The document discusses data structures and their importance in organizing large amounts of data efficiently. It describes primitive data structures like integers and characters that are supported at a low level. Non-primitive structures like arrays, files, stacks, queues, graphs and trees are built using primitive structures. Arrays store homogeneous data in contiguous memory locations and support random access. Stacks follow LIFO while queues follow FIFO for element insertion and removal. Formulas to calculate memory addresses of array elements are also provided.

1. UNIT-1
CS-303
DATA STRUCTURE

2. DATA STRUCTURE DEFINITION
• Whenever we want to work with a large amount of
data, then organizing that data is very important.
• If that data is not organized effectively, it is very
difficult to perform any task on that data.
• If it is organized effectively then any operation can
be performed easily on that data.
• Data structure is a method of organizing a large
amount of data more efficiently so that any
operation on that data becomes easy

3. TYPES OF DATA STRUCTURE

4. Primitive Data Structures
• These are the structures which are supported at the
machine level, they can be used to make non-
primitive data structures.
• Examples: Integer, float, character, pointers.
• The pointers, however don’t hold a data value,
instead, they hold memory addresses of the data
values. These are also called the reference data
types.

5. Non Primitive Data Structures
• The non-primitive data structures cannot be
performed without the primitive data structures.
Although, they too are provided by the system itself
yet they are derived data structures and cannot be
formed without using the primitive data structures.
• The Non-primitive data structures are further divided
into the following categories:

6. 1. Arrays
• Arrays are a homogeneous and contiguous collection
of same data types.
• They have a static memory allocation technique,
which means, if memory space is allocated for once,
it cannot be changed during runtime.
• If we do not know the memory to be allocated in
advance then array can lead to wastage of memory.

7. 2. Files
• A file is a collection of records.
• The file data structure is primarily used for managing
large amounts of data which is not in the primary
storage of the system.
• The files help us to process, manage, access and
retrieve or basically work with such data, easily.

8. 3.Lists
• The lists support dynamic memory allocation. The
memory space allocated, can be changed at run time
also. The lists are of two types:
• a) Linear Lists
• The linear lists are those which have the elements
stored in a sequential order. The insertions and
deletions are easier in the lists. They are divided into
two types:

9. Stacks

10. Stacks

11. Stacks
• The stack follows a “LIFO” technique for storing and
retrieving elements. The element which is stored at
the end will be the first one to be retrieved from the
stack. The stack has the following primary functions:
– Push(): To insert an element in the stack.
– Pop(): To remove an element from the stack.

12. Queues

13. Queues
• The queues follow “FIFO” mechanism for storing and
retrieving elements. The elements which are stored
first into the queue will only be the first elements to
be removed out from the queue.
• The “ENQUEUE” operation is used to insert an
element into the queue
• The “DEQUEUE” operation is used to remove an
element from the queue.

14. Non Linear Lists
The non linear lists do not have elements stored in a
certain manner. These are:
• Graphs: The Graph data structure is used to
represent a network.
• It comprises of vertices and edges (to connect the
vertices). The graphs are very useful when it comes
to study a network.
• Trees: Tree data structure comprises of nodes
connected in a particular arrangement and they
make search operations on the data items easy. The
tree data structures consists of a root node which is
further divided into various child nodes and so on.

15. Tree

16. Graph

17. Need of using Array
• In computer programming, the most of the cases
requires to store the large number of data of similar
type.
• To store such amount of data, we need to define a
large number of variables.
• It would be very difficult to remember names of all
the variables while writing the programs. Instead of
naming all the variables with a different name, it is
better to define an array and store all the elements
into it.

18. Example
• We have marks of a student in six different
subjects. The problem intends to calculate the
average of all the marks of the student.

19. Program without Array
#include <stdio.h>
void main ()
{
int marks_1 = 56, marks_2 = 78, marks_3 = 88,
marks_4 = 76, marks_5 = 56, marks_6 = 89;
float avg = (marks_1 + marks_2 + marks_3 + marks_4
+ marks_5 +marks_6) / 6 ;
printf(avg);
}

20. Array
• Arrays are defined as the collection of similar type of
data items stored at contiguous memory locations.
• Arrays are the derived data type in C programming
language which can store the primitive type of data
such as int, char, double, float, etc.
• Array is the simplest data structure where each data
element can be randomly accessed by using its index
number.

21. Array
• For example, if we want to store the marks of a
student in 6 subjects, then we don't need to define
different variable for the marks in different subject.
instead of that, we can define an array which can
store the marks in each subject at a the contiguous
memory locations.

22. How to declare an array?
Syntax:- data_type array_name[SIZE];
data_type is a valid C data type that must be common
to all array elements.
array_name is name given to array and must be a valid
C identifier.
SIZE is a constant value that defines array maximum
capacity.
Example :- int marks[5];

23. How to initialize an array?
There are two ways to initialize an array.
Static array initialization - Initializes all elements of
array during its declaration.
Dynamic array initialization - The declared array is
initialized some time later during execution of
program.

24. Static initialization of array
We define value of all array elements within a pair of
curly braces { and } during its declaration.
Values are separated using comma , and must be of
same type.
Example of static array initialization
int marks[500] = {90, 86, 89, 76, 91};
int marks[] = {90, 86, 89, 76, 91};

25. Dynamic initialization of array
You can assign values to an array element dynamically
during execution of program.
First declare array with a fixed size. Then use the
following syntax to assign values to an element
dynamically.
array_name[index] = some value;
Example to initialize an array dynamically
marks[0] = 90
Cin>>marks[1];

26. Basic Operations on Array
• Traverse − print all the array elements one by one.
• Insertion − Adds an element at the given index.
• Deletion − Deletes an element at the given index.
• Search − Searches an element using the given index
or by the value.
• Update − Updates an element at the given index.

27. Address Calculation in 1D Array
Array of an element of an array say “A[ I ]” is calculated
using the following formula:
Address of A [ I ] = B + W * ( I – LB )
Where,
B = Base address
W = Storage Size of one element stored in the array
(in byte)
I = Subscript of element whose address is to be
found
LB = Lower limit / Lower Bound of subscript, if not
specified assume 0 (zero)

28. Example:
Given the base address of an
array A[900…..1900] as 1400 and size of
each element is 4 bytes in the memory. Find
the address of B[1100] & B[1475].
A[I]=1400+4*(1475-900)

29. Example:
Solution:
The given values are: B = 1020, LB = 1300, W =
2, I = 1700
Address of A [ I ] = B + W * ( I – LB )
= 1020 + 2 * (1700 – 1300)
= 1020 + 2 * 400
= 1020 + 800
= 1820 [Ans]

30. Properties of the Array
• Each element is of same data type and carries a same
size i.e. int = 2 bytes.
• Elements of the array are stored at contiguous
memory locations
• Elements of the array can be randomly accessed
since we can calculate the address of each element
of the array with the given base address and the size
of data element.

31. Advantage of Array
1) Code Optimization: Less code to the access the data.
2) Ease of traversing: By using the for loop, we can
retrieve the elements of an array easily.
3) Ease of sorting: To sort the elements of the array, we
need a few lines of code only.
4) Random Access: We can access any element
randomly using the array.

32. Address Calculation in
2D Array
• While storing the elements of a 2-D array in memory,
these are allocated contiguous memory locations.
• Therefore, a 2-D array must be linearized so as to
enable their storage.
• There are two alternatives to achieve linearization:
Row-Major and Column-Major.

33. 2D Array

34. 2D Array

35. Address Calculation Formula
Row Major System:
Address of A [ I ][ J ] = B + W * [ N * ( I – Lr ) + ( J – Lc ) ]
Column Major System:
Address of A [ I ][ J ] = B + W * [( I – Lr ) + M * ( J – Lc )]
Where
B=Base Address W=Size of Each Element
N=No of Columns M=No of Rows

36. Address Calculation Formula
Lr =Lower Bound of row
Lc = Lower Bound of Column
I = Row subscript of element whose address is to be found
J = Column subscript of element whose address is to be found
Number of rows (M) will be calculated as = (Ur – Lr) + 1
Number of columns (N) will be calculated as = (Uc – Lc) + 1
LR =FIRST ROW FIRST VALUE
UR=FIRST ROW LAST VALUE
UC=LAST COLUMN LAST VALUE
LC=FIRST COLUMN LAST VALUE

37. Consider the linear array AAA(5:50),BBB(-5:10)
and CCC(18)
A) find the number of element in each array
B) suppose Base (AAA)=300 and w=4 words per
memory cell for AAA.find the address of
AAA[15],AAA[35] and AAA[55]

38. Example
• An array X [-15……….10, 15……………40] requires one
byte of storage. If beginning location is 1500
determine the location of X [15][20].

39. Solution
As you see here the number of rows and columns are
not given in the question.
So they are calculated as:
Number or rows say M = (Ur – Lr) + 1
= [10 – (- 15)] +1=26
Number or columns say N = (Uc – Lc) + 1
= [40 – 15)] +1 = 26

40. Solution
Column Major Wise Calculation of above equation
The given values are:
B = 1500 W = 1 byte I = 15
J = 20 Lr = -15 Lc = 15
M = 26
Address of A [ I ][ J ] = B + W * [ ( I – Lr ) + M * ( J – Lc ) ]
= 1500 + 1 * [(15 – (-15)) + 26 * (20 – 15)]
= 1500 + 1 * [30 + 26 * 5] = 1500 + 1 * [160]
= 1660 [Ans]

41. Solution
Row Major Wise Calculation of above equation
The given values are:
B = 1500 W = 1 byte I = 15
J = 20 Lr = -15 Lc = 15
N = 26
Address of A [ I ][ J ] = B + W * [ N * ( I – Lr ) + ( J – Lc ) ]
= 1500 + 1* [26 * (15 – (-15))) + (20 – 15)]
= 1500 + 1 * [26 * 30 + 5]
= 1500 + 1 * [780 + 5]
= 1500 + 785 = 2285 [Ans]

42. Example 2
• Each element of an array a[-20….20,10….35] requires
1 byte of storage .if the array is column major
implemented and the beginning of the array is at
location 500(base address).determine the address of
elementa[0,30]

43. Solution
Column Major Wise Calculation of above equation
The given values are:
B = 500 W = 1 byte I = 0
J = 30 Lr = -20 Lc = 10
Ur=20 Uc=35
M=Ur-Lr+1=20-(-20)+1=41
Address of A [ I ][ J ] = B + W * [ ( I – Lr ) + M * ( J – Lc ) ]
= 500 + 1 * [(0 – (-20)) + 41 * (30 – 10)]
= 500 + 1 * [20+41*20] = 500 + 1 * [840]
= 1340 [Ans]

44. Solution
Row Major Wise Calculation of above equation
The given values are:
B = 500 W = 1 byte I = 0
J = 30 Lr = -20 Lc = 10
Ur=20 Uc=35
N=Uc-Lc+1=35-10+1=26
Address of A [ I ][ J ] = B + W * [ N * ( I – Lr ) + ( J – Lc ) ]
= 500 + 1* [26 * (0 – (-20))) + (30 – 10)]
= 500 + 1 * [26 * 20 + 20]
= 500 + 1 * [540]
= 500 + 540 = 1040 [Ans]

45. Example 3
• Calculate the address of X[4,3]in a 2 d array
X[1….5,1….4] stored in row major order in main
memory .Assume the base address to be 1000 and
that each element requires 4 word of storage.

46. Solution
Column Major Wise Calculation of above equation
The given values are:
B = 1000 W = 4 byte I = 4
J = 3 Lr = 1 Lc =1
Ur=5 Uc=4
M=Ur-Lr+1=5-1+1=5
Address of A [ I ][ J ] = B + W * [ ( I – Lr ) + M * ( J – Lc ) ]
= 1000 + 4 * [(4– (1)) + 5 * (3 – 1)]
= 1000 + 4 * [3+5*2] = 1000 + 4 * [13]
= 1052 [Ans]

47. Solution
Row Major Wise Calculation of above equation
The given values are:
B = 1000 W = 4 byte I = 4
J = 3 Lr = 1 Lc =1
Ur=5 Uc=4
N=Uc-Lc+1=4-1+1=4
Address of A [ I ][ J ] = B + W * [ N * ( I – Lr ) + ( J – Lc ) ]
= 1000 + 4* [4 * (4– (1))) + (3 – 1)]
= 1000 + 4 * [4 * 3 + 2]
= 1000 + 4 * [14]
= 1000 + 56 = 1056 [Ans]

48. Example 4
• Each element of an array DATA[20][50]
requires 4 bytes of storage. Base Address of
DATA is 2000,determine the location of
DATA[10][10]when the array is stored as
• 1) Row Major
• 2) Column Major

49. Example 5
• A two dimensional array X[5][4] is stored a row wise
in the memory .The first element of the array is
stored at location 80 .find the memory location of
X[3][2] if the each elements of array requires 4
memory locations.

50. Algorithm
 Algorithm is a step-by-step procedure, which defines
a set of instructions to be executed in a certain order
to get the desired output.
 An algorithm is a sequence of unambiguous
instructions used for solving a problem, which can be
implemented (as a program) on a computer.
 Algorithms are generally created independent of
underlying languages, i.e. an algorithm can be
implemented in more than one programming
language.

51. Algorithm
• Algorithms are used to convert our problem solution
into step by step statements.
• These statements can be converted into computer
programming instructions which form a program.
• This program is executed by a computer to produce
a solution.
• Here, the program takes required data as input,
processes data according to the program instructions
and finally produces a result as shown in the
following picture.

52. Algorithm

53. Specifications of Algorithms
• Input - Every algorithm must take zero or more number
of input values from external.
• Output - Every algorithm must produce an output as
result.
• Definiteness - Every statement/instruction in an
algorithm must be clear and unambiguous (only one
interpretation).
• Finiteness - For all different cases, the algorithm must
produce result within a finite number of steps.
• Effectiveness - Every instruction must be basic enough
to be carried out and it also must be feasible.

54. Performance Analysis of an algorithm
• If we want to go from city "A" to city "B",
there can be many ways of doing this. We can
go by flight, by bus, by train and also by
bicycle.
• Depending on the availability and
convenience, we choose the one which suits
us.
• Similarly, in computer science, there are
multiple algorithms to solve a problem.

55. Performance Analysis of an algorithm
• When we have more than one algorithm to solve
a problem, we need to select the best one.
• Performance analysis helps us to select the best
algorithm from multiple algorithms to solve a
problem.
When there are multiple alternative algorithms to
solve a problem, we analyze them and pick the
one which is best suitable for our requirements.

56. Performance analysis of an algorithm
• Performance analysis of an algorithm is the process
of calculating space and time required by that
algorithm
• Space required to complete the task of that
algorithm (Space Complexity). It includes program
space and data space
• Time required to complete the task of that algorithm
(Time Complexity)

57. Space complexity
• Total amount of computer memory required by an
algorithm to complete its execution is called as space
complexity
Example 1
int square(int a)
{
return a*a;
}

58. Space complexity
• In the above piece of code, it requires 2 bytes of
memory to store variable 'a' and another 2 bytes of
memory is used for return value.
• That means, totally it requires 4 bytes of memory to
complete its execution. And this 4 bytes of memory
is fixed for any input value of 'a'. This space
complexity is said to be Constant Space Complexity.

59. Example 2
int sum(int A[ ], int n)
{
int sum = 0, i;
for(i = 0; i < n; i++)
sum = sum + A[i];
return sum;
}

60. Example 2 Solution
• In the above piece of code it requires
'n*2' bytes of memory to store array variable 'a[ ]'
2 bytes of memory for integer parameter 'n'
4 bytes of memory for local integer variables 'sum' and
'i' (2 bytes each)
2 bytes of memory for return value.
That means, totally it requires '2n+8' bytes of
memory to complete its execution. Here, the total
amount of memory required depends on the value of
'n'. As 'n' value increases the space required also
increases proportionately. This type of space
complexity is said to be Linear Space Complexity.

61. Time complexity
• The time complexity of an algorithm is the total
amount of time required by an algorithm to
complete its execution.
Example 1
int square(int a)
{
return a*a;
}

62. Time complexity
• In the above sample code, it requires 1 unit of
time to calculate a+b and 1 unit of time to return
the value.
• That means, totally it takes 2 units of time to
complete its execution.
• And it does not change based on the input values
of a and b. That means for all input values, it
requires the same amount of time i.e. 2 units.
• If any program requires a fixed amount of time
for all input values then its time complexity is said
to be Constant Time Complexity.

63. Dynamic Memory Allocation
.
Function Purpose
malloc Allocates the memory of requested size and
returns the pointer to the first byte of
allocated space
calloc Allocates the space for elements of an
array. Initializes the elements to zero and
returns a pointer to the memory.
realloc It is used to modify the size of previously
allocated memory space.
Free Frees or empties the previously allocated
memory space

64. The malloc Function
• The malloc() function stands for memory allocation.
• It is a function which is used to allocate a block of
memory dynamically.
• It reserves memory space of specified size and
returns the null pointer pointing to the memory
location.
• The pointer returned is usually of type void. It means
that we can assign malloc function to any pointer.
• Syntax
• ptr = (cast_type *) malloc (byte_size);

65. The malloc Function
• Here,
• ptr is a pointer of cast_type.
• The malloc function returns a pointer to the
allocated memory of byte_size.
• Example: ptr = (int *) malloc (50)
• When this statement is successfully executed, a
memory space of 50 bytes is reserved.
• The address of the first byte of reserved space is
assigned to the pointer ptr of type int.

66. The calloc Function
• The calloc function stands for contiguous allocation.
• This function is used to allocate multiple blocks of
memory.
• It is a dynamic memory allocation function which is
used to allocate the memory to complex data
structures such as arrays and structures.
• Malloc function is used to allocate a single block of
memory space while the calloc function is used to
allocate multiple blocks of memory space.

67. The calloc Function
• Each block allocated by the calloc function is of the
same size.
• Syntax:
• ptr = (cast_type *) calloc (n, size);
• The above statement is used to allocate n memory
blocks of the same size.
• After the memory space is allocated, then all the
bytes are initialized to zero.
• The pointer which is currently at the first byte of the
allocated memory space is returned.

68. The realloc Function
• Using the realloc() function, you can add more
memory size to already allocated memory.
• It expands the current block while leaving the
original content as it is.
• realloc stands for reallocation of memory.
• realloc can also be used to reduce the size of the
previously allocated memory.
• Syntax
• ptr = realloc (ptr,newsize);

69. The free Function
• The memory for variables is automatically
deallocated at compile time. In dynamic memory
allocation, you have to deallocate memory explicitly.
If not done, you may encounter out of memory error.
• The free() function is called to release/deallocate
memory

70. Linked List
• When we want to work with an unknown number of
data values, we use a linked list data structure to
organize that data.
• The linked list is a linear data structure that contains
a sequence of elements such that each element links
to its next element in the sequence.
• Each element in a linked list is called "Node".

71. • SINGLY LINKED LIST
• SINGLY CIRCULAR LINKED LIST
• DOUBLY LINKED LIST
• DOUBLY CIRCULAR LINKED LIST

72. Single Linked List
• Single linked list is a sequence of elements in which
every element has link to its next element in the
sequence.
• In any single linked list, the individual element is
called as "Node". Every "Node" contains two fields,
data field, and the next field.
• The data field is used to store actual value of the
node and next field is used to store the address of
next node in the sequence.

73. Single Linked List
Example:-

74. Operations on Single Linked List
• The following operations are performed on a Single
Linked List
• Insertion
• Deletion
• Display

75. Insertion
• In a single linked list, the insertion operation can be
performed in three ways.
• Inserting At Beginning of the list
• Inserting At End of the list
• Inserting At Specific location in the list

76. Abstract Data Types
• Abstract Data type (ADT) is a type (or class) for
objects whose behavior is defined by a set of value
and a set of operations.
The definition of ADT only mentions what operations
are to be performed but not how these operations
will be implemented.
• It does not specify how data will be organized in
memory and what algorithms will be used for
implementing the operations.

77. Abstract Data Types
• It is called “abstract” because it gives an
implementation-independent view. The process of
providing only the essentials and hiding the details is
known as abstraction.
• The user of data type does not need to know how
that data type is implemented,
• for example, we have been using Primitive values like
int, float, char data types only with the knowledge
that these data type can operate and be performed
on without any idea of how they are implemented.

78. List ADT
• A list contains elements of the same type arranged in
sequential order and following operations can be
performed on the list.
• get() – Return an element from the list at any given
position.
• insert() – Insert an element at any position of the list.
• remove() – Remove the first occurrence of any
element from a non-empty list.

79. List ADT
• removeAt() – Remove the element at a specified
location from a non-empty list.
• replace() – Replace an element at any position by
another element.
• size() – Return the number of elements in the list.
• isEmpty() – Return true if the list is empty, otherwise
return false.
• isFull() – Return true if the list is full, otherwise
return false.

80. Sparse Matrix
• In computer programming, a matrix can be defined
with a 2-dimensional array.
• Any array with 'm' columns and 'n' rows represent a
m X n matrix.
• There may be a situation in which a matrix contains
more number of ZERO values than NON-ZERO values.
Such matrix is known as sparse matrix.
• Sparse matrix is a matrix which contains very few
non-zero elements.

81. Sparse Matrix
• When a sparse matrix is represented with a 2-
dimensional array, we waste a lot of space to
represent that matrix.
• For example, consider a matrix of size 100 X 100
containing only 10 non-zero elements.
• In this matrix, only 10 spaces are filled with non-zero
values and remaining spaces of the matrix are filled
with zero.
• That means, totally we allocate 100 X 100 X 2 =
20000 bytes of space to store this integer matrix.

82. Sparse Matrix
• And to access these 10 non-zero elements we have
to make scanning for 10000 times.
• To make it simple we use the following sparse matrix
representation.
• A sparse matrix can be represented by using TWO
representations, those are as follows...
• Triplet Representation (Array Representation)
• Linked Representation

83. Triplet Representation
(Array Representation)
• In this representation, we consider only non-zero
values along with their row and column index values.
• In this representation, the 0th row stores the total
number of rows, total number of columns and the
total number of non-zero values in the sparse matrix.
• For example, consider a matrix of size 5 X 6
containing 6 number of non-zero values

84. Triplet Representation
(Array Representation)

85. Triplet Representation
(Array Representation)
• In above example matrix, there are only 6 non-zero
elements ( those are 9, 8, 4, 2, 5 & 2) and matrix size is
5 X 6.
• Here the first row in the right side table is filled with
values 5, 6 & 6 which indicates that it is a sparse matrix
with 5 rows, 6 columns & 6 non-zero values.
• The second row is filled with 0, 4, & 9 which indicates
the non-zero value 9 is at the 0th-row 4th column in
the Sparse matrix.
• In the same way, the remaining non-zero values also
follow a similar pattern

86. Linked Representation
• In linked representation, we use a linked list data
structure to represent a sparse matrix.
• In this linked list, we use two different nodes namely
header node and element node. Header node
consists of three fields and element node consists of
five fields

87. Linked Representation