2. Array in Data Structure

Data Structures 2
 An array is a list of a finite number of
homogenous (similar data) elements.
 This means that an array can store either all
integers, all floating point numbers, all
characters (string) or any other complex data
type but all of same type.
Arrays

• There are some important points :
– Arrays are always stored in successive memory
locations.
– An array can store multiple values which can be
referenced by a single name.
– Array name is actually a pointer to the first
location of the memory block allocated to the
name of the array
– An array either be an integer, character or floating
data type can be initialized only during declaration
time, and not afterwards
– There is no bound checking concept for arrays in
C.
Data Structures 3

• Elements of an array A may be denoted by the
subscript notation
A1, A2, …………… An
Or
A[1], A[2],………..A[n]
• The number n of elements is called the length
or size of the array.
• The number n in A[n] is called a subscript /
index and A[n] is called a subscript variable.
Data Structures 4

• Length / Size of array = UB – LB + 1
• Where UB is the largest index, called the
upper bound, and LB is the smallest index,
called the lower bound of the array
Length = UB where LB =1
Data Structures 5

Example
• An automobile company uses an array AUTO
to record the number of automobiles sold
each year from 1932 through 1984. Rather
than beginning the index set with 1, it is more
useful to begin the index set with 1932 so that
• AUTO[K] = number of automobiles sold in the
year K.
Data Structures 6

Answer
• LB = 1932 is the lower bound and UB= 1984 is
the upper bound of AUTO.
– Length = UB – LB +1
– Length = 1984 – 1932 + 1 = 53
• That is, AUTO contains 53 elements and its
index set consists of all integers from 1932
through 1984
Data Structures 7

One Dimensional Arrays
• A one dimensional array is one in which only
one subscript specification is needed to
specify a particular element of the array.
• One dimensional array can be declared as :
Data_type var_name [expression];
• Where data_type is the type of elements to be
stored in the array.
• Var_name specifies the name of array,
• Expression or subscript specifies the number
of values to be stored in the array.
Data Structures 8

• Example :
• Int num[5]
• The array will store five integer values, its
name is num
• num[0]=22, num[1] = 1, num[2]=30,
num[3]=9, num[4]=3
num
0 1 2 3 4
Data Structures 9
22 1 30 9 3

• The size of array can be determine as :
• Size = UB – LB +1
• For array num size will be :
• UB = 4
• LB = 0
• Size = 4 – 0 +1 = 5
• Size of array in bytes (i.e. amount of memory
array occupies)
Data Structures 10

• Size in bytes = size of array x size of base type
• E.g. if we have array
int num [5];
• Then size in bytes = 5 x 2 = 10 bytes
• Similarly, if we have array
float num[5];
• Then size in bytes = 5 x 4 = 20 bytes
Data Structures 11

Data Structures 12
Initializing Array
• int n[ 5 ] = { 1, 2, 3, 4, 5 };
– If not enough initializers, rightmost elements
become 0
–int n[ 5 ] = { 0 }
 All elements 0
– If too many initializers, a syntax error occurs
– C arrays have no bounds checking!!
• If size omitted, initializers determine it
 int n[ ] = { 1, 2, 3, 4, 5 };
– 5 initializers, therefore 5 element array

Data Structures 13
Addressing of element in 1-D Array
• Let A be a linear array stored in successive memory cells.
LOC(A[K]) = address of the element A[K] of the array A
1000
1001
1002
1003
• To calculate the address of any element of A formula is::
LOC(A[K])= Base (A) + w (K-lower bound)
• Base (A) =Address of the first element of A
• w =number of words per memory cell for the array
• K =any index of A

Example
• Consider the array AUTO in which records the
number of automobiles sold each year from 1932
through 1984. Suppose AUTO appears in memory as
pictured. That is Base(AUTO) = 200 and w = 4 words
per memory cell for AUTO Then
• LOC(AUTO[1932]) = 200,
• LOC(AUTO[1933])=204,
• LOC(AUTO[1934])=208,…..
• The address of the array element for the year K=1965
can be obtained by using formula.
Data Structures 14

Answer
• LOC(AUTO[1965]) = Base(AUTO) + w (1965 –
lower bound)
• LOC(AUTO[1965])= 200 + 4 (1965 – 1932)
• LOC(AUTO[1965]) = 332
Data Structures 15

Data Structures 16
Implementation of linear array
• Let LA be a linear array and stored in computer
memory .Now we want to count the no. of elements
of LA or print the contents of each element of LA,
this is done by traversing LA
• Also we want to insert an item in the linear array,
done by Insert operation
• Also we want to delete an item in the linear array,
done by Delete operation

Data Structures 17
Algo. For Traversing in linear array
• LA is a linear array with UB upper bound
and LB lower bound for traversing
we apply PROCESS to each element of LA
Steps are:
1.Set k=LB
2.Repeat step 3 and 4 while k<=UB
3.Apply PROCESS to LA [k]
4.k=k+1
5.exit

Data Structures 18
Inserting an element in array
• INSERT (LA, N, k, ITEM)
Here LA is a Linear array with N elements and K is a
positive integer. This algorithm inserts an element ITEM
into the Kth position in LA.
Steps are:
1.Set j=N
2. Repeats step 3 and 4 while j>=k
3.Set LA[j+1]=LA[j]
4.Set j=j-1
5.Set LA[k]=ITEM
6.Set N=N+1
7.Exit

Data Structures 19
Delete an element in array
• DELETE (LA, N, k, ITEM)
Here LA is a Linear array with N elements and K is a
positive integer. This algorithm deletes the Kth
element from LA.
Steps are:
1.Set ITEM = LA[k]
2. Repeat for j = k to N-1
Set LA[j] = LA[j+1]
3.Set N=N-1
4.Exit

Disadvantages of linear Array
• The number of elements, which is to be stored
in an array must be known first.
• These are static structures. Static in the sense
that memory is allocated at compilation time.
• When an array is declared, memory is
allocated to it. If memory is not filled
completely, the vacant place occupies
memory.
Data Structures 20

• The elements of the arrays are stored in
consecutive locations, the insertion and
deletion into the array are time consuming.
Because to insert/delete an item require
moving elements down to create a space for
new element or moving elements up to
occupy the space by the deleted element.
Data Structures 21

Data Structures 22
Two-Dimensional Array (Table)
Col 0 Col 1 Col 2 Col 3 Col 4
Row 0 a[0][0] a[0][1] a[0][2] a[0][3] a[0][4]
Row 1 a[1][0] a[1][1] a[1][2] a[1][3] a[1][4]
Row 2 a[2][0] a[2][1] a[2][2] a[2][3] a[2][4]
►short a[3][5]; // a table with 3 rows,5 columns
►The first dimension denotes the row-index,
which starts with 0.
►The second dimension denotes the column-
index, which also starts with 0.

Data Structures 23
Two-Dimensional Array
• A two dimensional array A is a collection of m * n
elements and each element is specified by pair of integers
(such as J, K) called subscripts
1<=J<= m AND 1<=K<=n
• Element of A with first subscript j and second subscript k
will denoted by
AJ,K or A[J,K]
• Two-dimensional arrays are called matrices in
mathematics and tables in business applications.
• The element A[J,K] appears in row J and column K
• A row is a horizontal list of elements and a column is a
vertical list of elements.

• A has 3 rows and 4 columns
Columns
1 2 3 4
Data Structures 24

• Suppose A is a two-dimensional m x n array. The
first dimension of A contains the index set
1,2…..m with lower bound 1 and upper bound m
• The second dimension of A contains the index set
1,2…..n with lower bound 1 and upper bound n.
• The length of a given dimension
Length = upper bound – lower bound + 1
• Size of array = m x n
Data Structures 25

Data Structures 26
EXAMPLE
• Let A be a 2D array such that
• A(2:5,-3:1)
• Here index set of dimensions 2,3,4,5 and -3,-
2,-1,0,1
• Length of first dimension is 5-2+1=4
• Length of second dimension is 1-(-3)+1=5
• Size of A=4*5=20

Representation of 2D array in memory
• There are two ways of representing 2D array
• Row major array
• Column major array
• Row major array : The elements are stored row by row
i.e. n elements of the first row stored in first n locations,
elements of the second row stored in next n locations
and so on.
• Column major array : The elements are stored column by
column i.e. m elements of the first column stored in first
m locations, elements of the second column stored in
next m locations and so on.
Data Structures 27

Data Structures 28
(1,1)
(1,2)
(1,3)
(2,1)
(2,2)
(2,3)
subscriptsA
Row 1
Row2
Row-major order

Data Structures 29
(1,1)
(2,1)
(3,1)
(2,1)
(2,2)
(2,3)
subscriptsA
column 1
column2
Column major order

Data Structures 30
Address calculation in 2D array
• Let a 2D array A of m*n size, to compute the LOC (A[j,k])
using the formula
LOC (A[j,k])=Base(A)+w[(N(j-1)+(k-1)]
(row major order)
Where Base(A) = address of 1st element of A, w = words per
memory cell and N = total no. of columns in array
LOC (A[j,k])=Base(A)+w[(M(k-1)+(j-1)]
(column major order)
Where Base(A) = address of 1st element of A, w = words per
memory cell and M = total no. of rows in array

Example
• Consider the 25 x 4 matrix array SCORE.
Suppose Base(SCORE) = 200 & w= 4. Calculate
the address of SCORE = [12,3] i.e. the 12th row
and 3 column using row major order and
column major order.
Data Structures 31

Answer
• M = 25,
• N = 4
• J = 12
• K = 3
• W = 4
• Base (SCORE) =200
Data Structures 32

Using Row-major order
• LOC (SCORE[12,3]) = Base (SCORE) + w [N(J-1) + (K-1)]
= 200 + 4 [4(12-1) + (3-1)]
= 200 + 4 [4(11) + 2)]
= 200 + 4 [44 +2]
= 200 + 4 [46]
= 200 + 184
= 384
Data Structures 33

Using Column Major Order
• LOC (SCORE[12,3]) = Base (SCORE) + w [M(K-1) + (J-1)]
= 200 + 4 [25(3-1) + (12-1)]
= 200 + 4 [25(2) + 11)]
= 200 + 4 [50 +11]
= 200 + 4 [61]
= 200 + 244
= 444
Data Structures 34

Multidimensional Array
• When array can be extended to any number of
dimensions. E.g. 3D array may be defined as :
• int a[2][4][3]
• Multidimensional array also called arrays of
arrays
• Suppose C is a 3D 2 x 4 x 3 array. Then B
contains 2 x 4 x 3 = 24 elements. These
elements appear in 3 layer called pages,
column and row.
Data Structures 35

2. Array in Data Structure

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2. Array in Data Structure