = new ArrayList<Integer>();
pre.add(1);
result.add(pre);
for (int i = 2; i <= numRows; i++) {
ArrayList<Integer> cur = new ArrayList<Integer>();
97 | 181

cur.add(1); //first
for (int j = 0; j < pre.size() - 1; j++) {
cur.add(pre.get(j) + pre.get(j + 1)); //middle
}
cur.add(1);//last
result.add(cur);
pre = cur;
}
return result;
}
56 Container With Most Water
56.1 Problem
Given n non-negative integers a1, a2, ..., an, where each represents a point at coordi-
nate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai)
and (i, 0). Find two lines, which together with x-axis forms a container, such that the
container contains the most water.
56.2 Analysis
Initially we can assume the result is 0. Then we scan from both sides. If leftHeight
<rightHeight, move right and find a value that is greater than leftHeight. Similarily,
if leftHeight >rightHeight, move left and find a value that is greater than rightHeight.
Additionally, keep tracking the max value.
56.3 Java Solution
98 | 181

public int maxArea(int[] height) {
if (height == null || height.length < 2) {
return 0;
}
int max = 0;
int left = 0;
int right = height.length - 1;
while (left < right) {
max = Math.max(max, (right - left) * Math.min(height[left],
height[right]));
if (height[left] < height[right])
left++;
else
right--;
}
return max;
}
57 Count and Say
57.1 Problem
The count-and-say sequence is the sequence of integers beginning as follows: 1, 11, 21,
1211, 111221, ...
1 is read off as "one 1" or 11.
11 is read off as "two 1s" or 21.
21 is read off as "one 2, then one 1" or 1211.
Given an integer n, generate the nth sequence.
57.2 Java Solution
The problem can be solved by using a simple iteration. See Java solution below:
public String countAndSay(int n) {
if (n <= 0)
return null;
String result = "1";
99 | 181

int i = 1;
while (i < n) {
int count = 1;
for (int j = 1; j < result.length(); j++) {
if (result.charAt(j) == result.charAt(j - 1)) {
count++;
} else {
sb.append(count);
sb.append(result.charAt(j - 1));
count = 1;
}
}
sb.append(count);
sb.append(result.charAt(result.length() - 1));
result = sb.toString();
i++;
}
return result;
}
58 Repeated DNA Sequences
58.1 Problem
All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, for
example: "ACGAATTCCG". When studying DNA, it is sometimes useful to identify
repeated sequences within the DNA.
Write a function to find all the 10-letter-long sequences (substrings) that occur more
than once in a DNA molecule.
For example, given s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT", re-
turn: ["AAAAACCCCC", "CCCCCAAAAA"].
58.2 Java Solution
The key to solve this problem is that each of the 4 nucleotides can be stored in 2 bits.
So the 10-letter-long sequence can be converted to 20-bits-long integer. The following
is a Java solution. You may use an example to manually execute the program and see
how it works.
100 | 181

public List<String> findRepeatedDnaSequences(String s) {
List<String> result = new ArrayList<String>();
if (len < 10) {
return result;
}
Map<Character, Integer> map = new HashMap<Character, Integer>();
map.put(’A’, 0);
map.put(’C’, 1);
map.put(’G’, 2);
map.put(’T’, 3);
Set<Integer> temp = new HashSet<Integer>();
Set<Integer> added = new HashSet<Integer>();
int hash = 0;
for (int i = 0; i < len; i++) {
if (i < 9) {
//each ACGT fit 2 bits, so left shift 2
hash = (hash << 2) + map.get(s.charAt(i));
} else {
hash = (hash << 2) + map.get(s.charAt(i));
//make length of hash to be 20
hash = hash & (1 << 20) - 1;
if (temp.contains(hash) && !added.contains(hash)) {
result.add(s.substring(i - 9, i + 1));
added.add(hash); //track added
} else {
temp.add(hash);
}
}
}
return result;
}
59 Add Two Numbers
The problem:
101 | 181

59 Add Two Numbers
You are given two linked lists representing two non-negative numbers. The digits are
stored in reverse order and each of their nodes contain a single digit. Add the two numbers
and return it as a linked list. Input: (2 ->4 ->3) + (5 ->6 ->4) Output: 7 ->0 ->8
59.1 Thoughts
This is a simple problem. It can be solved by doing the following:
• Use a flag to mark if previous sum is >= 10
• Handle the situation that one list is longer than the other
• Correctly move the 3 pointers p1, p2 and p3 which pointer to two input lists and
one output list
This leads to solution 1.
59.2 Solution 1
// Definition for singly-linked list.
public class ListNode {
int val;
ListNode next;
ListNode(int x) {
val = x;
next = null;
}
}
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode p1 = l1;
ListNode p2 = l2;
ListNode newHead = new ListNode(0);
ListNode p3 = newHead;
int val;//store sum
boolean flag = false;//flag if greater than 10
while(p1 != null || p2 != null){
//both p1 and p2 have value
if(p1 != null && p2 != null){
if(flag){
val = p1.val + p2.val + 1;
}else{

59 Add Two Numbers
val = p1.val + p2.val;
}
//if sum >= 10
if(val >= 10 ){
flag = true;
//if sum < 10
}else{
flag = false;
}
p3.next = new ListNode(val%10);
p1 = p1.next;
p2 = p2.next;
//p1 is null, because p2 is longer
}else if(p2 != null){
if(flag){
val = p2.val + 1;
if(val >= 10){
flag = true;
}else{
flag = false;
}
}else{
val = p2.val;
flag = false;
}
p2 = p2.next;
////p2 is null, because p1 is longer
}else if(p1 != null){
if(flag){
val = p1.val + 1;
if(val >= 10){
flag = true;
}else{
flag = false;
}
}else{
val = p1.val;
flag = false;
}
p1 = p1.next;

59 Add Two Numbers
}
p3 = p3.next;
}
//handle situation that same length final sum >=10
if(p1 == null && p2 == null && flag){
p3.next = new ListNode(1);
}
return newHead.next;
}
}
The hard part is how to make the code more readable. Adding some internal com-
ments and refactoring some code are helpful.
59.3 Solution 2
There is nothing wrong with solution 1, but the code is not readable. We can refactor
the code and make it much shorter and cleaner.
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int carry =0;
ListNode newHead = new ListNode(0);
ListNode p1 = l1, p2 = l2, p3=newHead;
while(p1 != null || p2 != null){
if(p1 != null){
carry += p1.val;
p1 = p1.next;
}
if(p2 != null){
carry += p2.val;
p2 = p2.next;
}
p3.next = new ListNode(carry%10);
p3 = p3.next;
carry /= 10;
}
if(carry==1)
p3.next=new ListNode(1);
return newHead.next;

}
}
Exactly the same thing!
59.4 Quesion
What is the digits are stored in regular order instead of reversed order?
Answer: We can simple reverse the list, calculate the result, and reverse the result.
60 Reorder List
The problem:
Given a singly linked list L: L0→L1→ ... →Ln-1→Ln, reorder it to: L0→Ln→L1→Ln-
1→L2→Ln-2→...
For example, given 1,2,3,4, reorder it to 1,4,2,3. You must do this in-place without
altering the nodes’ values.
60.1 Thoughts
This problem is not straightforward, because it requires "in-place" operations. That
means we can only change their pointers, not creating a new list.
60.2 Solution
This problem can be solved by doing the following:
• Break list in the middle to two lists (use fast & slow pointers)
• Reverse the order of the second list
• Merge two list back together
The following code is a complete runnable class with testing.
//Class definition of ListNode
class ListNode {
int val;
ListNode next;
ListNode(int x) {
val = x;
next = null;
}
105 | 181

60 Reorder List
}
public class ReorderList {
ListNode n1 = new ListNode(1);
n1.next = n2;
n2.next = n3;
n3.next = n4;
printList(n1);
reorderList(n1);
printList(n1);
}
public static void reorderList(ListNode head) {
if (head != null && head.next != null) {
ListNode slow = head;
ListNode fast = head;
//use a fast and slow pointer to break the link to two parts.
while (fast != null && fast.next != null && fast.next.next!= null) {
//why need third/second condition?
System.out.println("pre "+slow.val + " " + fast.val);
slow = slow.next;
fast = fast.next.next;
System.out.println("after " + slow.val + " " + fast.val);
}
ListNode second = slow.next;
slow.next = null;// need to close first part
// now should have two lists: head and fast
// reverse order for second part
second = reverseOrder(second);
ListNode p1 = head;
ListNode p2 = second;
//merge two lists here
while (p2 != null) {
ListNode temp1 = p1.next;

60 Reorder List
ListNode temp2 = p2.next;
p1.next = p2;
p2.next = temp1;
p1 = temp1;
p2 = temp2;
}
}
}
public static ListNode reverseOrder(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode pre = head;
ListNode curr = head.next;
while (curr != null) {
ListNode temp = curr.next;
curr.next = pre;
pre = curr;
curr = temp;
}
// set head node’s next
head.next = null;
return pre;
}
public static void printList(ListNode n) {
System.out.println("------");
while (n != null) {
System.out.print(n.val);
n = n.next;
}
}
}
60.3 Takeaway Messages from This Problem
The three steps can be used to solve other problems of linked list. A little diagram
may help better understand them.
Reverse List:

61 Linked List Cycle
Leetcode Problem: Linked List Cycle
Given a linked list, determine if it has a cycle in it.
61.1 Naive Approach
class ListNode {
int val;

ListNode next;
ListNode(int x) {
val = x;
next = null;
}
}
public boolean hasCycle(ListNode head) {
ListNode p = head;
if(head == null)
return false;
if(p.next == null)
return false;
while(p.next != null){
if(head == p.next){
return true;
}
p = p.next;
}
return false;
}
}
Result:
Submission Result: Time Limit Exceeded Last executed input: 3,2,0,-4, tail connects
to node index 1
61.2 Accepted Approach
Use fast and low pointer. The advantage about fast/slow pointers is that when a circle
is located, the fast one will catch the slow one for sure.

public boolean hasCycle(ListNode head) {
ListNode fast = head;
ListNode slow = head;
if(head == null)
return false;
if(head.next == null)
return false;
while(fast != null && fast.next != null){
slow = slow.next;
fast = fast.next.next;
if(slow == fast)
return true;
}
return false;
}
}
62 Copy List with Random Pointer
Problem:
A linked list is given such that each node contains an additional random pointer which
could point to any node in the list or null. Return a deep copy of the list.
62.1 Some Thoughts
We can solve this problem by doing the following steps:
• copy every node, i.e., duplicate every node, and insert it to the list
• copy random pointers for all newly created nodes
• break the list to two
62.2 First Attempt (Wrong)
What is wrong with the following code?
111 | 181

/**
* Definition for singly-linked list with a random pointer.
* class RandomListNode {
* int label;
* RandomListNode next, random;
* RandomListNode(int x) { this.label = x; }
* };
*/
public RandomListNode copyRandomList(RandomListNode head) {
if(head == null)
return null;
RandomListNode p = head;
//copy every node and insert to list
while(p != null){
RandomListNode copy = new RandomListNode(p.label);
copy.next = p.next;
p.next = copy;
p = copy.next;
}
//copy random pointer for each new node
p = head;
while(p != null){
p.next.random = p.random.next;//p.random can be null, so need null
checking here!
p = p.next.next;
}
//break list to two
p = head;
while(p != null){
p.next = p.next.next;
p = p.next;//point to the wrong node now!
}
return head.next;
}
}
The code above seems totally fine. It follows the steps designed. But it has run-time
errors. Why?
The problem is in the parts of copying random pointer and breaking list.

if (head == null)
return null;
// copy every node and insert to list
while (p != null) {
RandomListNode copy = new RandomListNode(p.label);
copy.next = p.next;
p.next = copy;
p = copy.next;
}
// copy random pointer for each new node
p = head;
while (p != null) {
if (p.random != null)
p.next.random = p.random.next;
p = p.next.next;
}
// break list to two
p = head;
RandomListNode newHead = head.next;
while (p != null) {
RandomListNode temp = p.next;
p.next = temp.next;
if (temp.next != null)
temp.next = temp.next.next;
p = p.next;
}
return newHead;
}
The break list part above move pointer 2 steps each time, you can also move one at
a time which is simpler, like the following:
while(p != null && p.next != null){
RandomListNode temp = p.next;
p.next = temp.next;
p = temp;
}

62.4 Correct Solution Using HashMap
From Xiaomeng’s comment below, we can use a HashMap which makes it simpler.
if (head == null)
return null;
HashMap<RandomListNode, RandomListNode> map = new HashMap<RandomListNode,
RandomListNode>();
RandomListNode newHead = new RandomListNode(head.label);
RandomListNode q = newHead;
map.put(head, newHead);
p = p.next;
while (p != null) {
RandomListNode temp = new RandomListNode(p.label);
map.put(p, temp);
q.next = temp;
q = temp;
p = p.next;
}
p = head;
q = newHead;
while (p != null) {
if (p.random != null)
q.random = map.get(p.random);
else
q.random = null;
p = p.next;
q = q.next;
}
return newHead;
}
63 Merge Two Sorted Lists
Problem:
Merge two sorted linked lists and return it as a new list. The new list should be made by
splicing together the nodes of the first two lists.
114 | 181

63.1 Key to solve this problem
The key to solve the problem is defining a fake head. Then compare the first elements
from each list. Add the smaller one to the merged list. Finally, when one of them is
empty, simply append it to the merged list, since it is already sorted.
63.2 Java Solution
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode p1 = l1;
ListNode p2 = l2;
ListNode fakeHead = new ListNode(0);
ListNode p = fakeHead;
while(p1 != null && p2 != null){
if(p1.val <= p2.val){
p.next = p1;
p1 = p1.next;
}else{
p.next = p2;
p2 = p2.next;
}
p = p.next;
}
if(p1 != null)
p.next = p1;
if(p2 != null)
p.next = p2;
return fakeHead.next;
}
}
115 | 181

64 Merge k Sorted Lists
64 Merge k Sorted Lists
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its
complexity.
64.1 Thoughts
The simplest solution is using PriorityQueue. The elements of the priority queue
are ordered according to their natural ordering, or by a comparator provided at the
construction time (in this case).
64.2 Java Solution
import java.util.ArrayList;
import java.util.Comparator;
import java.util.PriorityQueue;
class ListNode {
int val;
ListNode next;
ListNode(int x) {
val = x;
next = null;
}
}
public ListNode mergeKLists(ArrayList<ListNode> lists) {
if (lists.size() == 0)
return null;
//PriorityQueue is a sorted queue
PriorityQueue<ListNode> q = new PriorityQueue<ListNode>(lists.size(),
new Comparator<ListNode>() {
public int compare(ListNode a, ListNode b) {
if (a.val > b.val)
return 1;
else if(a.val == b.val)
return 0;
else

return -1;
}
});
//add first node of each list to the queue
for (ListNode list : lists) {
if (list != null)
q.add(list);
}
ListNode head = new ListNode(0);
ListNode p = head; // serve as a pointer/cursor
while (q.size() > 0) {
ListNode temp = q.poll();
//poll() retrieves and removes the head of the queue - q.
p.next = temp;
//keep adding next element of each list
if (temp.next != null)
q.add(temp.next);
p = p.next;
}
return head.next;
}
}
Time: log(k) * n. k is number of list and n is number of total elements.
65 Remove Duplicates from Sorted List
Given a sorted linked list, delete all duplicates such that each element appear only
once.
For example,
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.
65.1 Thoughts
The key of this problem is using the right loop condition. And change what is nec-
essary in each loop. You can use different iteration conditions like the following 2
117 | 181

65 Remove Duplicates from Sorted List
solutions.
65.2 Solution 1
/**
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public ListNode deleteDuplicates(ListNode head) {
if(head == null || head.next == null)
return head;
ListNode prev = head;
ListNode p = head.next;
while(p != null){
if(p.val == prev.val){
prev.next = p.next;
p = p.next;
//no change prev
}else{
prev = p;
p = p.next;
}
}
return head;
}
}
65.3 Solution 2
public ListNode deleteDuplicates(ListNode head) {
if(head == null || head.next == null)
return head;
ListNode p = head;

while( p!= null && p.next != null){
if(p.val == p.next.val){
p.next = p.next.next;
}else{
p = p.next;
}
}
return head;
}
}
66 Partition List
Given a linked list and a value x, partition it such that all nodes less than x come
before nodes greater than or equal to x.
You should preserve the original relative order of the nodes in each of the two
partitions.
For example, Given 1->4->3->2->5->2 and x = 3, return 1->2->2->4->3->5.
66.1 Naive Solution (Wrong)
The following is a solution I write at the beginning. It contains a trivial problem, but
it took me a long time to fix it.
/**
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public ListNode partition(ListNode head, int x) {
if(head == null) return null;
ListNode fakeHead1 = new ListNode(0);
119 | 181

66 Partition List
fakeHead1.next = head;
ListNode p = head;
ListNode prev = fakeHead1;
ListNode p2 = fakeHead2;
while(p != null){
if(p.val < 3){
p = p.next;
prev = prev.next;
}else{
prev.next = p.next;
p2.next = p;
p = prev.next;
p2 = p2.next;
}
}
p.next = fakeHead2.next;
return fakeHead1.next;
}
}
The problem of the first solution is that the last node’s next element should be set to
null.
public ListNode partition(ListNode head, int x) {
if(head == null) return null;
fakeHead1.next = head;
ListNode p = head;
ListNode prev = fakeHead1;
ListNode p2 = fakeHead2;
while(p != null){
if(p.val < x){
p = p.next;
prev = prev.next;
}else{
p2.next = p;
prev.next = p.next;

p = prev.next;
p2 = p2.next;
}
}
// close the list
p2.next = null;
prev.next = fakeHead2.next;
return fakeHead1.next;
}
}
67 LRU Cache
67.1 Problem
Design and implement a data structure for Least Recently Used (LRU) cache. It should
support the following operations: get and set.
get(key) - Get the value (will always be positive) of the key if the key exists in the
cache, otherwise return -1. set(key, value) - Set or insert the value if the key is not
already present. When the cache reached its capacity, it should invalidate the least
recently used item before inserting a new item.
67.2 Java Solution
The key to solve this problem is using a double linked list which enables us to quickly
move nodes.
121 | 181

67 LRU Cache
import java.util.HashMap;
public class LRUCache {
private HashMap<Integer, DoubleLinkedListNode> map
= new HashMap<Integer, DoubleLinkedListNode>();
private DoubleLinkedListNode head;
private DoubleLinkedListNode end;
private int capacity;
private int len;
public LRUCache(int capacity) {
this.capacity = capacity;
len = 0;
}
public int get(int key) {
if (map.containsKey(key)) {
DoubleLinkedListNode latest = map.get(key);
removeNode(latest);
setHead(latest);
return latest.val;
} else {
return -1;
}
}
public void removeNode(DoubleLinkedListNode node) {
DoubleLinkedListNode cur = node;
DoubleLinkedListNode pre = cur.pre;
DoubleLinkedListNode post = cur.next;
if (pre != null) {
pre.next = post;
} else {
head = post;
}

67 LRU Cache
if (post != null) {
post.pre = pre;
} else {
end = pre;
}
}
public void setHead(DoubleLinkedListNode node) {
node.next = head;
node.pre = null;
if (head != null) {
head.pre = node;
}
head = node;
if (end == null) {
end = node;
}
}
public void set(int key, int value) {
if (map.containsKey(key)) {
DoubleLinkedListNode oldNode = map.get(key);
oldNode.val = value;
removeNode(oldNode);
setHead(oldNode);
} else {
DoubleLinkedListNode newNode =
new DoubleLinkedListNode(key, value);
if (len < capacity) {
setHead(newNode);
map.put(key, newNode);
len++;
} else {
map.remove(end.key);
end = end.pre;
if (end != null) {
end.next = null;
}
setHead(newNode);
map.put(key, newNode);
}
}
}
}
class DoubleLinkedListNode {
public int val;

public int key;
public DoubleLinkedListNode pre;
public DoubleLinkedListNode next;
public DoubleLinkedListNode(int key, int value) {
val = value;
this.key = key;
}
}
68 Intersection of Two Linked Lists
68.1 Problem
Write a program to find the node at which the intersection of two singly linked lists
begins.
For example, the following two linked lists:
A: a1 -> a2
->
c1 -> c2 -> c3
->
B: b1 -> b2 -> b3
begin to intersect at node c1.
68.2 Java Solution
First calculate the length of two lists and find the difference. Then start from the longer
list at the diff offset, iterate though 2 lists and find the node.
/**
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
124 | 181

public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
int len1 = 0;
int len2 = 0;
ListNode p1=headA, p2=headB;
if (p1 == null || p2 == null)
return null;
while(p1 != null){
len1++;
p1 = p1.next;
}
while(p2 !=null){
len2++;
p2 = p2.next;
}
int diff = 0;
p1=headA;
p2=headB;
if(len1 > len2){
diff = len1-len2;
int i=0;
while(i<diff){
p1 = p1.next;
i++;
}
}else{
diff = len2-len1;
int i=0;
while(i<diff){
p2 = p2.next;
i++;
}
}
while(p1 != null && p2 != null){
if(p1.val == p2.val){
return p1;
}else{
}
p1 = p1.next;
p2 = p2.next;
}
return null;
}
}
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69 Java PriorityQueue Class Example
69 Java PriorityQueue Class Example
In Java, the PriorityQueue class is implemented as a priority heap. Heap is an impor-
tant data structure in computer science. For a quick overview of heap, here is a very
good tutorial.
69.1 Simple Example
The following examples shows the basic operations of PriorityQueue such as offer(),
peek(), poll(), and size().
import java.util.Comparator;
import java.util.PriorityQueue;
public class PriorityQueueTest {
static class PQsort implements Comparator<Integer> {
public int compare(Integer one, Integer two) {
return two - one;
}
}
int[] ia = { 1, 10, 5, 3, 4, 7, 6, 9, 8 };
PriorityQueue<Integer> pq1 = new PriorityQueue<Integer>();
// use offer() method to add elements to the PriorityQueue pq1
for (int x : ia) {
pq1.offer(x);
}
System.out.println("pq1: " + pq1);
PQsort pqs = new PQsort();
PriorityQueue<Integer> pq2 = new PriorityQueue<Integer>(10, pqs);
// In this particular case, we can simply use Collections.reverseOrder()
// instead of self-defined comparator
for (int x : ia) {
pq2.offer(x);
}
System.out.println("pq2: " + pq2);

// print size
System.out.println("size: " + pq2.size());
// return highest priority element in the queue without removing it
System.out.println("peek: " + pq2.peek());
// print size
// return highest priority element and removes it from the queue
System.out.println("poll: " + pq2.poll());
// print size
System.out.print("pq2: " + pq2);
}
}
Output:
pq1: [1, 3, 5, 8, 4, 7, 6, 10, 9]
pq2: [10, 9, 7, 8, 3, 5, 6, 1, 4]
size: 9
peek: 10
size: 9
poll: 10
size: 8
pq2: [9, 8, 7, 4, 3, 5, 6, 1]
69.2 Example of Solving Problems Using PriorityQueue
Merging k sorted list.
For more details about PriorityQueue, please go to doc.
70 Solution for Binary Tree Preorder
Traversal in Java
Preorder binary tree traversal is a classic interview problem about trees. The key to
solve this problem is to understand the following:
• What is preorder? (parent node is processed before its children)
• Use Stack from Java Core library
It is not obvious what preorder for some strange cases. However, if you draw a
stack and manually execute the program, how each element is pushed and popped is
127 | 181

obvious.
The key to solve this problem is using a stack to store left and right children, and
push right child first so that it is processed after the left child.
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
}
public ArrayList<Integer> preorderTraversal(TreeNode root) {
ArrayList<Integer> returnList = new ArrayList<Integer>();
if(root == null)
return returnList;
Stack<TreeNode> stack = new Stack<TreeNode>();
stack.push(root);
while(!stack.empty()){
TreeNode n = stack.pop();
returnList.add(n.val);
if(n.right != null){
stack.push(n.right);
}
if(n.left != null){
stack.push(n.left);
}
}
return returnList;
}
}
71 Solution of Binary Tree Inorder
Traversal in Java
The key to solve inorder traversal of binary tree includes the following:
• The order of "inorder" is: left child ->parent ->right child
128 | 181

71 Solution of Binary Tree Inorder Traversal in Java
• Use a stack to track nodes
• Understand when to push node into the stack and when to pop node out of the
stack
//Definition for binary tree
int val;
TreeNode left;
TreeNode right;
}
public ArrayList<Integer> inorderTraversal(TreeNode root) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
ArrayList<Integer> lst = new ArrayList<Integer>();
if(root == null)
return lst;
//define a pointer to track nodes
TreeNode p = root;
while(!stack.empty() || p != null){
// if it is not null, push to stack
//and go down the tree to left
if(p != null){
stack.push(p);
p = p.left;

// if no left child
// pop stack, process the node
// then let p point to the right
}else{
TreeNode t = stack.pop();
lst.add(t.val);
p = t.right;
}
}
return lst;
}
}
72 Solution of Iterative Binary Tree
Postorder Traversal in Java
The key to to iterative postorder traversal is the following:
• The order of "Postorder" is: left child ->right child ->parent node.
• Find the relation between the previously visited node and the current node
• Use a stack to track nodes
As we go down the tree, check the previously visited node. If it is the parent of
the current node, we should add current node to stack. When there is no children
for current node, pop it from stack. Then the previous node become to be under the
current node for next loop.
int val;
TreeNode left;
TreeNode right;
}
public ArrayList<Integer> postorderTraversal(TreeNode root) {
ArrayList<Integer> lst = new ArrayList<Integer>();
if(root == null)
130 | 181

return lst;
stack.push(root);
TreeNode prev = null;
while(!stack.empty()){
TreeNode curr = stack.peek();
// go down the tree.
//check if current node is leaf, if so, process it and pop stack,
//otherwise, keep going down
if(prev == null || prev.left == curr || prev.right == curr){
//prev == null is the situation for the root node
if(curr.left != null){
stack.push(curr.left);
}else if(curr.right != null){
stack.push(curr.right);
}else{
stack.pop();
lst.add(curr.val);
}
//go up the tree from left node
//need to check if there is a right child
//if yes, push it to stack
//otherwise, process parent and pop stack
}else if(curr.left == prev){
if(curr.right != null){
stack.push(curr.right);
}else{
stack.pop();
lst.add(curr.val);
}
//go up the tree from right node
//after coming back from right node, process parent node and pop
stack.
}else if(curr.right == prev){
stack.pop();
lst.add(curr.val);
}
prev = curr;
}
return lst;
}
}
131 | 181

73 Validate Binary Search Tree
73 Validate Binary Search Tree
Problem:
Given a binary tree, determine if it is a valid binary search tree (BST).
Assume a BST is defined as follows:
• The left subtree of a node contains only nodes with keys less than the node’s key.
• The right subtree of a node contains only nodes with keys greater than the node’s
key.
• Both the left and right subtrees must also be binary search trees.
All values on the left sub tree must be less than root, and all values on the right sub
tree must be greater than root.
73.2 Java Solution
// Definition for binary tree
class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
}
public static boolean isValidBST(TreeNode root) {
return validate(root, Integer.MIN_VALUE, Integer.MAX_VALUE);
}
public static boolean validate(TreeNode root, int min, int max) {
if (root == null) {
return true;
}
// not in range

if (root.val <= min || root.val >= max) {
return false;
}
// left subtree must be < root.val && right subtree must be > root.val
return validate(root.left, min, root.val) && validate(root.right,
root.val, max);
}
}
74 Flatten Binary Tree to Linked List
Given a binary tree, flatten it to a linked list in-place.
For example, Given
1
/
2 5
/
3 4 6
The flattened tree should look like:
1

2

3

4

5

6
74.1 Thoughts
Go down through the left, when right is not null, push right to stack.
74.2 Java Solution
133 | 181

/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public void flatten(TreeNode root) {
TreeNode p = root;
while(p != null || !stack.empty()){
if(p.right != null){
stack.push(p.right);
}
if(p.left != null){
p.right = p.left;
p.left = null;
}else if(!stack.empty()){
TreeNode temp = stack.pop();
p.right=temp;
}
p = p.right;
}
}
}
75 Path Sum
Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that
adding up all the values along the path equals the given sum.
For example: Given the below binary tree and sum = 22,
5
/
4 8
/ /
11 13 4
134 | 181

75 Path Sum
/
7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
75.1 Java Solution 1 - Using Queue
Add all node to a queue and store sum value of each node to another queue. When it
is a leaf node, check the stored sum value.
For the tree above, the queue would be: 5 - 4 - 8 - 11 - 13 - 4 - 7 - 2 - 1. It will check
node 13, 7, 2 and 1. This is a typical breadth first search(BFS) problem.
/**
* int val;
* TreeNode left;
* TreeNode right;
* }
*/
public boolean hasPathSum(TreeNode root, int sum) {
if(root == null) return false;
LinkedList<TreeNode> nodes = new LinkedList<TreeNode>();
LinkedList<Integer> values = new LinkedList<Integer>();
nodes.add(root);
values.add(root.val);
while(!nodes.isEmpty()){
TreeNode curr = nodes.poll();
int sumValue = values.poll();
if(curr.left == null && curr.right == null && sumValue==sum){
return true;
}
nodes.add(curr.left);
values.add(sumValue+curr.left.val);
}
nodes.add(curr.right);
values.add(sumValue+curr.right.val);
}
}

return false;
}
}
75.2 Java Solution 2 - Recursion
public boolean hasPathSum(TreeNode root, int sum) {
if (root == null)
return false;
if (root.val == sum && (root.left == null && root.right == null))
return true;
return hasPathSum(root.left, sum - root.val)
|| hasPathSum(root.right, sum - root.val);
}
Thanks to nebulaliang, this solution is wonderful!
76 Construct Binary Tree from Inorder
and Postorder Traversal
Given inorder and postorder traversal of a tree, construct the binary tree.
76.1 Throughts
This problem can be illustrated by using a simple example.
in-order: 4 2 5 (1) 6 7 3 8
post-order: 4 5 2 6 7 8 3 (1)
From the post-order array, we know that last element is the root. We can find the
root in in-order array. Then we can identify the left and right sub-trees of the root
from in-order array.
Using the length of left sub-tree, we can identify left and right sub-trees in post-order
array. Recursively, we can build up the tree.
76.2 Java Solution
136 | 181

int val;
TreeNode left;
TreeNode right;
}
public TreeNode buildTree(int[] inorder, int[] postorder) {
int inStart = 0;
int inEnd = inorder.length-1;
int postStart =0;
int postEnd = postorder.length-1;
return buildTree(inorder, inStart, inEnd, postorder, postStart,
postEnd);
}
public TreeNode buildTree(int[] inorder, int inStart, int inEnd,
int[] postorder, int postStart, int postEnd){
if(inStart > inEnd || postStart > postEnd)
return null;
int rootValue = postorder[postEnd];
TreeNode root = new TreeNode(rootValue);
int k=0;
for(int i=0; i< inorder.length; i++){
if(inorder[i]==rootValue){
k = i;
break;
}
}
root.left = buildTree(inorder, inStart, k-1, postorder, postStart,
postStart+k-(inStart+1));
// Becuase k is not the length, it it need to -(inStart+1) to get the
length
root.right = buildTree(inorder, k+1, inEnd, postorder,
postStart+k-inStart, postEnd-1);
// postStart+k-inStart = postStart+k-(inStart+1) +1
return root;
}
}
137 | 181

77 Convert Sorted Array to Binary
Search Tree
Given an array where elements are sorted in ascending order, convert it to a height
balanced BST.
77.1 Thoughts
Straightforward! Recursively do the job.
77.2 Java Solution
class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
}
public TreeNode sortedArrayToBST(int[] num) {
if (num.length == 0)
return null;
return sortedArrayToBST(num, 0, num.length - 1);
}
public TreeNode sortedArrayToBST(int[] num, int start, int end) {
if (start > end)
return null;
int mid = (start + end) / 2;
TreeNode root = new TreeNode(num[mid]);
root.left = sortedArrayToBST(num, start, mid - 1);
root.right = sortedArrayToBST(num, mid + 1, end);
return root;
}
}
138 | 181

78 Convert Sorted List to Binary Search Tree
78 Convert Sorted List to Binary Search
Tree
Given a singly linked list where elements are sorted in ascending order, convert it to a
height balanced BST.
78.1 Thoughts
If you are given an array, the problem is quite straightforward. But things get a little
more complicated when you have a singly linked list instead of an array. Now you no
longer have random access to an element in O(1) time. Therefore, you need to create
nodes bottom-up, and assign them to its parents. The bottom-up approach enables us
to access the list in its order at the same time as creating nodes.
78.2 Java Solution
class ListNode {
int val;
ListNode next;
ListNode(int x) {
val = x;
next = null;
}
}
class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
}
static ListNode h;
public TreeNode sortedListToBST(ListNode head) {
if (head == null)
return null;

h = head;
int len = getLength(head);
return sortedListToBST(0, len - 1);
}
// get list length
public int getLength(ListNode head) {
int len = 0;
ListNode p = head;
while (p != null) {
len++;
p = p.next;
}
return len;
}
// build tree bottom-up
public TreeNode sortedListToBST(int start, int end) {
if (start > end)
return null;
// mid
int mid = (start + end) / 2;
TreeNode left = sortedListToBST(start, mid - 1);
TreeNode root = new TreeNode(h.val);
h = h.next;
TreeNode right = sortedListToBST(mid + 1, end);
root.left = left;
root.right = right;
return root;
}
}
79 Minimum Depth of Binary Tree
Given a binary tree, find its minimum depth.
The minimum depth is the number of nodes along the shortest path from the root
node down to the nearest leaf node.
140 | 181

79 Minimum Depth of Binary Tree
79.1 Thoughts
Need to know LinkedList is a queue. add() and remove() are the two methods to
manipulate the queue.
79.2 Java Solution
/**
* int val;
* TreeNode left;
* TreeNode right;
* }
*/
public int minDepth(TreeNode root) {
if(root == null){
return 0;
}
LinkedList<TreeNode> nodes = new LinkedList<TreeNode>();
LinkedList<Integer> counts = new LinkedList<Integer>();
nodes.add(root);
counts.add(1);
while(!nodes.isEmpty()){
TreeNode curr = nodes.remove();
int count = counts.remove();
nodes.add(curr.left);
counts.add(count+1);
}
nodes.add(curr.right);
counts.add(count+1);
}
if(curr.left == null && curr.right == null){
return count;
}
}
return 0;
}

}
80 Binary Tree Maximum Path Sum
Given a binary tree, find the maximum path sum.
The path may start and end at any node in the tree.
For example: Given the below binary tree,
1
/
2 3
Return 6.
80.1 Thoughts
1) Recursively solve this problem 2) Get largest left sum and right sum 2) Compare to
the stored maximum
class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
}
//store max value
int max;
public int maxPathSum(TreeNode root) {
max = (root == null) ? 0 : root.val;
findMax(root);
return max;
}
142 | 181

public int findMax(TreeNode node) {
if (node == null)
return 0;
// recursively get sum of left and right path
int left = Math.max(findMax(node.left), 0);
int right = Math.max(findMax(node.right), 0);
//update maximum here
max = Math.max(node.val + left + right, max);
// return sum of largest path of current node
return node.val + Math.max(left, right);
}
}
We can also use an array to store value for recursive methods.
public int maxPathSum(TreeNode root) {
int max[] = new int[1];
max[0] = Integer.MIN_VALUE;
calculateSum(root, max);
return max[0];
}
public int calculateSum(TreeNode root, int[] max) {
if (root == null)
return 0;
int left = calculateSum(root.left, max);
int right = calculateSum(root.right, max);
int current = Math.max(root.val, Math.max(root.val + left, root.val +
right));
max[0] = Math.max(max[0], Math.max(current, left + root.val + right));
return current;
}
}
143 | 181

81 Balanced Binary Tree
81 Balanced Binary Tree
Given a binary tree, determine if it is height-balanced.
For this problem, a height-balanced binary tree is defined as a binary tree in which
the depth of the two subtrees of every node never differ by more than 1.
81.1 Thoughts
A typical recursive problem for solving tree problems.
81.2 Java Solution
class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
}
public boolean isBalanced(TreeNode root) {
if (root == null)
return true;
if (getHeight(root) == -1)
return false;
return true;
}
public int getHeight(TreeNode root) {
if (root == null)
return 0;
int left = getHeight(root.left);
int right = getHeight(root.right);
if (left == -1 || right == -1)
return -1;
if (Math.abs(left - right) > 1) {
return -1;

}
return Math.max(left, right) + 1;
}
}
82 Symmetric Tree
82.1 Problem
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its
center).
For example, this binary tree is symmetric:
1
/
2 2
/ /
3 4 4 3
But the following is not:
1
/
2 2

3 3
82.2 Java Solution - Recursion
This problem can be solve by using a simple recursion. The key is finding the con-
ditions that return false, such as value is not equal, only one node(left or right) has
value.
public boolean isSymmetric(TreeNode root) {
if (root == null)
return true;
return isSymmetric(root.left, root.right);
}
public boolean isSymmetric(TreeNode l, TreeNode r) {
145 | 181

if (l == null && r == null) {
return true;
} else if (r == null || l == null) {
return false;
}
if (l.val != r.val)
return false;
if (!isSymmetric(l.left, r.right))
return false;
if (!isSymmetric(l.right, r.left))
return false;
return true;
}
83 Clone Graph Java
LeetCode Problem:
Clone an undirected graph. Each node in the graph contains a label and a list of its
neighbors.
146 | 181

83 Clone Graph Java
83.1 Key to Solve This Problem
• A queue is used to do breath first traversal.
• a map is used to store the visited nodes. It is the map between original node and
copied node.
It would be helpful if you draw a diagram and visualize the problem.

83 Clone Graph Java
/**
* Definition for undirected graph.
* class UndirectedGraphNode {
* int label;
* ArrayList<UndirectedGraphNode> neighbors;
* UndirectedGraphNode(int x) { label = x; neighbors = new
ArrayList<UndirectedGraphNode>(); }
* };
*/
public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
if(node == null)
return null;
LinkedList<UndirectedGraphNode> queue = new
LinkedList<UndirectedGraphNode>();
HashMap<UndirectedGraphNode, UndirectedGraphNode> map =
new
HashMap<UndirectedGraphNode,UndirectedGraphNode>();
UndirectedGraphNode newHead = new UndirectedGraphNode(node.label);
queue.add(node);
map.put(node, newHead);

while(!queue.isEmpty()){
UndirectedGraphNode curr = queue.pop();
ArrayList<UndirectedGraphNode> currNeighbors = curr.neighbors;
for(UndirectedGraphNode aNeighbor: currNeighbors){
if(!map.containsKey(aNeighbor)){
UndirectedGraphNode copy = new
UndirectedGraphNode(aNeighbor.label);
map.put(aNeighbor,copy);
map.get(curr).neighbors.add(copy);
queue.add(aNeighbor);
}else{
map.get(curr).neighbors.add(map.get(aNeighbor));
}
}
}
return newHead;
}
}
84 How Developers Sort in Java?
While analyzing source code of a large number of open source Java projects, I found
Java developers frequently sort in two ways. One is using the sort() method of Col-
lections or Arrays, and the other is using sorted data structures, such as TreeMap and
TreeSet.
84.1 Using sort() Method
If it is a collection, use Collections.sort() method.
// Collections.sort
List<ObjectName> list = new ArrayList<ObjectName>();
Collections.sort(list, new Comparator<ObjectName>() {
public int compare(ObjectName o1, ObjectName o2) {
return o1.toString().compareTo(o2.toString());
}
});
If it is an array, use Arrays.sort() method.
149 | 181

84 How Developers Sort in Java?
// Arrays.sort
ObjectName[] arr = new ObjectName[10];
Arrays.sort(arr, new Comparator<ObjectName>() {
}
});
This is very convenient if a collection or an array is already set up.
84.2 Using Sorted Data Structures
If it is a list or set, use TreeSet to sort.
// TreeSet
Set<ObjectName> sortedSet = new TreeSet<ObjectName>(new
Comparator<ObjectName>() {
}
});
sortedSet.addAll(unsortedSet);
If it is a map, use TreeMap to sort. TreeMap is sorted by key.
// TreeMap - using String.CASE_INSENSITIVE_ORDER which is a Comparator that
orders Strings by compareToIgnoreCase
Map<String, Integer> sortedMap = new TreeMap<String,
Integer>(String.CASE_INSENSITIVE_ORDER);
sortedMap.putAll(unsortedMap);
//TreeMap - In general, defined comparator
Map<ObjectName, String> sortedMap = new TreeMap<ObjectName, String>(new
Comparator<ObjectName>() {
}
});
sortedMap.putAll(unsortedMap);
This approach is very useful, if you would do a lot of search operations for the
collection. The sorted data structure will give time complexity of O(logn), which is
lower than O(n).
84.3 Bad Practices
There are still bad practices, such as using self-defined sorting algorithm. Take the
code below for example, not only the algorithm is not efficient, but also it is not

readable. This happens a lot in different forms of variations.
double t;
for (int i = 0; i < 2; i++)
for (int j = i + 1; j < 3; j++)
if (r[j] < r[i]) {
t = r[i];
r[i] = r[j];
r[j] = t;
}
85 Solution Merge Sort LinkedList in Java
LeetCode - Sort List:
Sort a linked list in O(n log n) time using constant space complexity.
85.1 Keys for solving the problem
• Break the list to two in the middle
• Recursively sort the two sub lists
• Merge the two sub lists
This is my accepted answer for the problem.
package algorithm.sort;
class ListNode {
int val;
ListNode next;
ListNode(int x) {
val = x;
next = null;
}
}
public class SortLinkedList {
// merge sort
public static ListNode mergeSortList(ListNode head) {
if (head == null || head.next == null)
return head;
151 | 181

// count total number of elements
int count = 0;
ListNode p = head;
while (p != null) {
count++;
p = p.next;
}
// break up to two list
int middle = count / 2;
ListNode l = head, r = null;
ListNode p2 = head;
int countHalf = 0;
while (p2 != null) {
countHalf++;
ListNode next = p2.next;
if (countHalf == middle) {
p2.next = null;
r = next;
}
p2 = next;
}
// now we have two parts l and r, recursively sort them
ListNode h1 = mergeSortList(l);
ListNode h2 = mergeSortList(r);
// merge together
ListNode merged = merge(h1, h2);
return merged;
}
public static ListNode merge(ListNode l, ListNode r) {
ListNode p1 = l;
ListNode p2 = r;
ListNode fakeHead = new ListNode(100);
ListNode pNew = fakeHead;
while (p1 != null || p2 != null) {
if (p1 == null) {
pNew.next = new ListNode(p2.val);
p2 = p2.next;
pNew = pNew.next;
} else if (p2 == null) {

p1 = p1.next;
pNew = pNew.next;
} else {
if (p1.val < p2.val) {
// if(fakeHead)
p1 = p1.next;
pNew = pNew.next;
} else if (p1.val == p2.val) {
pNew.next.next = new ListNode(p1.val);
pNew = pNew.next.next;
p1 = p1.next;
p2 = p2.next;
} else {
p2 = p2.next;
pNew = pNew.next;
}
}
}
// printList(fakeHead.next);
return fakeHead.next;
}
n1.next = n2;
n2.next = n3;
n3.next = n4;
n4.next = n5;
n5.next = n6;
n1 = mergeSortList(n1);
printList(n1);
}
public static void printList(ListNode x) {
if(x != null){

System.out.print(x.val + " ");
while (x.next != null) {
System.out.print(x.next.val + " ");
x = x.next;
}
}
}
}
Output:
2 3 3 4 4 5
86 Quicksort Array in Java
Quicksort is a divide and conquer algorithm. It first divides a large list into two
smaller sub-lists and then recursively sort the two sub-lists. If we want to sort an array
without any extra space, Quicksort is a good option. On average, time complexity is
O(n log(n)).
The basic step of sorting an array are as follows:
• Select a pivot, normally the middle one
• From both ends, swap elements and make all elements on the left less than the
pivot and all elements on the right greater than the pivot
• Recursively sort left part and right part
public class QuickSort {
int[] x = { 9, 2, 4, 7, 3, 7, 10 };
printArray(x);
int low = 0;
int high = x.length - 1;
quickSort(x, low, high);
printArray(x);
}
public static void quickSort(int[] arr, int low, int high) {
154 | 181

if (arr == null || arr.length == 0)
return;
if (low >= high)
return;
//pick the pivot
int middle = low + (high - low) / 2;
int pivot = arr[middle];
//make left < pivot and right > pivot
int i = low, j = high;
while (i <= j) {
while (arr[i] < pivot) {
i++;
}
while (arr[j] > pivot) {
j--;
}
if (i <= j) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
i++;
j--;
}
}
//recursively sort two sub parts
if (low < j)
quickSort(arr, low, j);
if (high > i)
quickSort(arr, i, high);
}
public static void printArray(int[] x) {
for (int a : x)
System.out.print(a + " ");
}
}
Output:
9 2 4 7 3 7 10 2 3 4 7 7 9 10
The mistake I made is selecting the middle element. The middle element is not
(low+high)/2, but low + (high-low)/2. For other parts of the programs, just follow the
155 | 181

87 Solution Sort a linked list using insertion sort in Java
algorithm.
87 Solution Sort a linked list using
insertion sort in Java
Insertion Sort List:
Sort a linked list using insertion sort.
This is my accepted answer for LeetCode problem - Sort a linked list using insertion
sort in Java. It is a complete program.
Before coding for that, here is an example of insertion sort from wiki. You can get
an idea of what is insertion sort.
Code:
class ListNode {
int val;
ListNode next;
ListNode(int x) {
val = x;
next = null;
}
}
public class SortLinkedList {
public static ListNode insertionSortList(ListNode head) {

87 Solution Sort a linked list using insertion sort in Java
if (head == null || head.next == null)
return head;
ListNode newHead = new ListNode(head.val);
ListNode pointer = head.next;
// loop through each element in the list
while (pointer != null) {
// insert this element to the new list
ListNode innerPointer = newHead;
ListNode next = pointer.next;
if (pointer.val <= newHead.val) {
ListNode oldHead = newHead;
newHead = pointer;
newHead.next = oldHead;
} else {
while (innerPointer.next != null) {
if (pointer.val > innerPointer.val && pointer.val <=
innerPointer.next.val) {
ListNode oldNext = innerPointer.next;
innerPointer.next = pointer;
pointer.next = oldNext;
}
innerPointer = innerPointer.next;
}
if (innerPointer.next == null && pointer.val > innerPointer.val) {
innerPointer.next = pointer;
pointer.next = null;
}
}
// finally
pointer = next;
}
return newHead;
}

n1.next = n2;
n2.next = n3;
n3.next = n4;
n4.next = n5;
n5.next = n6;
n1 = insertionSortList(n1);
printList(n1);
}
public static void printList(ListNode x) {
if(x != null){
System.out.print(x.val + " ");
while (x.next != null) {
System.out.print(x.next.val + " ");
x = x.next;
}
}
}
}
Output:
2 3 3 4 4 5
88 Maximum Gap
88.1 Problem
Given an unsorted array, find the maximum difference between the successive ele-
ments in its sorted form.
Try to solve it in linear time/space. Return 0 if the array contains less than 2 ele-
ments. You may assume all elements in the array are non-negative integers and fit in
the 32-bit signed integer range.
88.2 Java Solution 1 - Sort
A straightforward solution would be sorting the array first (O(nlogn) and then finding
the maximum gap. The basic idea is to project each element of the array to an array of
158 | 181

88 Maximum Gap
buckets. Each bucket tracks the maximum and minimum elements. Finally, scanning
the bucket list, we can get the maximum gap.
The key part is to get the interval:
From: interval * (num[i] - min) = 0 and interval * (max -num[i]) = n
interval = num.length / (max - min)
See the internal comment for more details.
88.3 Java Solution 2 - Bucket Sort
We can use a bucket-sort like algorithm to solve this problem in time of O(n) and space
O(n).
class Bucket{
int low;
int high;
public Bucket(){
low = -1;
high = -1;
}
}
public int maximumGap(int[] num) {
if(num == null || num.length < 2){
return 0;
}
int max = num[0];
int min = num[0];
max = Math.max(max, num[i]);
min = Math.min(min, num[i]);
}
// initialize an array of buckets
Bucket[] buckets = new Bucket[num.length+1]; //project to (0 - n)
for(int i=0; i<buckets.length; i++){
buckets[i] = new Bucket();
}
double interval = (double) num.length / (max - min);
//distribute every number to a bucket array
int index = (int) ((num[i] - min) * interval);
if(buckets[index].low == -1){
buckets[index].low = num[i];
buckets[index].high = num[i];

}else{
buckets[index].low = Math.min(buckets[index].low, num[i]);
buckets[index].high = Math.max(buckets[index].high, num[i]);
}
}
//scan buckets to find maximum gap
int result = 0;
int prev = buckets[0].high;
for(int i=1; i<buckets.length; i++){
if(buckets[i].low != -1){
result = Math.max(result, buckets[i].low-prev);
prev = buckets[i].high;
}
}
return result;
}
89 Iteration vs. Recursion in Java
89.1 Recursion
Consider the factorial function: n!=n*(n-1)*(n-2)*...*1
There are many ways to compute factorials. One way is that n! is equal to n*(n-1)!.
Therefore the program can be directly written as:
Program 1:
int factorial (int n) {
if (n == 1) {
return 1;
} else {
return n*factorial(n-1);
}
}
In order to run this program, the computer needs to build up a chain of multipli-
cations: factorial(n) → factorial(n-1) → factorial(n-2) → ... → factorial(1). Therefore,
the computer has to keep track of the multiplications to be performed later on. This
type of program, characterized by a chain of operations, is called recursion. Recursion
can be further categorized into linear and tree recursion. When the amount of infor-
mation needed to keep track of the chain of operations grows linearly with the input,
160 | 181

the recursion is called linear recursion. The computation of n! is such a case, because
the time required grows linearly with n. Another type of recursion, tree recursion,
happens when the amount of information grows exponentially with the input. But we
will leave it undiscussed here and go back shortly afterwards.
89.2 Iteration
A different perspective on computing factorials is by first multiplying 1 by 2, then
multiplying the result by 3, then by 4, and so on until n. More formally, the program
can use a counter that counts from 1 up to n and compute the product simultaneously
until the counter exceeds n. Therefore the program can be written as:
Program 2:
int factorial (int n) {
int product = 1;
for(int i=2; i<n; i++) {
product *= i;
}
return product;
}
This program, by contrast to program 2, does not build a chain of multiplication. At
each step, the computer only need to keep track of the current values of the product
and i. This type of program is called iteration, whose state can be summarized by
a fixed number of variables, a fixed rule that describes how the variables should be
updated, and an end test that specifies conditions under which the process should
terminate. Same as recursion, when the time required grows linearly with the input,
we call the iteration linear recursion.
89.3 Recursion vs Iteration
Compared the two processes, we can find that they seem almost same, especially in
term of mathematical function. They both require a number of steps proportional to
n to compute n!. On the other hand, when we consider the running processes of the
two programs, they evolve quite differently.
In the iterative case, the program variables provide a complete description of the
state. If we stopped the computation in the middle, to resume it only need to supply
the computer with all variables. However, in the recursive process, information is
maintained by the computer, therefore "hidden" to the program. This makes it almost
impossible to resume the program after stopping it.
89.4 Tree recursion
As described above, tree recursion happens when the amount of information grows
exponentially with the input. For instance, consider the sequence of Fibonacci num-

bers defined as follows:
By the definition, Fibonacci numbers have the following sequence, where each num-
ber is the sum of the previous two: 0, 1, 1, 2, 3, 5, 8, 13, 21, ...
A recursive program can be immediately written as:
Program 3:
int fib (int n) {
if (n == 0) {
return 0;
} else if (n == 1) {
return 1;
} else {
return fib(n-1) + fib(n-2);
}
}
Therefore, to compute fib(5), the program computes fib(4) and fib(3). To computer
fib(4), it computes fib(3) and fib(2). Notice that the fib procedure calls itself twice at
the last line. Two observations can be obtained from the definition and the program:
• The ith Fibonacci number Fib(i) is equal to phi(i)/rootsquare(5) rounded to the
nearest integer, which indicates that Fibonacci numbers grow exponentially.
• This is a bad way to compute Fibonacci numbers because it does redundant
computation. Computing the running time of this procedure is beyond the
scope of this article, but one can easily find that in books of algorithms, which is
O(phi(n)). Thus, the program takes an amount of time that grows exponentially
with the input.
On the other hand, we can also write the program in an iterative way for computing
the Fibonacci numbers. Program 4 is a linear iteration. The difference in time required
by Program 3 and 4 is enormous, even for small inputs.
Program 4:
int fib (int n) {
int fib = 0;
int a = 1;
for(int i=0; i<n; i++) {
fib = fib + a;
a = fib;
}
return fib;
}

However, one should not think tree-recursive programs are useless. When we con-
sider programs that operate on hierarchically data structures rather than numbers,
tree-recursion is a natural and powerful tool. It can help us understand and design
programs. Compared with Program 3 and 4, we can easily tell Program 3 is more
straightforward, even if less efficient. After that, we can most likely reformulate the
program into an iterative way.
90 Edit Distance in Java
From Wiki:
In computer science, edit distance is a way of quantifying how dissimilar two strings
(e.g., words) are to one another by counting the minimum number of operations required
to transform one string into the other.
There are three operations permitted on a word: replace, delete, insert. For example,
the edit distance between "a" and "b" is 1, the edit distance between "abc" and "def" is
3. This post analyzes how to calculate edit distance by using dynamic programming.
90.1 Key Analysis
Let dp[i][j] stands for the edit distance between two strings with length i and j, i.e.,
word1[0,...,i-1] and word2[0,...,j-1]. There is a relation between dp[i][j] and dp[i-1][j-1].
Let’s say we transform from one string to another. The first string has length i and
it’s last character is "x"; the second string has length j and its last character is "y". The
following diagram shows the relation.
163 | 181

90 Edit Distance in Java
• if x == y, then dp[i][j] == dp[i-1][j-1]
• if x != y, and we insert y for word1, then dp[i][j] = dp[i][j-1] + 1
• if x != y, and we delete x for word1, then dp[i][j] = dp[i-1][j] + 1
• if x != y, and we replace x with y for word1, then dp[i][j] = dp[i-1][j-1] + 1
• When x!=y, dp[i][j] is the min of the three situations.
Initial condition: dp[i][0] = i, dp[0][j] = j
90.2 Java Code
After the analysis above, the code is just a representation of it.
public static int minDistance(String word1, String word2) {
int len1 = word1.length();
int len2 = word2.length();
// len1+1, len2+1, because finally return dp[len1][len2]
int[][] dp = new int[len1 + 1][len2 + 1];
for (int i = 0; i <= len1; i++) {
dp[i][0] = i;
}
for (int j = 0; j <= len2; j++) {
dp[0][j] = j;
}

//iterate though, and check last char
for (int i = 0; i < len1; i++) {
char c1 = word1.charAt(i);
for (int j = 0; j < len2; j++) {
char c2 = word2.charAt(j);
//if last two chars equal
if (c1 == c2) {
//update dp value for +1 length
dp[i + 1][j + 1] = dp[i][j];
} else {
int replace = dp[i][j] + 1;
int insert = dp[i][j + 1] + 1;
int delete = dp[i + 1][j] + 1;
int min = replace > insert ? insert : replace;
min = delete > min ? min : delete;
dp[i + 1][j + 1] = min;
}
}
}
return dp[len1][len2];
}
91 Single Number
The problem:
Given an array of integers, every element appears twice except for one. Find that single
one.
91.1 Thoughts
The key to solve this problem is bit manipulation. XOR will return 1 only on two
different bits. So if two numbers are the same, XOR will return 0. Finally only one
number left.
91.2 Java Solution
public int singleNumber(int[] A) {
165 | 181

int x=0;
for(int a: A){
x = x ^ a;
}
return x;
}
}
The question now is do you know any other ways to do this?
92 Single Number II
92.1 Problem
Given an array of integers, every element appears three times except for one. Find that
single one.
92.2 Java Solution
This problem is similar to Single Number.
public int singleNumber(int[] A) {
int ones = 0, twos = 0, threes = 0;
twos |= ones & A[i];
ones ^= A[i];
threes = ones & twos;
ones &= ~threes;
twos &= ~threes;
}
return ones;
}
93 Twitter Codility Problem Max Binary
Gap
Problem: Get maximum binary Gap.
166 | 181

For example, 9’s binary form is 1001, the gap is 2.
93.1 Thoughts
The key to solve this problem is the fact that an integer x & 1 will get the last digit of
the integer.
93.2 Java Solution
public static int solution(int N) {
int max = 0;
int count = -1;
int r = 0;
while (N > 0) {
// get right most bit & shift right
r = N & 1;
N = N >> 1;
if (0 == r && count >= 0) {
count++;
}
if (1 == r) {
max = count > max ? count : max;
count = 0;
}
}
return max;
}
System.out.println(solution(9));
}
}
94 Number of 1 Bits
167 | 181

94.1 Problem
Write a function that takes an unsigned integer and returns the number of ’1’ bits it
has (also known as the Hamming weight).
For example, the 32-bit integer ’11’ has binary representation 00000000000000000000000000001011,
so the function should return 3.
94.2 Java Solution
public int hammingWeight(int n) {
int count = 0;
for(int i=1; i<33; i++){
if(getBit(n, i) == true){
count++;
}
}
return count;
}
public boolean getBit(int n, int i){
return (n & (1 << i)) != 0;
}
95 Reverse Bits
95.1 Problem
Reverse bits of a given 32 bits unsigned integer.
For example, given input 43261596 (represented in binary as 00000010100101000001111010011100),
return 964176192 (represented in binary as 00111001011110000010100101000000).
Follow up: If this function is called many times, how would you optimize it?
Related problem: Reverse Integer
95.2 Java Solution
public int reverseBits(int n) {
for (int i = 0; i < 16; i++) {
n = swapBits(n, i, 32 - i - 1);
}
168 | 181

return n;
}
public int swapBits(int n, int i, int j) {
int a = (n >> i) & 1;
int b = (n >> j) & 1;
if ((a ^ b) != 0) {
return n ^= (1 << i) | (1 << j);
}
return n;
}
96 Permutations
Given a collection of numbers, return all possible permutations.
For example,
[1,2,3] have the following permutations:
[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].
We can get all permutations by the following steps:
[1]
[2, 1]
[1, 2]
[3, 2, 1]
[2, 3, 1]
[2, 1, 3]
[3, 1, 2]
[1, 3, 2]
[1, 2, 3]
Loop through the array, in each iteration, a new number is added to different loca-
tions of results of previous iteration. Start from an empty List.
public ArrayList<ArrayList<Integer>> permute(int[] num) {
//start from an empty list
169 | 181

96 Permutations
result.add(new ArrayList<Integer>());
//list of list in current iteration of the array num
ArrayList<ArrayList<Integer>> current = new
for (ArrayList<Integer> l : result) {
// # of locations to insert is largest index + 1
for (int j = 0; j < l.size()+1; j++) {
// + add num[i] to different locations
l.add(j, num[i]);
ArrayList<Integer> temp = new ArrayList<Integer>(l);
current.add(temp);
//System.out.println(temp);
// - remove num[i] add
l.remove(j);
}
}
result = new ArrayList<ArrayList<Integer>>(current);
}
return result;
}
We can also recursively solve this problem. Swap each element with each element
after it.
public ArrayList<ArrayList<Integer>> permute(int[] num) {
permute(num, 0, result);
return result;
}
void permute(int[] num, int start, ArrayList<ArrayList<Integer>> result) {
if (start >= num.length) {
ArrayList<Integer> item = convertArrayToList(num);
result.add(item);
}
for (int j = start; j <= num.length - 1; j++) {

swap(num, start, j);
permute(num, start + 1, result);
}
}
private ArrayList<Integer> convertArrayToList(int[] num) {
for (int h = 0; h < num.length; h++) {
item.add(num[h]);
}
return item;
}
private void swap(int[] a, int i, int j) {
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
97 Permutations II
Given a collection of numbers that might contain duplicates, return all possible unique
permutations.
For example, [1,1,2] have the following unique permutations:
[1,1,2], [1,2,1], and [2,1,1].
97.1 Basic Idea
For each number in the array, swap it with every element after it. To avoid duplicate,
we need to check the existing sequence first.
public ArrayList<ArrayList<Integer>> permuteUnique(int[] num) {
permuteUnique(num, 0, result);
return result;
}
171 | 181

97 Permutations II
private void permuteUnique(int[] num, int start,
ArrayList<ArrayList<Integer>> result) {
if (start >= num.length ) {
ArrayList<Integer> item = convertArrayToList(num);
result.add(item);
}
for (int j = start; j <= num.length-1; j++) {
if (containsDuplicate(num, start, j)) {
permuteUnique(num, start + 1, result);
}
}
}
private ArrayList<Integer> convertArrayToList(int[] num) {
for (int h = 0; h < num.length; h++) {
item.add(num[h]);
}
return item;
}
private boolean containsDuplicate(int[] arr, int start, int end) {
for (int i = start; i <= end-1; i++) {
if (arr[i] == arr[end]) {
return false;
}
}
return true;
}
private void swap(int[] a, int i, int j) {
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
Use set to maintain uniqueness:
public static ArrayList<ArrayList<Integer>> permuteUnique(int[] num) {
ArrayList<ArrayList<Integer>> returnList = new
returnList.add(new ArrayList<Integer>());

Set<ArrayList<Integer>> currentSet = new HashSet<ArrayList<Integer>>();
for (List<Integer> l : returnList) {
for (int j = 0; j < l.size() + 1; j++) {
l.add(j, num[i]);
ArrayList<Integer> T = new ArrayList<Integer>(l);
l.remove(j);
currentSet.add(T);
}
}
returnList = new ArrayList<ArrayList<Integer>>(currentSet);
}
return returnList;
}
Thanks to Milan for such a simple solution!
98 Permutation Sequence
The set [1,2,3,. . . ,n] contains a total of n! unique permutations.
By listing and labeling all of the permutations in order, We get the following se-
quence (ie, for n = 3):
"123"
"132"
"213"
"231"
"312"
"321"
Given n and k, return the kth permutation sequence.
Note: Given n will be between 1 and 9 inclusive.
98.1 Thoughts
Naively loop through all cases will not work.
public String getPermutation(int n, int k) {
// initialize all numbers
173 | 181

98 Permutation Sequence
ArrayList<Integer> numberList = new ArrayList<Integer>();
for (int i = 1; i <= n; i++) {
numberList.add(i);
}
// change k to be index
k--;
// set factorial of n
int mod = 1;
for (int i = 1; i <= n; i++) {
mod = mod * i;
}
String result = "";
// find sequence
for (int i = 0; i < n; i++) {
mod = mod / (n - i);
// find the right number(curIndex) of
int curIndex = k / mod;
// update k
k = k % mod;
// get number according to curIndex
result += numberList.get(curIndex);
// remove from list
numberList.remove(curIndex);
}
return result.toString();
}
}
public String getPermutation(int n, int k) {
boolean[] output = new boolean[n];
StringBuilder buf = new StringBuilder("");
int[] res = new int[n];
res[0] = 1;
for (int i = 1; i < n; i++)
res[i] = res[i - 1] * i;
for (int i = n - 1; i >= 0; i--) {

int s = 1;
while (k > res[i]) {
s++;
k = k - res[i];
}
for (int j = 0; j < n; j++) {
if (j + 1 <= s && output[j]) {
s++;
}
}
output[s - 1] = true;
buf.append(Integer.toString(s));
}
return buf.toString();
}
}
99 Generate Parentheses
Given n pairs of parentheses, write a function to generate all combinations of well-
formed parentheses.
For example, given n = 3, a solution set is:
"((()))", "(()())", "(())()", "()(())", "()()()"
99.1 Java Solution
Read the following solution, give n=2, walk though the code. Hopefully you will
quickly get an idea.
public List<String> generateParenthesis(int n) {
ArrayList<String> result = new ArrayList<String>();
ArrayList<Integer> diff = new ArrayList<Integer>();
result.add("");
diff.add(0);
for (int i = 0; i < 2 * n; i++) {
ArrayList<String> temp1 = new ArrayList<String>();
175 | 181

ArrayList<Integer> temp2 = new ArrayList<Integer>();
for (int j = 0; j < result.size(); j++) {
String s = result.get(j);
int k = diff.get(j);
if (i < 2 * n - 1) {
temp1.add(s + "(");
temp2.add(k + 1);
}
if (k > 0 && i < 2 * n - 1 || k == 1 && i == 2 * n - 1) {
temp1.add(s + ")");
temp2.add(k - 1);
}
}
result = new ArrayList<String>(temp1);
diff = new ArrayList<Integer>(temp2);
}
return result;
}
Solution is provided first now. I will come back and draw a diagram to explain the
solution.
100 Reverse Integer
LeetCode - Reverse Integer:
Reverse digits of an integer. Example1: x = 123, return 321 Example2: x = -123, return
-321
100.1 Naive Method
We can convert the integer to a string/char array, reverse the order, and convert the
string/char array back to an integer. However, this will require extra space for the
string. It doesn’t seem to be the right way, if you come with such a solution.
100.2 Efficient Approach
Actually, this can be done by using the following code.
public int reverse(int x) {
176 | 181

100 Reverse Integer
//flag marks if x is negative
if (x < 0) {
x = 0 - x;
flag = true;
}
int res = 0;
int p = x;
while (p > 0) {
int mod = p % 10;
p = p / 10;
res = res * 10 + mod;
}
if (flag) {
res = 0 - res;
}
return res;
}
100.3 Succinct Solution
This solution is from Sherry, it is succinct and it is pretty.
public int reverse(int x) {
int rev = 0;
while(x != 0){
rev = rev*10 + x%10;
x = x/10;
}
return rev;
}
100.4 Handle Out of Range Problem
As we form a new integer, it is possible that the number is out of range. We can use
the following code to assign the newly formed integer. When it is out of range, throw
an exception.
try{
result = ...;
}catch(InputMismatchException exception){
System.out.println("This is not an integer");

}
Please leave your comment if there is any better solutions.
101 Palindrome Number
Determine whether an integer is a palindrome. Do this without extra space.
101.1 Thoughts
Problems related with numbers are frequently solved by / and
Note: no extra space here means do not convert the integer to string, since string
will be a copy of the integer and take extra space. The space take by div, left, and right
can be ignored.
101.2 Java Solution
public boolean isPalindrome(int x) {
//negative numbers are not palindrome
if (x < 0)
return false;
// initialize how many zeros
int div = 1;
while (x / div >= 10) {
div *= 10;
}
while (x != 0) {
int left = x / div;
int right = x % 10;
if (left != right)
return false;
x = (x % div) / 10;
div /= 100;
}
return true;
}
}
178 | 181

102 Pow(x, n)
102 Pow(x, n)
Problem:
Implement pow(x, n).
This is a great example to illustrate how to solve a problem during a technical in-
terview. The first and second solution exceeds time limit; the third and fourth are
accepted.
102.1 Naive Method
First of all, assuming n is not negative, to calculate x to the power of n, we can simply
multiply x n times, i.e., x * x * ... * x. The time complexity is O(n). The implementation
is as simple as:
public double pow(double x, int n) {
if(x == 0) return 0;
if(n == 0) return 1;
double result=1;
for(int i=1; i<=n; i++){
result = result * x;
}
return result;
}
}
Now we should think about how to do better than O(n).
102.2 Recursive Method
Naturally, we next may think how to do it in O(logn). We have a relation that xn̂ =
xˆ
(n/2) * xˆ
(n/2) * xˆ
(n
public static double pow(double x, int n) {
if(n == 0)
return 1;
if(n == 1)
return x;

102 Pow(x, n)
int half = n/2;
int remainder = n%2;
if(n % 2 ==1 && x < 0 && n < 0)
return - 1/(pow(-x, half) * pow(-x, half) * pow(-x, remainder));
else if (n < 0)
return 1/(pow(x, -half) * pow(x, -half) * pow(x, -remainder));
else
return (pow(x, half) * pow(x, half) * pow(x, remainder));
}
In this solution, we can handle cases that x <0 and n <0. This solution actually takes
more time than the first solution. Why?
102.3 Accepted Solution
The accepted solution is also recursive, but does division first. Time complexity is
O(nlog(n)). The key part of solving this problem is the while loop.
if (n == 0)
return 1;
if (n == 1)
return x;
int pn = n > 0 ? n : -n;// positive n
int pn2 = pn;
double px = x > 0 ? x : -x;// positive x
double result = px;
int k = 1;
//the key part of solving this problem
while (pn / 2 > 0) {
result = result * result;
pn = pn / 2;
k = k * 2;
}
result = result * pow(px, pn2 - k);
// handle negative result
if (x < 0 && n % 2 == 1)
result = -result;
// handle negative power
if (n < 0)
result = 1 / result;

102 Pow(x, n)
return result;
}
102.4 Best Solution
The most understandable solution I have found so far.
public double power(double x, int n) {
if (n == 0)
return 1;
double v = power(x, n / 2);
if (n % 2 == 0) {
return v * v;
} else {
return v * v * x;
}
}
if (n < 0) {
return 1 / power(x, -n);
} else {
return power(x, n);
}
}

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