Lecture 2 DataStrucure - complexity , IS.pptx

Complexity of an algorithm
CS201, DS, B.Tech CSE/IT/EE 3rd Semester, 2nd Year 1
• Sometimes, there are more than one way to solve a problem.
• We need to learn how to compare the performance of different
algorithms and choose the best one to solve a particular problem.
• While analyzing an algorithm, we mostly consider time complexity and
space complexity.
• Time complexity of an algorithm quantifies the amount of time taken by
an algorithm to run as a function of the length of the input.
• Space complexity of an algorithm quantifies the amount of space or
memory taken by an algorithm to run as a function of the length of the
input.
• Time & space complexity depends on lots of things like hardware,
operating system, processors, etc.
• However, we don't consider any of these factors while analyzing the
algorithm. We will only consider the execution time of an algorithm.

Unit 1 : Calculating Time Complexity
Q: What are the number of elements, If first element is at 1 and last element
is at 10 ?
Q: What are the no. of elements, if first element is at 527 and last element is
at 9346 ?
Length of array : UB – LB + 1

Unit 1 : Time space trade off
• Calculate Time Complexity of Algorithm
• Time Complexity is most commonly estimated by counting the number
of elementary functions performed by the algorithm.
• Since the algorithm's performance may vary with different types of
input data,
• Hence for an algorithm we usually use the worst-case Time complexity
of an algorithm because that is the maximum time taken for any input
size.

Unit 1 : Time space trade off
• Execution Time Cases
•There are three cases which are usually used to compare various data
structure's execution time in a relative manner.
 Worst Case − This is the scenario where a particular data structure
operation takes maximum time it can take. If an operation's worst case
time is ƒ(n) then this operation will not take more than ƒ(n) time
where ƒ(n) represents function of n.
 Average Case − This is the scenario depicting the average execution
time of an operation of a data structure. If an operation takes ƒ(n) time
in execution, then m operations will take mƒ(n) time.
 Best Case − This is the scenario depicting the least possible execution
time of an operation of a data structure. If an operation takes ƒ(n) time
in execution, then the actual operation may take time as the random
number which would be maximum as ƒ(n).

Case of : Insertion sort & related questions
• Q1: What are the number of comparison required to insert 40? // fig : 1
• Hint : j = 4 , i = 3
• Answer : 2
• Q2: How many times the while loop execute?
• Answer : 2
• // for i= 3 condition is true
• // for i=2 condition is false
• Q3 : What are the number of comparison required to insert 25 ? // fig :2
• Hint : j = 5, i = 4
• Answer : 4
• Answer: 4
• //for i=4,3,2 condition is true and for i= 1 condition is false
• Q5 : What are the number of comparison required to insert 10 ? //fig : 2
• Hint : j = 5, i=4
• Answer : 4
• Answer : 5 // for i = 4,3,2,1 condition is true & for i=0 condition is false
20 30 50
1 2 3
20 30 40 50
1 2 3 4
Algorithm IS (a,n)
{
for ( j=2 to n)
{ // j -> the position of element
which we want to insert in
sorted
// i -> the no. of elements
already in sorted array
// i -> max no of comparisons
Key = a[j] ;
i = j – 1;
while ( i > 0 && key < a[i])
{
Shift a[i] to +1
position &
decrement i;
}
Place key at its correct
position
}
}

Complexity : Insertion sort
• Algorithm IS (a,n)
• { Line 1. // this algo will sort the
elements of array a[], having n elements
• Line 2. for ( j=2 to n)
• {
• Line 3. Key = a[j] ;
• Line 4. i = j – 1;
• Line 5. while ( i > 0 && key <a[i])
• {
• Line 6. a[i+1] = a[i];
• Line 7. i = i -1 ;
• }
• Line 8. a[i+1] = key;
• }
• }
1. Comment 0
Line
No
How to find complexity Time taken
2 No of time the for loop execute (N-2+1) + 1=
n-1+1=n
3. Loop -1 N-1
4. Loop -1 N-1
5. While loop depends on comparison of
key, key is dependent on J, i.e a2, a3,
a4, a5 also loop depend on I, ie no of
sorted elements, so loop vary from 2 to
n . We assume that the while loop
execute times , where j is from 2 to 5
ie.
T2 + T3 + T4 +T5
6. While Loop -1
7. While Loop -1
8. For loop -1 N-1
Total complexity = sum of all time
T(n) = n + (n-1) + (n-1) + + + + n-1 --- eqn: 1

Analysis of Insertion Sort
Elements are in increasing order
• Case 1 : When the elements are in increasing order
• // tj => is the number of times the while loop execute
• j =2 , key 30, i=1
• While(1>0 & key < a1) // condition false
• => t2 = 1
• J=3, key = 40, i= 2
• While(2>0 & key< a2) // condition false
• => t3 = 1
• J=4, key = 50, i=3
• While (3>0 & key <a3) // condition false
• => t4 = 1
• J=5, key =60, I = 4
• While(4>0 & key <a4) // condition false
• => t5 = 1
20 30 40 50 60
1 2 3 4 5
• { // this algo will sort the elements
of array a[], having n elements
• for ( j=2 to n)
• {
• Key = a[j] ;
• i = j – 1;
• while ( i > 0 && key < a[i])
• {
• a[i+1] = a[i];
• i = i -1 ;
• }
• a[i+1] = key;
• }
• }

Elements are in increasing order
• = 1 + 1 + 1 + 1 …………(n-2+1)
• = 1 + 1 + …….(n-1) = n-1
• 0
• // (n-1) – (n-1)= 0
T(n) = n + (n-1) + (n-1) + + + + n-1 --- eqn: 1
T(n) = n + (n-1) + (n-1) + (n-1) + 0 + 0 + (n-1)
T(n) = An + B where A and B are constants
T(n) = Ɵ (n) // theta(n)
Complexity when elements are in increasing order

Elements are in decreasing order
• Case 2 : When the elements are in decreasing order
• // tj => is the number of times the while loop execute
• j =2 , key 50, i=1
• While(1>0 & key < a1) // condition true
• Shift 60 to second position, decrement i
• While (0>0 // condition false
• But again while loop execute,
• hence we can say that for j= 2 while loop execute twice
• J=3, key = 40, i= 2
• While(2>0 & key< a2) // condition true
• Shift 60 to 3rd
position and decrement i
• While ( 1>0 & key <a1) //condition true
• Shift 50 to 2nd
position, decrement i
• While( 0 > 0 // condition false
• Hence for j= 3 while loop executes thrice
Similarly for j=4 while loop executes 4 times
And so on
60 50 40 30 20
1 2 3 4 5
• { // this algo will sort the elements
of array a[], having n elements
• for ( j=2 to n)
• {
• Key = a[j] ;
• i = j – 1;
• while ( i > 0 && key < a[i])
• {
• a[i+1] = a[i];
• i = i -1 ;
• }
• a[i+1] = key;
• }
• }
50 60
1 2
40 50 60
1 2 3

Elements are in decreasing order
• = 2 +3 + 4 + ………… + n
• = { n(n+1)/2 } - 1 // sum of n natural numbers is n(n+1)/2
• [{ n(n+1)/2 } - 1 ] – (n-1) = n(n-1)/2
T(n) = n + (n-1) + (n-1) + + + + n-1 --- eqn: 1
T(n) = n + (n-1) + (n-1) + [{n(n+1)/2} – 1] + n(n-1)/2 + n(n-1)/2 + (n-1)
T(n) = An2
+ Bn +C where A,B and C are constants
T(n) = O (n2
) // Big-oh(n2
)
Complexity when elements are in decreasing order

Complexity of Insertion sort
• T(n) = O (n2
) // Big-oh(n2
)
•Complexity when elements are in decreasing order
Total complexity
T(n) = n + (n-1) + (n-1) + + + + n-1
T(n) = O (n2
) // Big-oh(n2
)
Complexity when elements are in decreasing order
Worst Case
T(n) = Ɵ (n) // theta(n)
Complexity when elements are in increasing order
Best Case
General Equation for Insertion Sort

Lecture 2 DataStrucure - complexity , IS.pptx

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