1. Array Variable: Placeholder/Structure to store a basic unit of data.
Example:
1 integer variable stores single integer data.
1 float variable stores single float data.
Question is:
In a computer, where is that value assigned to a variable stored?
In memory, possibly Random Access Memory (RAM)
But where exactly in memory?
On some physical address in memory because,
Memory is divided in physical locations and,
Each location has a particular address associated with it.

2. Variable
1000
1001
1002
1003
1004
1005
1006
1007
1008
1009 5
(a)
int a = 5;
char n = ‘z’; z
(n)

3. Array Variable: Sometimes in our program, Example: 2 solutions:
There might be a need to store multiple data of similar type.
Example:
Roll numbers of 10 students.
Marks of DFS of 10 students.
2 solutions:
Create different variables each having a different name.
int num1, num2, num3 etc.
Create a collection of variables referred by a common name.
int num[3]
This looks much better solution.

4. Array
1000
1001
1002
1003
1004
1005
1006
1007
1008
1009 5
(a)
int a = 5;
char n = ‘z’; 11
(num)
int num[3] = {11, 10, 20}; 10
Ordered 20
Finite
Same data type /
Homogeneous z
(n)

5. Array
Array:
Finite, Ordered collection of Homogeneous data elements where,
Ordering is maintained by storing all elements in,
Continuous / Contiguous locations of memory.
Properties:
Finite:
Contains only a limited (finite) number of elements.
Ordered:
All elements are stored one by one in continuous / contiguous locations of computer memory.
Homogeneous:
All the elements of an array are of the same data type.
Example:
An array of integers to store the roll numbers of all students in a class.
An array of strings to store the names of all villagers in a village.

6. Array (Terminologies)
Size / Length / Dimension
Lower Bound (L)
int A[5] = {50, 25, 40, 10, 35}
Base Address (M)
Upper Bound (U)
Pascal
Fortran C
Position
(P)
Values
Address
Index (i)
Index (i)
Index (i) Ap
A[i] 1 -3 1 50
1000
(A) 2 -2 2 1 25
1001 3 -1 3 2 40
1002 4 4 3 10
1003 5 1 5 4 35
1004
C Language: int A[5]
C Language: int A[0...4]
Fortran Language: int A[5]
Fortran Language: int A[1...5]
Pascal Language: int A[-3...1]
Pascal Language: int A[-3...1]

7. Array (Terminologies)
float A[0...5] = {50.50, 25.50, 40.50, 10.00, 35.00}
Values
Address
Values
Address
50.50
1000
2 address locations
1001
2 bytes
50.50
1000
25.50
1002
25.50
1002
1003
40.50
1004
40.50
1004
Word Size (W)
1005
10.00
1006
10.00
1006
35.00
1008
1007
35.00
1008
1009

8. Array Terminology: Size: Type: Base / Base Address (M):
Number of elements in an array.
Also called:
Length, Dimension.
Example:
int A[5]
Size is 5.
Type:
Kind of data type it is meant for.
Type is int.
Base / Base Address (M):
Address of the memory location where the first element of the array is located.
Denoted by symbol M.

9. Array Terminology: Index (i):
Subscript by which the elements in an array can be referenced / accessed.
Always an integer value.
Lower Bound (L):
Starting index of an array.
Denoted by symbol L.
Upper Bound (U):
Ending index of an array.
Denoted by symbol U.
Example:
A[1]: Element located on index 1.
A[-3]: Element located at index -3.

10. Array Terminology: Range of Indices: Position (P): Word Size (W):
Depends on the programming language in which the array is created.
In C,
Index always starts from 0 upto Size-1.
In Fortran,
Index always starts from 1 to Size.
In Pascal,
Index can have user-defined integer values.
Position (P):
Relative rank of an element in the array.
Does not depend on the programming language.
Example:
A1:First element in the array.
A5:Fifth element in the array.
Word Size (W):
Size of one element in the array.
Denoted by symbol W.

11. Array (Answer the Questions)
Data:
Array A[0...8]
M = 1000
W = 4B
Position
Index
Array A
Address
Questions: 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8
1000
1004
1008
1012
1016
1020
1024
1028
1032
What is the size of A? 9
What is the position of A[4]? 5
What is the index of A8? 7
What is the address of A[4]?
1016
What is the address of A2?
1004
Total memory occupied by A?
9*4 = 36B
Which element is on address 1028?
A8 or A[7]

12. Array (Answer the Questions)
Data:
Array A[-5...1]
M = 2000
W = 2B
Position
Index
Array A
Address
Questions: 1 2 3 4 5 6 7 -5 -4 -3 -2 -1 1
2000
2002
2004
2006
2008
2010
2012
What is the size of A? 7
What is the position of A[-4]? 2
What is the index of A6?
What is the address of A[-1]?
2008
What is the address of A8?
Incorrect
What is the address of A[2]?
Incorrect
Which element is on address 2004?
Element with position 3 / Element with index -3.

13. Array (Answer the Questions)
Data:
Array A[ ]
M = 1517
W = 4B
Questions:
What is the size of A?
Wait a minute. I’m calculating on fingers.
What is the position of A[-4]?
One second.
What is the index of A61?
You’re confusing me.
What is the address of A[-1]?
Stop it please.
What is the address of A[-54]?
Incorrect.
Atleast, got 1 right answer.
Very difficult to answer because the array seems to be very big.
It is difficult to represent the array on paper.
Equations are needed to answer these questions.

14. Array (Formula to calculate Size)
Size = U – L + 1
Array A[0...4]
L = 0
U = 4
Array A[1...5]
L = 1
U = 5
Array A[-3...2]
L = -3
U = 2
Index
Index
Index 1 2 3 4 L 1 2 3 4 5 -3 -2 -1 1 2 U
Size(A) = 5
Size(A) = 5
Size = U – L + 1
Size(A) = U + 1
Size(A) = U
Size = 2 – (-3) + 1
Size(A) = U + 1 -
Size(A) = U
+ 1 -1
Size = = 6
Size(A) = U L
Size(A) = U L

15. Array (Formula to calculate Index from Position & vice-versa)
Index = i = L + P - 1
Array A[0...4]
L = 0, U = 4
Array A[1...5]
L = 1, U = 5
Array A[ ]
L = -31, U = 20
Position
(P)
Index
(i)
Position
(P)
Index
(i)
Position
(P)
Index
(i) 1 2 3 4 5 1 2 3 4 1 2 3 4 5 1 2 3 4 5 1 2 3 4
-31
-30
-29
-28
Find index of 4th element.
Find index of element
located at position 4.
Find index of A4.
Index(A1) = 0
Index(A1) = 1
Index(A2) = 1
= 2 - 1
Index(A2) = 2
Index(A3) = 2
= 3 - 1
Index(A3) = 3
Index(Ap) = P - 1
Index(Ap) = P
Index = i = L + P - 1
Index(Ap) = P – 1 +
Index(Ap) = P
-1 + 1
Index = i =
Index(Ap) = P – 1 + L
Index(Ap) = P – 1 + L
Index = i = -28

16. Array (Formula to calculate Address from Index / Position)
Address = M + ( i – L ) * W
Array A[0...4]
L = 0, U = 4, M=1000
Array A[1...5]
L = 1, U = 5, M=1000
Index
(i)
Index
(i)
Address
Address 1 2 3 4
1000
1001
1002
1003
1004 1 2 3 4 5
1000
1001
1002
1003
1004
Will Word Size (W) matter?
Yes
Address(A[0]) = 1000
Address(A[1]) = 1000
Address(A[1]) = 1001 =
Address(A[2]) = 1001 =
Address(A[2]) = 1002 =
Address(A[3]) = 1002 =
Address(A[i]) = i
= M + i
Address(A[i]) = (i – 1)
Address(A[i]) = M + (i - 0)
Address(A[i]) = M + (i – 1)
Address(A[i]) = M + ( i – L )
* W
Address(A[i]) = M + ( i – L )
* W

17. Array (Formula to calculate Address from Index)
Address = M + ( i – L ) * W
Array A[-5...1], L = -5, U = 1, M = 1000, W = 4
Index
(i)
What is the address of element
located at index -1?
What is the address of A[-1]?
Address -5 -4 -3 -2 -1 1
1000
1004
1008
1012
1016
1020
1024
Address = M + (i - L) * W
Address = (-1 – (-5)) * 4
Address = (-1 + 5) * 4
Address = (4) * 4
Address =
Address = 1016

18. Array Formulas / Equations: Size = U – L + 1 Index or i = L + P – 1
U: Upper Bound of the array.
L: Lower Bound of the array.
Index or i = L + P – 1
P: Position of element in the array.
Address = M + (i – L) * W
M: Base address of the array.
i: Index of the element.
L: Lower bound of the array.
W: Word size of the array.
Also called:
Indexing Formula
Could also be expressed in the form of Position ‘P’ as:
Address = M + (P – 1) * W
Because (i – L) = (P – 1)

19. Array (Indexing Formula)
Address
Index L
L+1
L+2 . i
U-2
U-1 U
Array in a
program
Array in
memory M .
Address = M + (i – L) * W
Logical View
Physical View

20. Array
Question-1:
Suppose, an array A[ ] is stored in a memory whose starting address is 459. Assume that the word size for each element is 2. Then obtain the following:
(a) How many number of elements are there in the array A?
(b) How much memory is required to store the entire array?
(c) What is the address location for A[50]?
(d) What is the address location of 10th element?
(e) Which element is located at address 599?

21. Array Answer-1: Data: L = -15 U = 64 M = 459 W = 2
Size = U – L + 1
i = L + P – 1
Address = M + (i – L) * W
Array
Answer-1:
Data:
L = -15
U = 64
M = 459
W = 2
(a) How many number of elements are there in the array A?
Size = U – L + 1
Size = 64 – (-15) + 1
Size =
Size = 80
80 elements are there in the array A.

22. Size = U – L + 1
i = L + P – 1
Address = M + (i – L) * W
Array
Data:
L = -15,
U = 64,
M = 459,
W = 2
Answer-1:
(b) How much memory is required to store the entire array?
Total memory required = Size * W
Total memory required = 80 * 2
Total memory required = 160
(c) What is the address location for A[50]?
i = 50
Address = M + (i – L) * W
Address = (50 – (-15)) * 2
Address = ( ) * 2
Address = (65) * 2
Address =
Address = 589

23. Array Answer-1: (d) What is the address location of 10th element?
Size = U – L + 1
i = L + P – 1
Address = M + (i – L) * W
Array
Data:
L = -15,
U = 64,
M = 459,
W = 2
Answer-1:
(d) What is the address location of 10th element?
P = 10
Index or i = L + P - 1
i =
i =
i = -6
Address = M + (i – L) * W
Address = (-6 – (-15)) * 2
Address = ( ) * 2
Address = (9) * 2
Address =
Address = 477

24. Array Answer-1: (e) Which element is located at address 599?
Size = U – L + 1
i = L + P – 1
Address = M + (i – L) * W
Array
Data:
L = -15,
U = 64,
M = 459,
W = 2
Answer-1:
(e) Which element is located at address 599?
Address = 599
Address = M + (i – L) * W
Index or i = L + P - 1
599 = (i – (-15)) * 2
55 = P - 1
599 = (i + 15) * 2
= P
599 = i + 30
P = 71
599 – 459 – 30 = 2i
2i = 110
i = 110 / 2
i = 55
71st element with index 55 is located on address 599. /
A71 is on address 599. /
A[55] is on address 599.

25. Size = U – L + 1
i = L + P – 1
Address = M + (i – L) * W
Array
Question-2:
Suppose, an array A[ ] is stored in a memory whose starting address is Assume that the word size for each element is 4. Then obtain the following:
(a) How many number of elements are there in the array A?
(b) How much memory is required to store the entire array?
(c) What is the address location for A[1]?
(d) What is the address location of 53rd element?
(e) Which element is located at address 1076?

26. Increment all the values by 1.
Array
Index
Array A
Question:
Increment all the values by 1.
Some Process/Procedure/Function
needs to be done on the array. 1 2 3 4 11 20 9 18 15
What is a ‘Process’ then?
‘Process’ is a sequence of steps that is executed
to get the work done.
Example: Process of making Tea.
In Computer terminology, such process is called:
‘ALGORITHM’ / ‘PSEUDOCODE’.
When an Algorithm is implemented in a programming language,
it is called a:
Values
(Marks of DFS
of 5 students)
‘PROGRAM’

27. Array Algorithm v/s Program: Algorithm: Program:
It is not dependent on the programming language.
It does not have to follow the syntax of any programming language.
Should be easily understandable.
Program:
It is dependent on the programming language.
It has to follow the syntax of the programming language in which it is implemented.
Might not be easily understandable.

28. Array Algorithm: TraverseArray Travel through the array.
Visit each and every element of the array.
Question:
Increment all the values by 1.
Index
Array A
Steps: 11
i = 0
A[0] = A[0] + 1
For i = 0 to 4
While i <= 4, do 1 20
A[1] = A[1] + 1
A[ i ] = A[ i ] + 1
A[ i ] = A[ i ] + 1 2 9
A[2] = A[2] + 1
i = i + 1
A[3] = A[3] + 1 3 18
EndWhile
EndFor
A[4] = A[4] + 1 4 15
Same thing is being done
repetitively just using
different values.
Better to do it using a Loop.

29. Increment all the values by 1.
Array
TRACING
i = 0
Algorithm:
TraverseArray
Iteration-1:
Condition
0 <= 4
True
Question:
Increment all the values by 1.
A[0] = 11+1 = 12
i = = 1
Index
Array A
Steps:
Iteration-2:
Condition
1 <= 4
True
A[1] = 20+1 = 21 11
i = 0
i = = 2
While i <= 4, do 1 20
Iteration-3:
Condition
2 <= 4
True
A[2] = 9+1 = 10
A[ i ] = A[ i ] + 1 2 9
i = = 3 3 18
i = i + 1
Iteration-4:
Condition
3 <= 4
True
A[3] = 18+1 = 19 4 15
EndWhile
i = = 4
Iteration-5:
Condition
4 <= 4
True
Check whether this algorithm
is working for this array or not.
A[4] = 15+1 = 16
i = = 5
How to check whether an algorithm
is working or not?
Iteration-6:
Condition
5 <= 4
False

30. Array Algorithm: TraverseArray Steps: Index Index Array A i = 0 i = L-2 11
While i <= 4, do
While i <= U, do -1 1 20
A[ i ] = A[ i ] + 1
Process( A[ i ] ) 2 9
i = i + 1
i = i + 1 1 3 18
EndWhile
EndWhile 2 4 15
Question:
Will this algorithm work for any kind of array?
Whether C, Fortran, Pascal?
Whether Small array, Large array etc?
Question:
Will this algorithm be able to do any kind of processing?

31. Algorithm: TraverseArray
Iteration-1:
Condition: -2<=1
True
Process(A[-2])
/ Process(11)
Algorithm: TraverseArray
i = = -1
Iteration-2:
Trace the algorithm TraverseArray
on the following array.
Condition: -1<=1
True
Process(A[-1])
i = = 0
Iteration-3:
Index
Array A
Steps:
Condition: 0<=1
True
i = L
Process(A[0]) -2 11 i L
While i <= U, do
i = = 1 i -1 20
Iteration-4:
Process( A[ i ] )
Condition: 1<=1
True i 9
i = i + 1
Process(A[1]) 1 18 i U
i = = 2
EndWhile
Iteration-5:
i = 2
Condition: 2<=1
False
Stop

32. Array Algorithm: Input: Output: Data Structure: TraverseArray.
Array A with elements.
Output:
According to Process().
Data Structure:
Array A[L...U]

33. Algorithm: TraverseArray
Steps:
i = L
While i <= U, do
Process( A[ i ] )
i = i + 1
EndWhile
Stop

34. Array Algorithm: SearchArray Question: Search for an element and tell
whether it is present in the array or not.
Index
Array A 11 9
Search is successful.
Return 2 (Index of 9). 1 20 22
Search is unsuccessful.
Return NULL. 2 9
Question:
Will TraverseArray algorithm be used here? 3 18 4 15

35. Algorithm: SearchArray
i = L
(i <= U)
Steps:
KEY = 9
i = 0
, found = 0
, location = NULL
While (i <= 4)
&& (found == 0), do
Index
Array A
If (A[ i ] == KEY)
Process (A[ i ])
If( A[ i ] == 9 ), then
Question: 11
found = 1
Is it the most optimized / efficient? 1 20
location = i
Question: 2 9
EndIf
Will it work for any array?
i = i + 1 3 18
EndWhile
Question: 4 15
Will it work for any search value?
If (found == 0), then
print “Search is unsuccessful.”
Else
print “Search is successful.”
EndIf
Return(location)

36. Trace Algorithm: SearchArray
KEY = 20
Steps:
i = -1
, found = 0
, location = NULL
i = L, found = 0, location = NULL
While (i <= U) && (found == 0), do
Iteration-1:
If( A[ i ] == KEY ), then
Condition:
-1<=2 && 0==0
True
Index
Array A
If (11 == 20)
False
found = 1 -1 11
location = i
i = = 0 20
Iteration-2:
EndIf
Condition:
0<=2 && 0==0
True 1 9
i = i + 1
If (20 == 20)
True 2 18
EndWhile
found = 1
If (found == 0), then
location = 0
print “Search is unsuccessful.”
i = = 1
Else
Iteration-3:
print “Search is successful.”
Condition:
1<=2 && 1==0
False
EndIf
Search is successful
Return(location)
Return (0)

37. Array Algorithm: Input: Output: Data Structure: SearchArray.
Array A with elements.
KEY: Element to be searched.
Output:
On Success, appropriate message and return Index/Location of KEY in array A.
On Failure, appropriate message and return NULL.
Data Structure:
Array A[L...U]

38. Algorithm: SearchArray
Steps:
i = L, found = 0, location = NULL
While (i <= U) && (found == 0), do
If( A[ i ] == KEY ), then
found = 1
location = i
EndIf
i = i + 1
EndWhile
If (found == 0), then
print “Search is unsuccessful.”
Else
print “Search is successful.”
Return(location)
Stop

39. to be polluted / corrupted.
Array
Algorithm:
InsertArray
Question:
Insert an element ’16’ in this array.
Question:
At which location / index, the element is to be inserted?
Index
Array A
Index
Array A
Index
Array A 11 11 11 1 20 1 20 1 20 2 9 2 2 9 3 18 3 18 3 18 4 15 4 15 4
Possible.
Not possible.
Possible.
Last index /
location
is NULL.
Array is
already full.
However, array seems
to be polluted / corrupted.

40. Array Algorithm: InsertArray Question:
Insert an element ’16’ in this array.
Question:
At which location / index, the element is to be inserted? 1
Index
Array A
Index
Array A
Solution-1:
Solution-2: 11 11
Steps:
Steps: 1 20 16
A[1] = 16 1 20
A[4] = A[1] 16
Stop
A[1] = 16 2 9 2 9
Stop 3 3 18 18 4 4 20
Does not work.
Does not work.
It should be insertion, not updation.
Original sequence is changed,
hence polluted.

41. Array Algorithm: InsertArray Question:
Insert an element ’16’ in this array.
Question:
At which location / index, the element is to be inserted? 1
Index
Array A 11
It seems that the only solution is:
Shifting down of elements to make place for the new element. 1 20
Question: Will downward shifting start from Top or Bottom?
i.e. Will 20 be shifted down first or 18 will be shifted down first? 2 20 9 3 18 9 4 18

42. Array Algorithm: InsertArray Question:
Insert an element ’16’ in this array.
Question:
At which location / index, the element is to be inserted? 1
Index
Array A
Index
Array A
Downward Shifting
starting from Top.
Downward Shifting
starting from Bottom. 11 11
Elements will be lost.
All elements are preserved. 1 20 1 20
So it cannot work.
So it can work. 2 9 2 20 20 9 3 18 3 18 9 4 4 18

43. Array Algorithm: InsertArray LOCATION KEY Question:
Insert an element ’16’ in this array.
Question:
At which location / index, the element is to be inserted? 1
While (i > 1)
Index
Array A
Steps:
i = 4
i = U
A[4] = A[3] 11
While (i >= 2)
While (i > LOCATION) 1
A[3] = A[2]
A[ i ] = A[ i – 1]
A[ i ] = A[ i – 1] 20 16
i = i – 1
i = i – 1 2 20 9
A[2] = A[1]
EndWhile
EndWhile 3 18 9 4 18
A[1] = 16
A[1] = 16
A[LOCATION] = KEY

44. Array Algorithm: InsertArray Steps: If (A[U] != NULL), then
Index
Array A
Algorithm:
InsertArray 11
Question:
Insert KEY = ’16’ in this array at LOCATION = ‘1’. 1 20
Steps:
If (A[U] != NULL), then
print “Array is full. Insertion not possible.”
Else
i = U
While (i > LOCATION)
A[ i ] = A [ i – 1 ]
i = i – 1
EndWhile
A[LOCATION] = KEY
EndIf
Stop 2 9 3 18 4 15
Index
Array A
Controls the
downward shifting
of elements. 11 1 20 2 9 3 18 4

45. Trace Algorithm: InsertArray-2 10
KEY = 20, LOCATION = -1
Index
Array A
i = 1 -1 30
Steps:
If (A[U] != NULL), then
print “Array is full.
Insertion not possible.”
Else
i = U
While (i > LOCATION)
A[ i ] = A [ i – 1 ]
i = i – 1
EndWhile
A[LOCATION] = KEY
EndIf
Stop
Iteration-1: -2 10
Condition: 1 > -1
True 40 -1
A[ 1 ] = A[ 0 ] //A[1] = 40 30 1 40
i = 1 – 1 = 0 40 -2 10 1 50
Iteration-2:
Condition: 0 > -1
True -1 30
Index
Array A
A[ 0 ] = A[ -1 ] //A[0] = 30
i = 0 – 1 = -1 30 -2 10 1 40
Iteration-3: -1 30
Condition: -1 > -1
False 10 -2 40 -1 20 1
A[ -1 ] = 20 30 40 1

46. Array Algorithm: Input: Output: Data Structure: InsertArray.
Array A with elements.
KEY: Element to be inserted.
LOCATION: Index where KEY is to be inserted.
Output:
On Success, Element with value KEY inserted at index LOCATION.
On Failure, appropriate message to be displayed.
Data Structure:
Array A[L...U]

47. Algorithm: InsertArray
Steps:
If (A[U] != NULL), then
print “Array is full. Insertion not possible.”
Else
i = U
While(i > LOCATION)
A[ i ] = A[ i – 1 ]
i = i – 1
EndWhile
A[LOCATION] = KEY
EndIf
Stop

48. How to convert any algorithm into a ‘C’ program?
Array
How to convert any algorithm into a ‘C’ program?

49. Array 1 2 3 4 Algorithm: DeleteArray Question:
Delete an element from the array.
Question:
What is the value of the element to be deleted?
Array A
Index 60
Element is not there in the array. It cannot be deleted. 1 2 3 4 10 20
Search is successful. 20 is available at index/location 1. 30 20
A[1] = NULL
Empty location is in between the other elements of the array.
It should be at the tail end of the array. 40 30 50 40
So it leaves the array polluted / corrupted.
Can the last element ’50’ come to take place of
deleted element ’20’ directly? 50
No. Original sequence of elements will get changed.
It will pollute / corrupt the array.
Question: Will the
upward shifting start
from Bottom / Top?
Hence the remaining values needs to shifted upwards
so that the empty space automatically moves downwards.

50. Array 1 2 3 4 1 2 3 4 Algorithm: DeleteArray Question:
Delete an element from the array.
Question:
What is the value of the element to be deleted? 20
Array A
Array A
Index
Index
Solution-2:
Solution-1:
(Upward shifting
starting from Top)
(Upward shifting
starting from Bottom) 1 2 3 4 1 2 3 4 10 10
i = SearchArray(A, 20)
// i = 1 20
i = SearchArray(A, 20)
// i = 1 30 20 30 30 40
An element is lost.
All elements are
preserved. 40 50
So this cannot work. 40 50
So this can work. 50 50

51. Array 1 2 3 4 KEY Algorithm: DeleteArray Question:
Delete an element from the array.
Question:
What is the value of the element to be deleted? 20
While(i < 4), do
Array A
Steps:
Index
i = SearchArray(A, 20)
// i = 1
i = SearchArray(A, 20)
// i = 1
i = SearchArray(A, KEY) 1 2 3 4 10
While(i <= 3), do
While(i < U), do
A[ 1 ] = A[ 2 ] 30 20
A[ i ] = A[ i + 1 ]
A[ i ] = A[ i + 1 ]
A[ 2 ] = A[ 3 ] 30 40
i = i + 1
i = i + 1
A[ 3 ] = A[ 4 ] 40 50
EndWhile
EndWhile 50
A[ 4 ] = NULL
A[ 4 ] = NULL
A[ U ] = NULL

52. Array 1 2 3 4 Steps: i = SearchArray(A, KEY) If( i == NULL ), then
Index
Array
Array A 1 2 3 4 60
Algorithm:
DeleteArray 70
Question:
Delete an element with KEY = ’20’ from the array. 80
Steps: 90
i = SearchArray(A, KEY)
100
If( i == NULL ), then
print “KEY not found. Deletion not possible.”
Else
Array A
Index
While(i < U), do
Controls the Upward Shifting
of elements
starting from Top. 1 2 3 4 10
A[ i ] = A[ i + 1 ]
i = i + 1 20
EndWhile 30
A[ U ] = NULL 40
EndIf 50
Stop

53. Trace Algorithm: DeleteArray
KEY = ’22’
i = -1
Array A
Index -2 -1 1 11
Condition:
-1 == NULL
False 33 -2 -1 1
Steps: 44
Iteration-1: 33
Condition:
-1 < 1
True 55
i = SearchArray(A, KEY) 44
A[ -1 ] = A[ 0 ] //A[-1] = 33
If( i == NULL ), then 66
i = = 0
print “KEY not found.
Deletion not possible.” -2 -1 1 11 77
Iteration-2:
Else 33
Condition:
0 < 1
True
While(i < U), do 44
A[ 0 ] = A[ 1 ] //A[0] = 44
Array A
Index
A[ i ] = A[ i + 1 ] 44
i = = 1
i = i + 1 -2 -1 1 11
Iteration-3:
EndWhile
Condition:
1 < 1
False -2 -1 1 11 22 33
A[ U ] = NULL
A[1] = NULL 33 44
EndIf 44
Stop

54. Array Algorithm: Input: Output: Data Structure: DeleteArray.
Array A with elements.
KEY: Element to be deleted.
Output:
On Success, Element with value KEY deleted from the array.
On Failure, appropriate message to be displayed.
Data Structure:
Array A[L...U]

55. Algorithm: DeleteArray
Steps:
i = SearchArray(A, KEY)
If( i == NULL), then
print “KEY not found Deletion not possible.”
Else
While(i < U)
A[ i ] = A[ i +1 ]
i = i + 1
EndWhile
A[U] = NULL
EndIf
Stop