Data structure and algorithm.(dsa)

Processing of data items by linear fashion is linear data structure e.g. array,
stack, queue, linked list.
o/p
i/p
In case of stack:- -Last input is the first output
1 input process is called push.
2 Output process is called pop.
In case of queue - First input is the first output.
3 Input is called rear.
4 Output is called front.
In case of linked list
start
info
info
mation
info
mation
Data Structure:-
It is a wave to organize the data in some way so we can do the operations
on these data in effective way. Data structure may be organized in many different
ways. The logical or mathematical model of a particular organization of data is
called data structure. Choice of data model depends on two considerations.
It must be rich enough in structure to mirror the actual relationship of the
data in the real world.
The structure should be simple enough that one can effectively process the
data when necessary.
Data structure can be classified as:-
1. Linear data structure:-

First part consists of information and second part consists of pointer to
second block.
2. Non Linear data structure:-
A data structure in which insertion and deletion is not possible in a
linear fashion is called non –linear data structure. E.g. tree, graph.
student
Student_Id Name address Branch
Electro
nices
Computer
F_name
BCA BITBEareastreetL-name
- It is hierarchical in nature.
Graph:-
Illam
Biratnagar
Kathmandu
Nepalgunj
Birgunj
fig. airlines flights
- Contents relationship between pair of elements which is not in
hierarchical in nature.
3. Abstract Data type (ADT):-
ADT is used to specify the logical properties of the data type. It is a set of
operation which is called with the component of the element of that abstract data
type.
e. g. list as ADT the term ‘ADT’ refers to basic mathematical
Component concept that defines the data type.
Item
Operations
-Insertion

- Deletion
- Search
- Display
Here, item is component of ADT.
An ADT consists of two parts.
(a) value definition
i. Definition clause
ii. Condition clause
(b) operator definition
# ADT for Relation:-
/*value of definition */
Abstract type def <integer, integer> RTIONAL // value definition
Condition RATIONAL [1] = 0; // denominator is not equal to zero;
condn
.
/* operator definition * /
Abstract RATIONAL make rational (a, b)
Int a,b;
Precondition b! = 0;
Post condition make rational [0] == a;
Make rational [1] == b;
Abstract RATIONAL add (a, b)
RATIONAL a,b;
Past condition add [1] == a[1] * b[1];
Add [0] == a[a] * b[1] + b[0] * a[1];
Abstract RATIONAL mult (a,b)
RATIOANL a, b;
Post condition mult [0] ==a[a] * b[0];
Mult [1] == a[1] *b[1];
Abstract RATIONAL equal (a,b)
RATIONAL a,b;
Post condition equal == (a[0] * b[1] == b [0] * a[1]);

Goals of data structure:-
Data structure involves two complementary goals:-
(1) identify and develop useful mathematical entities and operations and
determine what class of problems can be solved by using these entities and
operations.
(2) Determine representation fro those abstract entities and implement the
abstract operations on these concrete representations.
Properties of Data structure:-
- Generalization:-
Entity generalizes the primitive data type (integer, real, float)
- Encapsulation:-
All the operation of that type can be localized to one section of program.

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Chapter:- 2
Stack:- Stack is an ordered collection of items in to which new items may be
inserted and from which items may be delete at one end, called tope of the
stack.
Two operations are possible on stack:-
(1) push:- Add item in to stack.
(2) Pop :- delete the item from stack.
Top =-1 A A
B
A
B
C
A
B
A A
D
A
(a) Empty (b) Push A (c ) Push B (d) Push C (f) Pop B (g) Pus D (h) Pop D(e) Pop C
From figure we see that the items inserted at last is the first item to be
poped or deleted. Thus stack follows last in first out (LIFO) or first in last out
(FILO) policy.
Stack of ADT:-
Abstract typedef << dtype>> STACK(datatype); //define datatype (stack type element)
Abstract empty (s) // for empty function.
STACK (eltype) s;
Past condition empty == (len(s)== 0); // condition for empty.
Abstract eltype pop (s) //function declaration.
STACK (eltype ) S; // variable declaration.
Precondition empty (S)== FALSE; //condition for fulfill.
Past condition pop == first (S’); /length in pree condition.
S= = subh(S”,1, len(S’)-1); // S’- original length, 1- deleting item,
len(S)1- post condition .-
push
abstract push (S, elt) // add new element elt.
STACK (eltype) S; // new variable declearation.
Post condition S== <elt>+S’; //

Stack Implementation:-
(i) Array implementation
(ii) Linked list.
Array Implementation:-
In Array we can push elements one by one from 0th
position 1st
position
…….. n-1th
position. Any element can be added or deleted at any place and we can
push or pop the element from the top of the stack only.
1. When there is no place for adding the element in the array, then this is
called stack overflow. So first we check the value of top with size of array.
2. When there is no element in stack, then value of top will be -1. so we check
the value of top before deleting the element of stack.
5 10 15
Stack_array [0] [1] [2] [3] [4] [5] [6]
Here, Stack is implemented with stack array, size of stack array is 7 and value
of top is 2.
Operation of Stack:-
Push Operation:-
If (top == (max-1)]
Printf(“stack over flow”);
Else
{
Top == top + 1;
Stack_arr[top] = pushed –item;
}
Pop Operation:-
If (top ==-1)
Printf(“stack underflow”);
Else
{
Printf(“Poped element is %d”,stack_arr[top]);

Top= top – 1;
}
Algorithm for push & Pop
PUSH:-
Let stack [Max size] is an array for implementing the stack
1. [check for stack overflow?]
If top = maxsize-1, then print overflow & exit
2. set top = top + 1 (increase top by 1)
3. set stack [top] = item (inserts item in new top position )
4. exit.
POP:-
1. check for the stack underflow
If Top <0 then
Print stack underflow and exit
Else
[Remove the top element]
Set item = stack [ top]
2. Decrement the stack top.
3. Return the deleted item from the stack.
4. Exit.
Application of Stack:-
1 Conversion of an expression from infix to post fix.
2 Evaluation of an arithmetic expression from post fix expression.
Precedence:-
$ (Power), %( remainder) - 5
*(mul), /(div) - 4
+ (add), -(sub) - 3
( -2
) - 1

Conversion of Infix to post fix:-
Algorithm:-
1. Add a unique symbol # in to stack and at the end of array infix.
2. Scan symbol of array infix fro left to right.
3. If the symbol is left parenthesis ‘(‘then add in to the stack.
4. If symbol is operand then add it to array post fix.
5. (i) If symbol is operator then pop the operator which have same precedence
or higher precedence then the operator which occurred.
(ii) add the popped operator to array post fix.
(iii) Add the scanned symbol operator in to stack.
6. (i) If symbol is right parenthesis ‘)’ then pop all the operators from stack
until left parenthesis ‘(‘ in stack.
(ii) Remove left parenthesis ‘(‘ from stack.
7. If symbol is # then pop all the symbol form stack and add them to array
post fix except #.
8. Do the same process until # comes in scanning array infix.
Change Infix to post fix:-
Q. 1. A + B – C + D
1 A+ B – C + D #
Step Symbol Operator in stack post fix expression
1 A # A
2 + #- A
3 B #+ AB
4 - #- AB+
5 C #- AB+ C
6 + #+ AB +C -
7 D #+ AB + C -D
8 # # AB+C-D+

Q.2.A* (B+C $D) – E $F * (G/H)
A*(B+C$D) – E$ F*(G/H)
Step Symbol Operator in stack post fix expression
1 A
#
A
2 *+ #* A
3 ( #+ AB
4 B #- AB+
5 + #- AB+ C
6 C #+ AB +C -
7 $ #+ AB + C -D
8 D # AB+C-D+
9 ) #* ABCD $+
10 - #- ABCD$+*
11 E #- ABCD$+*E
12 $ #-$ ABCD $ +
*E
13 F #-$ ABCD$
+*EF
14 * #-* ABCD$ +*EF$
15 ( #-*( ABCD$+*EF$
16 G #-*( ABCD$+*EF$G
17 / #-*(/ ABCD$+*EF$G
18 # #-*(/ ABCD$+ *EF$GH/
19 ) #-* ABCD$ +*EF$GH/*-
20 # # ABCD$+*EF$GH/*-
Hence, the required post fix expression is ABCD$ +* EF $ GH/*-
Evaluation of post fix expression:-
In this case stack contents the operands. In stead of operators. Whenever
any operator occurs in scanning we evaluate with last two elements of stack.

Algorithm:-
(i) Add the unique symbol # at the end of array post fix.
(ii) Scan the symbol of array post fix one by one from left to right.
(iii) If symbol is operand, two push in to stack.
(iv) If symbol is operator, then pop last two element of stack and evaluate it as
[top- 1] operator [top] & push it to stack.
(v) Do the same process until ‘#’ comes in scanning.
(vi) Pop the element of stack which will be value of evaluation of post fix
arithmetic expression.
Postfix expression ABCD $+* EF$GHI * -
Evaluate postfix expression where A = 5, B = 5, C = 4, D = 2 E = 2, F= 2 ,
G = 9, H= 3 now, 4, 5, 2, $ , +, *, 2, 2, $, 9, 3, /, *, - , #
Step Symbol Operand in stack
1 4 4
2 5 4, 5
3 4 4,5,4
4 2 4,5,4,2
5 $ 4,5,16
6 + 4,21
7 * 84
8 2 84,2
9 2 84, 2,2
10 $ 84,4
11 9 84,4,9
12 3 84,4,9,3
13 / 84, 4, 3
14 * 84,12
15 - 72
The required value of postfix expression is 72

For checking:-
A*(B+C$D) – E$F*(G/H)
4*(5+42
)-22
*(9/3 )
4* (5+16)-4*3
84-12
72
Q. Convert the post fix:-
((A+B)*C – (D-E) 4F+G
Evaluate post fix where A=1, B=2, C= 3, D= 4, E= 3, F=2, G=1.
Verify it by evaluating it’s infix expression.
Step Symbol Operand in stack Post fix operation
1 ( #(
2 ( #(( A
3 + #((+ A
4 B #((+ AB
5 ) #() AB+
6 * #(* AB+
7 C #(* AB+C
8 - #(- AB+C*
9 ( #(-( AB+C*
10 D #(-(- AB+C*D
11 - #(-(- AB+C*D
12 E #(-(- AB+C*DE
13 ) #(- AB+C*DE-
14 ) # AB+C*DE--
15 $ #$ AB+C*DE--
16 ( #$( AB+C*DE--
17 F #$( AB+C*DE - -f
18 + #$(+ AB+C*DE--f
19 G #$(+ AB+C*DE - - FG
20 ) # AB+C*DE- - FG+ $
The required postfix expression is AB+C*DE - - FG +$
Now 1,2, +3, *4, 3, -, -, 2, 1, +, $, #

Step Symbol Operand in stack
1 1 1
2 2 1,2
3 + 3
4 3 3,3
5 * 9
6 4 9,4
7 3 9,4,3,1
8 - 3,1
9 - 8
10 1 8,2
11 1 8,2,1
12 $ 8,3
13 $ 24
For checking
The given infix expression is
((A+B)*C-(D-e))$(F+G)
= ((1+2)*3-(4-3))4(2+1)
= (3*3 -1) $3
= (9-1)$3
= 8 $ 9
= 24

Chapter:- 3
Queue:-
A queue is an ordered collections of items from which items may be
deleted at one end called the front of the queue and in to which items may be
inserted at the other end called rear of the queue.
Deletion ⇒ ⇐ Insertion
Front rear
e.g.
queue – arr [5] [0] [1] [2] [3] [4]
front =-1 (a) empty queue.
Rear =-1
5
Front =0 fig:- adding an item in queue
Rear =0
5 1 0
Front =0 fig:- adding an item in queue
Rear = 1
5 1 0 15
Front =0
Rear =2 fig:- adding an item in queue.
1 0 15
Front = 1 fig:- deleting an element from queue
Rear =2
1 0 15 20
Front = 1
Rear = 3 fig:- adding an element in queue
from figure we see that item are inserted at rear end and deleted at front
end. Thus queue follows FIFO policy. (First in first out).

Queue as an A-DT
Abstract typedef <<eltype>>QUEUE (eltype); //queue type element.
Abstract empty (q)
QUEUE (eltype) q;
Post condition empty = = (len (q) = =0);
Abstract eltype delete (q)
QUEUE(eltype ) q;
Pre condition empty (q) = = FALSE
Post condition remove = = front (q’);
q = = sub(q’, 1, len(q’-1);
abstract insert (q, elt)
QUEUE (eltype) q;
Eltype elt ;
Post condition insert = rear (q’);
Q = q’+<elt>;
Queue Implementation :-
8 array implementation
9 linked list implementation
Array implementation of Queue:-
A queue top two pointers front and rear pointing to the front and rear
element of the queue.
- when there is no place for adding elements in queue, then this is called queue
overflow.
- when there is no element for deleting from queue, then value of front and rarer
will be -1 or front will be greater then rarer.
[0] [1] [2] [3] [4]
5 10
Front t =1
Rear =1

Operation in queue
1. Add operation :-
If (rear = = Max -1)
Printf(“queue overflow”);
Else
{
If (front = = -1)
Front = 0;
Rear = rear + 1;
Queue_arr[rear]= added-item;
}
Delete operation:-
If (front = = -1)|| (front >rear)
{
Printf(“Queue under flow”);
Return;
}
Else
Printf(“Element detected from queue”, queue_arr[fornt]);
Front = front +1;
}
Drawbacks:-
[0] [1] [2] [3] [4]
5 10 1 5
Front =2
Rear = 4
Here we see that the rear is at the last position of array and front is not at
the zeroth
position. But we can not add any element in the queue because the rear
is at the n-1 position.
There are two spaces for adding the elements in queue. But we can not add

any element in queue because are is at the last position of array. One way is to
shift all the elements of array to left and change the position of front and rear but it
is not practically good approach. To overcome this type of problem we use the
concept of circular queue.
Algorithm to insert an element in a queue Deletion
Step:-1 Step:-1
[check overflow condition] [check underflow
condition]
If rear (=) > [Max -1] if front =-1
o/p :”over flow” o/p:”under flow & return;
return;
Step:-2 Step:-2
[increment rear pointer] Remove an element
Rear = rear +1 value: Q[Front]
Step:-3 Step:-3
[insert an element ] [check for empty queue]
Q[rear]= value if front = = rear
Front =-1
Step:-4 Rear = -1
[set front pointer] else
If front =-1 front = front +1
Step:-5 Step:-4
Return Return [value]
Circular Queue:-
[0] [1] [2] [3] [4]
1 0 20 3 0
A circular queue is one in which the insertion of new element is done at the very
first location of the queue if the last location of the queue is full.
e.g consider C queue_arr[4]

a. [0] [1] [2] [3]
5 10 1 5
from = 1 rear = 3.
initial queue
b.
10 1 5
front = 2
fig. deletion in queue rear = 3
c.
10 1 53 5
front =0 rear = 1
fig. addition in queue.
d
10 1 52 035
rear =1
front =2
fig. addition in queue.
e
1 52 035
front =1
rear = 3
fig. deletion in queue.
f
2 035
front = 1
rear =1
fig. deletion in queue
g
2 0
front = 1
rear =1
h rear = 1
Front =-1

Insersation:-
If ((front == 0 && rear = Max -1 ) || (front = = rear +1))
{
Printf(“Queue overflow”);
Return;
}
If (front = = - 1)
{
Front =0;
Rear = 0;
}
Else
If(rear == Max -1)
Rear =0;
Else
Rear = rear +1;
(queue_arr[rear] = added_item;
Operation
Deletion:-
If (front = = -1
{ printf(“queue underflow”);
Return;
}
Else
Printf(“element deleted from queue is %d”);
(queue_arr[front];
If (front == rear)
{
Front =-1;
Rear = -1;
}
Else

5 10 15 8
left =2 right =5
10 15 8
left =3 right =5
10 15 8 20
left =3 right =6
If (front = = Max – 1)
Front = 0;
Else
Front = front +1;
Dequeue (Double ended queue)
In dequeue we can add or delete the element from both sides i.e. front end
or form the rear end. It is of two types.
(i) I/P restricted
(ii) O/P
In input restricted dequeue element can be added at only one end but we
can delete the element from both sides.
In output restricted dequeue eleme4tn can be added from both side but
deletion is allowed only at one end.
Operation in Dequeue:-
We assume a circular array fro operation of addition or deletion.
[0] [1] [2] [3] [4] [5] [6] [7]
5 10 15
Left = 2
Right = 4
We maintain two pointers right & left which indicate positions of dequeue. Here,
left pointer is at position 2 and right pointer is at position 4.
For I/P restricted:-
Add 8 in the queue from right.
Delete the element from left of the queue
Add the element 20 in the queue

10 15 8 20 16
left =3 right =7
10 15 8 20
left =3 right =6
10 15 8 20 30
left =3 right =7
10 15 8 20 30
left =0 right =3
n
10 15 8 20 30
left =1 right =3
12 35
10 15 8 20 30
left =2 right =3
12 35 6
Add the element is in queue
Delete the element form right of queue.
Add the element 80 in the queue
Add the element 12 in the queue.
Add the element 45 in the queue. No it overflows because right pointer will
become equal to the left pointer after adding an element.
Right addition:-
If(left = = 0 && right == Max -1) || (left = = right +1))
{
Return;
}
If (left==-1)
{
Left =0;
Right =0;
}
Else
If (right = = Max -1)

Right = 0;
Else
Right = right +1;
Dedequeue_arr[right] = added_item;
Left Addition:-
If (( left == 0 && right == Max-1) || (left == right +1))
{
Return;
}
If (left==-1)
{
Left =0;
Right =0;
}
Else
If (left == 0)
Left = Max -1;
Else
Left = left -1;
Dequeue_arr[left] = added_item;
Delete left:-
If (left==-1)
{
Printf(“Queue flow”);
Return;
}
Printf(“Element deletd from queue is “%d”, dequeue_arr[left]. // point delete
item
If (left == right ) // make empty
{
Left ==-1;

Right =-1;
}
Else
If(left== Max -1) //move from zero.
Left =0;
Else
Left = left +1; // increase space.
Delete Right:-
If (left ==-1)
{
Printf(“Quieue underflow”);
Return;
}
Printf(“element delected from queue is %d”, deque_arr[right]);
If(left==right)
{
Left =-1;
Right =-1;
}
Else
If(right==0)
Right =Max-1;
Else
Right = right [-1;
}

Chapter:- 4
A linear list is an ordered set consisting of a umber of elements to which
addition w& deletion can be made. A linear list displays the relationship of
physical adjenci. The first element of a list is called the hear of list & the last is
called the tail of the list. The next element of the head of the list is called it’s
successes. The previous element to the tail (if it is not head of the list) is called it’s
predessor. Clearly a head doesn’t have as predessor & a tail doesn’t have a
successor. Any other element of the list has both one successor & one predessor.
10 20 30 40 50
head
Operations perform in list:-
1. Traversing an array list
2. Searching an element in the list
3. Insertion of an element in the list
4. Deletion of an element in the list.
Array Implementation:-
Take an array of size 10 which has 5 elements
10 20 30 40 50
[0] [1] [2] [3] [4] [5] [6] [7] [8] [9]
⇒ Traversing an array list:-
Here each element can be found through indeed no. of array & is
incremented by 1.
When index =0.
Then,
Arr[10] = 10 ⇒ determines 1st
element in array
Index = index +1
Then
Arr[index]= 20

In this way we can transverse each element of array by incrementing
the index by 1 until index is not greater than no. of elements.
⇒⇒⇒⇒ Searching in an array list:-
For searching an element, we first travers the array list and while traversing
, we compare each element of array with the given element.
Int I, item;
For (i=0;i<n;i++)
‘{
If(item== arr[i])
Return(i+1);
}
Insertion of an element in the list:-
-two ways:-
(i) insertion at end
(ii) insertion in between.
(i) Insertion at end
10 20 30 40 50
60 elemetn inserted at 6th position
Set the array index to the total no of elements & then insert the element
Index = Total no of element (i.e 5)
Arr[index] = value of inserted element
2. Insertion in between
60 elemetn inserted at 4th position
10 20 30 40 50
Shift right one position all array elements from last array element to the
array element before which we want to insert the element.
Int tem, item, position;
If (n = = max)
{

Printf(“list overflow”);
Return;
}
If (position >n+1)
{
Printf(“enter position less than or equal to %d”n+1);
Return;
}
If(position = = n+1) /*insertion at the end */
{
Arr [n]= item;
N= n+1;
Return;
}
Temp = n-1 //insertion in between
While (temp>=positon-1)
{
Arr[temp+1] = arr [temp];
Temp - -
}
Arr [positon -1] = item;
n = n+1;
Deletion of an element in the list:-
deletion of the last element
deletion in between
Deletion of the last element:-
10 20 30 40 50
Deleted element
Traverse the array last and if the item is last item of the array, then delete
that element & decrease the total no of element by 1,

Deletion in between:-
10 20 30 40 50
Deleted element
First traverse the array list & compare array element with the element
which is to be deleted, then shift left one position from the next element to
the last element of array and decrease the total no. of elements by 1.
Int temp, position , item, n;
If (n= = 0) // present or not element
{
Printf(“list underflow”);
Return;
}
If (item == arr [n-1] //deletion at the end.
{
n = n- 1;
return;
}
Temp = position -1;
While (temp < = n -1)
{
Arr [temp] = arr [temp +1]
Temp + + ;
}
n = n -1;
Advantage of list:-
⇒⇒⇒⇒ Easy to compute the address of the array through index 4 can be access the
array element through index.
Disadvantage:-
⇒⇒⇒⇒ Use of contiguous list which is time consuming.
⇒⇒⇒⇒ As array size is declared we can’t take elements more then array size.
⇒⇒⇒⇒ If the elements are less than the size of array, then there is wastage of memory.

⇒⇒⇒⇒ Too many shift operation on each insertion and deletion.
To overcome this problem we use liked list or DLL.
Index to prefix conversion :-
Rules:-
1. Parenthesize the expression starting from left to right.
2. During parenthesizing the expression, the operands associated with
operator having higher precedence are first parenthesized.
e.g. B * C is first parenthesized before A+B in A+B *C
3. the sub expression which has been converted in to prefix is to be treated as
single operand.
4. once the expression is converted to postfix from remove the parenthesis.
Question:- (A*B + (C/D) –F
(A*B + (C/D) –F
((A*B + (C/D)) – F)
((A*B+W) – F) [ W = /CD let]
(( *AB + W) –F)
((X +W)-F [ X = *AB]
(+XW – F)
(Y –F) [ Y = +XW
-YF
-+XWF
- + ABWF
- + * AB/CDF.
Which is required prefix expression.
Dynamic memory allocation:-
In an array it is necessary to declare the size of array. This creates two
possible problems.
If the records that are stored is less than the size of array, then there is a
wastage of memory.

If we want to store more records then size of array, we can’t store.
To overcome these problem we use dynamic memory allocation. The
functions used for dynamic memory The functions used for dynamic memory
allocation and deallocaton are :-
Malloc()
Calloc()
Free()
Realloc()
Malloc():-
This function is used to allocate memory space. The malloc () function
reserves a memory space of specified size and gives the strating address to the
pointer variable.
Syntax:-
Ptr = (data type * ) malloc (specified size)
Type of pointer size required to reserve in memory.
Example:-
Ptr = (int *) malloc (10);
Struct student
{
Int roll_no;
Char name [30];
Float percentage;
};
Struct student * st_ptr.
St_ptr = (struct student *) malloc size of (struct student);

Calloc():-
The caloc () function is used to allocate multiple blocks of memory. This
has two arguments.
Syntax:-
Ptr = (data type *) calloc (memory block, size )
Example:-
Ptr = (int * ) calloc (5, 2)
Struc record
{
Char name [10];
Int age;
Float sal;
};
Int to_record =100;
Ptr = (struct record *) calloc (tol_record, size of (record));
Free():-
This function is used to deallocate the previously allocated memory using
mulloc () or calloc() function .
Syntax:-
Free (ptr);
Realloc():-
This function is used to resize the size of memory block which is already
allocated. It is used in two condition.
⇒⇒⇒⇒ If the allocated memory block is insufficient for current application.
⇒⇒⇒⇒ If the allocated memory is much more thasn what is required by the
current application.
Syntax:-
To allocate memory; we use
Ptr = (char *) mallc(6);
To reallocate memory;
Ptr = (char * ) realloc (ptr, B)

Chapter:- 5
A link list is a collection of elements called nodes each node hap two parts,
1st
part contains the information field & the 2nd
part contains the address of the
next node. The address part of the last node of linked list will have null value.
start Informaton
linked
address part of the node which contains the adderess of the next node
info. part of the node
null
address part of the node which contains
the address of the next node.
Infro. Part of the node.
In liked list one member is a pointer that points to a structure itself.
Syntax:-
Struct node
{
Int data;
Struct node * link;
};
Here, member of the structure struct node * link points to the structures itself.
Implementation:-
Traversing a linked list:-
In liked list start is a pointer which points to the 1st
element of the list for
processing next element we assign the address of the next element to the pointer
(ptr) as:
Ptr = ptr – link;
Each element of the liked list can travel through this assignment until ptr has null
address. So, the linked list can be traversed as:-
Which (ptr ! = NULL)
Ptr = ptr – link;

Searching in to a linked list:-
For searching the element first traverse the linked list and with traversing
compare the information part of each element with the given element it can be
written as:
While (ptr!=NULL)
{
If (ptr→ info = = data)
Printf(“Item %d found at position %d”, data, pos);
Else
Ptr = ptr → link;
Insertion in to a linked list:-
Insertion at beginning
Insertion in between
null
start
tamp
temp
Here, temp is a pointer points to the node that has to be inserted ie.
Temp → infro = data;
Start points to the 1st
elements of linked list for insertion at beginning we
assign the value of start to the linked part of the inserted node as
Temp → infro = start; and start can be re assigned as
Start = temp;

Insertion in between:-
null
start
tamp
a
First we traverse the linked list for obtaining the node after which we want
to insert the element. We obtain pointer 9 which points to the element after which
we have to insert new node. For inserting the element after the node, we assign
link part of that node to the link part of inserted node & the address of the inserted
node is placed into the link part of the previous node. This can be written as
Temp → infro = data;
Temp → link = q → link;
q → link = temp;
Deletion in to a linked list:-
null
start
tamp
Here, start points to the 1st
element of linked list. If element to be deleted is the 1st
element of linked list then we assign the value of start to temp as
Temp = start;
temp points to the 1st
node which has to be deleted & assign link part of the
deleted node to start as
start = star – link
Since start points to the 1st
element of linked list so, start→ link will point
to the second element of linked list,l now we free the element to be deleted. ie.
Free (temp)

null
start
tamp
If the element is other then the 1st
element of linked list then we give the link part
of the deleted node to the link part of the previous node & is given by
Temp = q → link;
q → link = temp → link
Free (temp);
If the node to be deleted is last node of linked list then we can write
Temp = q → link;
q → link = NULL;
Free (temp);
Stack implementation using linked list:-
null
Top
tamp
Push operation:-
For pushing the element on stack we follow the insertion operation of
linked list. i.e. we add the element at the start of the list this can be written as
Temp = q → link = pushed item;
q → link = top;
top = temp;
Here, top always points to the 1st
node of the linked list. Last pushed item will

become 1st
node of linked list. As there is use of linked list we don’t check
overflow condition.
Pop operation:-
null
Top
tamp
For this we delete the first element of linked list.
If (top = = NULL)
Printf(“Stack is empty”);
Else
Temp = top;
Printf(“popped item is %d”, temp→info);
Top = top→ link;
Free(temp);
Queue implementation using linked list:-
null
Front
Rear
Temp
For adding the element in queue, we add the element at the end of the list.
Here, front will point to the 1st
node of the linked list & rear will point to the last
node of the linked list this can be written as
Temp → info = added_item;
Temp → link = NULL
If (front == NULL)

Front = temp;
Else
Rear → link = temp;
Rear = temp;
Deletion operation:-
For deleting the element of a queue we delete the 1st
node of linked list .
null
Front
Rear
Temp
If (front = = NULL)
Printf(“queue underflow”);
Else
{
Temp = fornt;
Printf(“Deleted element is %d”, temp→ info);
Front = front → link;
Free (temp);
};
Double linked list (DLL):-
In single linked list we can traverse only in one direction because each node
has address of next node only. Suppose we are in the middle of linked list & we
want to operation with just previous node, then we have no way to go in previous
node. And we’ll again traverse from starting node which is time consuming./ To
overcome this problem we use double linked list.
In DLL each node has address of previous & next node along with data.
Start
NULL
Information
part of node
contain address of
previous node
NULL
Contain address
of next node

Each node has to contain the address of information of previous node. The data
structure for DLL is
Struct node
{
Struc node * previous;
Int info;
Struct node * next;
}
Here, struct node * previous is pointer to structure which will contain the address
of previous node & struct node * next will contain the address of next node in list.
Thus we can travel in both the direction.
Operation
1. Traversing a double linked list:-
Start points to the 1st
element of the list and we assign the value of start to ptr
so ptr so ptr points to the 1st
node of the list. For processing the next element
we assign the address of next node to ptr as
Ptr = ptr → next
Now, ptr has address of next node. Thus, we can traverse each element of list
through this assignment until ptr has null value which the next part value of
last element . it can traverse as:
While (ptr!= NULL)
Ptr = ptr→ next;
2. Insertion:-
(a) Insertion of beginning:
Start
NULL

For insertion at beginning we assign the value of start to the next part of
inserted node and address of inserted node to the previous part of stat as:
temp→ next= start;
start → prev = temp
now, inserted node points to the next node which is beginning node of DLL
and ‘prev’ part of second node will point to ht new inserted node. Now inserted
node is the first node of DLL so start will be re assign as
start = temp;
now, start will point to the inserted node which is the 1st
node DLL. Assign
NULL to the previous part of inserted node as ‘prev’ node of the first node is
null.
Temp → prev = NULL.
Thus whole process can re assign as
Temp → next = start;
Start → prev = temp;
Start → temp;
temp→ prev = Null.
Insertion in between:-
start
temp
a
First we traverse the DLL for obtaining the node after which we want to
insert the element. For inserting the element after the node we assign the address
of inserted d node to the ‘prev’ part of next node. Then we assign the next part of
previous node to next part of inserted node. Address of previous node will be

assign to pre. Part of inserted node and the address of inserted node will be
assigned to tyeh next part of prevous node. The whole process can be re assigned
as
q→ next → prev = temp;
Temp → next = q → next;
temp→ prev = q;
q →next = temp;
Deletion
Deletion at beginning:-
Start
tamp
Starts points to the 1st
node of DLL. It node to be deleted is the4 1st
of list
then we assign the value of start to temp as
Temp = start ;
Assign next part of deleted node to start as
Start = start→ next;
Since, start points to the 1st
node of linked list so start → next will point to
the second node of list. Then null will be assigned to start point ot previous.
Start → previous = Null
Now, we can free the node to be deleted which is pointed by temp as;
Free(temp)
The whole process can be re assigned as
Temp = start;
Start = start → next;
Start → prev = NULL
Free (temp)

a temp
If the element is other then the 1st
element of linked list then we
assign the next part of deleted node to next part of previous node & address of
previous node to ‘prev’ part to next node this can be written as
Temp = q →next; //select node i.e. pointed.
a→ next → prev =q;
temp → next → prev = q;
free (temp);
Deletion of the last node:-
if node to be deleted is last node then we’ll just free the last node & next
part of second last will be null. This can be written as
temp =q → next;
free (temp);
q →next = Null.
Priority queue:-
In priority queue every of queue has some priority and based on that
priority it wil be processed so the element of more priority will be process before
the element which has less priority. If two elements have same priority. Then in
this case it follows FIFO policy.
Structure:-
Struct pq {
Int priority;
Int info;
Struct pq * link;
}

front
start
1 info 2 info 3 info 4 info
3 info
Implementation:-
Temp → info = added – item
Temp → priority = item – priority;
If (front = = NULL || item – priority < front → priority)
{
Temp → link = front;
Front = temp;
Else
{
While(q → link ! = Null && q →link →priority < = item – priority)
q = q→ link;
temp → link = q → link;
q → link = temp;
}
Deletion:-
Delete operation wil be the deletion of 1st
element of list because it has
more priority then other elements of queue.
If (front = = NULL)
Printf(“Queue underflow”))
Else
{
Temp = front;
Printf(“deleted item is %d”, temp →info):
Front = front → link
Free (temp);
}

Chapter – 6
Recursion :-
Recursion is defined as defining any thing interns of itself. Recursion is
used to solve problems involving iteration s in reverse order.
Recursive functions:-
A function calls itself repeatedly.
Recursive program will not continue infinitely. A recursive procedure must
have the following two properties.
a. There must be the certain criteria, called base criteria fro which the
procedures does not call itself.
b. Each time the procedure does call itself, it must be closer to base
criteria.
Syntax:-
Main()
{
----------
-----------
Function();
__________
}
Function()
{_____________
Function();
}
/* Find factorial of any number */
Main(()
{
Int no, fact;
Printf(“Enter no”);
Scanf(“%d”, &no);
Fact = factl(no);

Printf(“Factorial no = %d”, fact);
}
Int factl(int n)
{
If (n = = 0)
Return(1);
Else
Result = n* fa t (n-1));
Return (result);
}
Output process:-
no =4
result = 24
after every call function will return the value to the previous function
fact l(u) = u* fact l(3)
fact l(3) = 3 * fctl(2)
factl(2) = 1 * fact(0)
fact(0) =1-
Here, steps for calling function are finished new each function will return the
value to it’s previous functions:
Factl(2) = 2 * fact (1) =2
Factl(3) = 3 * fact (2) = 6
` factl(4) = 4 * fct (3) = 4*6 = 24
Fact(1) = 1
2 * fact (1) 2 * fact(1) 2 *1 = 2
3*fact(2) 4*fact(3) 4 * fact(2) 3 * fact (2) 3*2 = 6
4*fact (3) 4*fct (3) 4*fact(3) 4*fact(3) 4* factl(3)` 4 * factl(3) 4*6 =24=
24

Need for recursion:-
There must be a terminating condition for problem which you want to solve
with recursion. This condition is called the base criteria for that problem.
There must be an if condition in the recursion routing, this if clause specifies
the terminating condition for the recursion.
Reversal in the order of execution is the characteristic of every recursion.
Problem.
Every time a new recursive calls is made, a new memory space is allocate to
each automatic variables used by the recursive routine.
The duplicated values of the local variables of a recursive call are pushed on to
the stack with it’s respective call & all these value are available to the
respective function when it is popped off from the stack.
Disadvantage:-
It consume more storage space because the recursive calls along with
automatic variable are stored in stack.
The computer may run out of memory if recursive calls are not cheked.
It is not more efficient in turms of speed & execution time.
Doesn’t offer any concrete advantage over non recursive functions.
If proper precautions are not taken, recursion may result non terminating
iterations.
Removal of recursion through iteration:-
Using recursion:-
Int factorial (int no)
{
Int fact = 1;
If (no>1)
Fact = no * factorial (n -1);
Return fact;
Without using recursion:-
Int factorial (int no)
{

Int fact =1
For(int i=no; i>1;I - -)
Fact = fact * I;
Return fact;
;
Tower of Hanoi (TOH):-
It is a game problem. In this case the disk is moved from one 0pillar to
another pillar with the use of temporary pillar.
Rule:-
We can move only one disk from one pillar to another at a time.
Large disk can ‘t be can not be placed on smaller disk.
source temporary destination
n = 2
n = 3n = 1
Step:- I
2
3
Step:- II
1 2 3
Step:- II
1 2
3

Step:- IV
12
3
Step:- V
123
Step:- VI
13
Step:- VII
1
2
3
Algorithm:-
Move upper 0-1 disk from source to temporary.
move largest disk from source to destination.
move n -1 disk from temporary to destination.
Problem:-
Proc(N -1, S, D, T)
Proc(1, S, T, D)
Proc(N-1, T, S, D)]
For n = 1
Proc(1, S, T, D)
Proc(1, S, D, T)
Proc (3, S,T,D) then, proc(2,S,D,T) proc(1,D, S,T)
Proc(1,S, T, D) S →D
Proc(2, S, T, D) proc(1, T, D,S)
Proc(1, T, S, D)
Proc(1, T, D, S)

proc(4, S, T, D)
proc(3, S, T, D)
proc(1, S, T, D)..........................................................................................S
proc(3 , S, T, D)
proc(2, S, T, D)
proc(1, S, T, D) .................................................S T
proc(2, S, T, D)
proc(1 , S, T, D)................................... .................T
proc(2, S, T, D)
proc(2 , S, T, D)
proc(1, S, T, D)................ S T
proc(1, S, T, D) .................S D
proc(1, T,S, D) ..................T D
proc(1,D, T, S) ....................D S
proc(1, S, T, D) .....................S T
proc(1,T, S, D) ......................T D
proc(1, T, D, S) ......................T S
proc(1, D, T, S) .......................D S
proc(1, T, S, D)......................T D
proc(1 , S, T, D).................... S D
proc(1 , S, T, D) .....................S T
proc(1,D, T, S) ......................D T
Translation from prefix to postfix using recursion:-
Algorithm:-
⇒⇒⇒⇒ In the prefix string is a single variable, it is it’s own postfix equivalent.
⇒⇒⇒⇒ Let operator be the first operator of the postfix string.
⇒⇒⇒⇒ Find the first operand opend1, of the string, convert it to postfix call it post 1.
⇒⇒⇒⇒ Find the second and 2, convert & call it post 2.
⇒⇒⇒⇒ Concatenate post 1, post 2 & op (operand)
Convert into postfix from the given string ** + AB - - + CDEF
Step Symbol operator postfix
1 * OP1
2 * OP2
3 + OP3
4 A Post + AB + ⇒⇒⇒⇒post 21
5 B Post 32
6 - OP4
7 + OP5
8 C Post 51 ⇒⇒⇒⇒ CD + Post 41 ⇒⇒⇒⇒ ⇒⇒⇒⇒post 22
9 D Post 52
10 E Post 42
11 F post 12

Chapter :-7
Definition:-
A tree is as non linear data structure in which items are arranged in a sorted
sequence, It is used to represent hierarchial relationship existing among several
data items. Each node of a tree may or may not pointing more than one node.
Theoretic definition of tree:-
It is a finite se of one or more data items (nodes) such that
there is a special data item called the root.
It’s remaining data items are portioned into no of mutually exclusive subsets,
each of which is it self a tree and they are called subtrees.
A
B
E F
J
G
C
D
H
K
I
M
L
Fig. Tree
Tree terminology:-
Root:-
The first node in a hierarchial arrangement of data items is root.
Node:-
Each data item in a tree is called a node.
Degree of a node:- It is the no. of subtrees of a node in a given tree from fig.
Here, degree of node A = 3
Degree of node D = 2
Degree of node F = 1
Degree of node I = 3

Degree of a tree:-
It is the maximum degree of nodes in a given tree here degree of tree is 3.
Terminal node:-
A node with degree 0 is terminal node here, SIGHXLM are terminal node.
Non terminal node:-
Any node except the root whose degree is not zero is call non terminal
node.
Siblings:-
The children nodes of a given parent nodes are called siblings. E and F are
siblings of parent node B and K, L, M are siblings of parents node I & so on.
Level:-
The entire tree structure is leveled in such a way that the root rode is always
at level zero. Then it’s immediate children are at level 1 and there immediate
children are at level 2 and so on up to the terminal nodes. Here, the level of a tree
is 3.
Path:-
It is a sequence of consecutive edge from the source node to the destination
node.
Edge:-
Is it is a connecting line of two nodes i.e. line drawn from one node to
another.
Depth or height:-
It is the maximum label of may node in a given tree here, the depth of a tree is
4.
Forest:-
It is a set of disjoint trees. In a given tree if we remove it’s rot then it
becomes forest.
Binary tree:-
A binary tree is a finite set of data items which is either empty or consist of

a single item called the root and two disjoint binary trees the lefts subs tree and
right subtree. In binary tree the maximum degree of any node is almost 2 i.e. each
node having 0, 1. or 2 degree node.
A
B
C
D
F
K
M
L
E
G
Strictly Binary tree:-
If every non terminal node in a binary tree consist of anon empty left sub
tree and right sub tree then such tree is called strictly binary tree.
A
B
D
K
L
E
G
c
F
N
M
H
I
Completely binary tree:-
Completely binary tree of depth ‘d’ is the binary tree of depth d that
contains exactly 21` nodes at each level l between zero and d i.e. in a complete
binary tree there is exactly one node at level zero, 2 node at level 1, 4 node at
level 2 and so on.

Extended binary tree:- (2-tree)
A binary tree is called extended binary tree if every node of tree has zero or
two children. It is called 2- tree.
Algebraic expression representation tree.
+
a
b
Any algebraic expression can be represented in binary two in which left
and right child represent operand of he expression and parent of the child
repressing the operator.
e.g, a + b
Algorithm:-
Note the order of precedence. All expression in parenthesis one to be ealuated
first.
Exponential will come next.
Division & multiplication will be the next in order of precedence.
Subtraction and addition will be the last to be proceeded.
+
a b
*C -
*
d e

+
a b
*C - (d*e)
(a+b)* C – (d * e)
*
+ c
a b
*
d e
-
* *
`
a
b
+ c d e
(ii) A + (B-C) * D & (E * F) (iii) (A+B*C) & (( B+ B) * C)
-
b c
*
e f
A + D &
-
B C
&
D *
E F
A + *

+
A
-
*
&
D *
B
C
C
F
*
B
B
B
B
B
B
B B
A +
A
B
D
E
F
G
C
H
Representation of binary Tree:-
A
B C
D E G H
F
Structure:-
Struc node
{
Char data:
Node * lchild;

Node * rchild;
}
Binary Tree Traversal:
While traversing a tree root is denoted by ‘N’ left subtree as ‘R’. Then there
will be six combination of traversal i.e. NRL, NLR, LRN< LNR< RNL & RLN
NLR pre ordered traversal
LAN post order traversal
LNR Inorder
Pre ordered traversal:- NLR
Visit the root node.
Traverse the left sib tree of root in pre order,”
Traverse the right subtree of root in pre order.
Pre order (ptr)
Struct node * ptr;
{
If (ptr ! = NULL)
{
Printf(“%d”, ptr→ data);
Pre order (ptr → lchild);
Pre order (ptr → rchild);
}}
A
B
D
E
I
F
C
G
J
Preorder:-
A B D E I C F G I

(ii) in order Traversal (LNR)
Travere the left subtree of root in order.
Visit the root.
Traverse the right subtree of root in order.
In order (ptr)
Struct node * ptr;
{
If (pre! = NULL)
{
In order (pre→ lchild)
Printf(“%C”, ptr →data);
In ordre (ptr→ rchild);
}}
A
B
D
E
I
F
C
G
J
Post order traversal:- (LRN)
Traverse the left subtree of root in post order.
Traverse the right subtree of root in post order.
Visit the root.
Post order (ptr)
Strct node * ptr;
{
If(pt! = NULL)
{
Post order (ptr →l child);
Post order (ptr → r child);
Printf(“%c”, ptr→ data);
}

A
A
B A
B
D
}
Post order:-
D I E B F J G C A
Question:- construct a binary tree from given algebraic expression and show it’s
traversals.
(a – b * c ) / (a + e/f)
A - /(a + e/f)
Preorder:-
/ - a * bc + d/e f
Inorder: a – b * / d + e – f
Post order: a b c * - d e f/ + /
Creating the tree from preorder and in order traversal:-
In pre order traversal, scan the nodes one by one and keep them inserting in
tree.
In order traversal, put the cross mark over the node which has been
inserted.
To insert a node in it’s proper position in the tree we look at that node the
in order traversal & insert it according to it’s position with respect to
the crossed nodes:-
Preorder:- A B D G H E I C F J K
Inorder:- G D H B E I A C J F K
Step:-1 Inset A
Pre: A B D G H E I C F J K
Inorder:- G G D H B E I A C J F K
Step:-2 Inset B
Preorder: A B D G H E I C F J K

A
B
D
G
A
B
D
G
H
A
B
D
G
H
E
A
B
D
G
H
E
I
A
B
D
G
H
E
I
C
A
B
D
G
H
E
I
C
F
Inorder:- G D H B E I A C I F K
Step:-2 Inset D
Preorder: A B D G H E I C F J K
Step:-3 Inset G
Step:- 4 Inset H
Inorder:- G D H B E I A C J A K
Step:- 5 Inset E
Inorder:- G D H B E I A C J F K
Step:- 6 Inset I
Inorder:- G D H B F I A C J F K
Step:- 7 Inset C
Inordre:- G D H B E I A C I F K

A
B
D
G
H
E
I
C
F
J
K
Step:- 8 Inset C
Preorder:- A B D G H E I C F I K
Inorer:- G D H B E I A C I C K
Step:- 9 Inset I
Inorer:- G D H B E I A C I F K
Binary search tree:-
A binary search tree is a binary tree in which each node has value greater
than every node of left sub tree and les than every node of right sub tree.
Question:- construct the binary tree from given data item.
50 30 60 22 38 55 34
50
30
22
38
34
55
60
Construct the binary tree:-
20 17 6 8 10 20 7 18 13 12 5
6

20
17
6
18
8
10
137
12
5
Delete operation in binary tree:-
Leaves:- easiest to delete.
- set pointer from it’s parent (if any ) to NULL.
Node with one child; fairly easy.
- Replace the node being deleted by it’s in order successor.
e.g
62
50
60
75
65
90
80
85
Delete 85:-
Since 85 has no children so, we can delete it., simply by giving NULL
value to it’s parents’ right pointer thus the tree becomes
62
50
60
75
65
90
80
Delete item 50:-
Since, 50 has only one child so we can delete it by giving the address of
right child to it’s parents left pointer thus the tree becomes.

62
60
75
65
90
80
Delete items 75:-
In order : 60 62 65 75 80 90
62
60
80
65
90
Since, 75 has two children so 1st
we delete the item which is in order
successor of 75. Here, 80 is the in order successor of 75 we delete so by giving
NULL value to it’s parents left pointer after that we replace item 75 with item 80.
We give the address of the left and right pointer of 75 to the left and right pointer
of 80 & address of 80 to the right pointer of parent 75 which is 62. Thus the
required tree is shown above.
AVL Tree “Balanced Tree”
“G.M adel’s son, vel’ skill & E.M land is”
A AVL tree is a binary search tree where, height of left & right subtree at
any node will be with maximum difference 1 each node of AVL tree has a
balanced factor. Balance factor of a node is defined as the difference between the
left subtree & right subtree of a node.
i.e. balanced factor = Height of left subtree – height of right sub tree.
A node is called right heavy or right high if height of it’s right sub tree is
one more than height of it’s left sub tree. A node is called left heavy or high if
height of it’s left sub tree is one more than height of it’s right sub tree. A node is
called balanced, if heights of left or right sub tree are equal. The balanced factor
will be 1 for left high & -1 for right high & 0 for balanced node. Thus in an AVL
tree each node can have only three values of balance factor which are 1, 0 & -1.

18
15
20
1*9
24
16
Procedure for insertion of a node in AVL tree:-
insert the node at it’s proper place as in binary search tree.
Calculate the balanced factors of all the nodes on the path starting from the
inserted nodes to the root node. If the value of balance factor of each node
on this path is -1 or 1 then, the tree is balanced. If the balanced factor of nay
node is more than 1 than the tree becomes unbalanced . The nod3e which is
nearest to the inserted node 4 has absolute value of balance factor greater
than 1, then it is marked as pivot node.
If the tree has become unbalanced after the insertion, then the tree is rotated
about the pivot node to achieve all the properties of AVL treep.
AVL rotations:-
Left to left rotation:-
15
15
10
5
12
16
20
24
Insert 2.
Pivot node
1 5
1 5
1 0
5
1 2
1 6
2 0
2 4
2
1
0
2
2
- 1
If the node is inserted on the left side of left subtree than we perform left to

left rotation.
18
10
5
12
15
20
24
2
16
1
0
1
-1
0
Right to right rotation:-
15
15
10
16
20
35
38
26
Insert 39
15
15
10
16
20
35
38
26
39
-1
-1
-2
-2
1
If the node is inserted on the right side of right sub tree than we perform right
to right rotation.
18
15
10
16
20
24
12
5

15
15
10
16
35
38
26
39
19
Left to right rotation:-
If the node is inserted on the right side of left sub tree then we perform left
to right rotation.
In this left to right rotation, the operation involves two steps 1st
we
perform right to right from next to pivot node then perform left to left.
1 8
1 5
1 2
1 0
1 4
1 6
2 0
2 4
5
18
12
10
14
15
20
24
5
16
Right to left rotation:-
18
15
10
19
20
35
26 38
18
15
10
19
20
35
26 38
24
-2
-2
1
insert 24
18
15
1
19
20
26
26 35
-2
-2
38
-1
-1

50
40
50
40
35
40
35
50
58
48
40
35
50
58
40
35
48
42
50
58
Right to left
rotation
Rotate right to Right:-
1 8
1 5
1 0
1 9
2 6
2 4
3 5
3 8
2 0
Construct an AVL tree from given data.
I/Ps : 50, 40, 35, 58, 48, 42, 60, 30, 33, 32
Insert 50
50
Insert 40
Insert 35
Insert 58
Insert 48
Insert 42

Step – I
48
35
42
48
50
58
Step –ii
35
40
48
42
50
58
Insert 60
50
48
40
35
42
58
60
Perform R R 35
48
58
6040
42 50
Insert 30
60
5 8
5042
40
35
48
30
Insert 33
48
40
35
30
33
42 50
58
60
2
2
2
-1
Step I Step II

33
40
48
48
58
60
50
3530
48
48
48
48
48
48 48
48
48
60
58
48
33
30
25
35
42
5040
Insert 25
48
60
40
58
5042
35
33
30
25
Huffman Tree:-
0
1
2
2
2
1
2
3 3
4
43 3
Internalnode
Externalnode

W1 W2
Fig:- extended binary tree
Total path length of internal nodes:-
PI = 0 + 1 + 2 + 1 + 1 + 2 + 3 = 9
Total path length of external nodes:
PE = 2 + 3 + 3 + 2 + 4 + 4 + 4 + 3 = 21
If external node has weight w, then weighted path length for external node is
P = W1P1 + W2 P2 + ……….WnPn
Algorithm:-
Let us take there are n weights W1, W2, ………., ………., Wn
Take two minimum weights & create a subtree, suppose W1 & W2 are first two
min weight then subtree will be
Now the remaining weight will be W1 + W2, W3, W4…….Wn
Create all sub tree at the last weight.
Q. Construct an extended binary tree by huff man
method.
A B C d E F G
Weight 16 11 7 20 25 5 16
Step:1 Taking two nodes with minimum weight 7 & 5,
12
7
5
Step:2 taking two nodes with minimum weight 11 & 12
12
7
5
23
11
Now the elements are 16, 23, 20, 25, 16

Step:3 Taking two nodes with minimum weight 16 & 16
32
16
16
Now,
The elements are 32, 23, 20, 25
Step:4
12
7
5
43
11
20
23
Now the elements are 32, 43, 25
Step:5
57
16
32 25
16
32
57
100
43
23
25
12
20
16 16 11
5 7
Total path length:-
PI = 0 + 1 + 2 + 2 + 3 + 4 = 12
PE = 2 + 3 + 3 + 2 + 4 + 4 + 3 = 21

B – Tree “Balanced Tree”
It is also known as balanced sort tree.
The height of the tree must be kept to a minimum.
There must be no empty sub trees. Above the leaves of the tree.
The leaves of the tree must all be the same level.
All nodes except the leaves must have at least some minimum no of
children.
B – Tree of order n can be defined as:-
Each node has at least n+2 & maximum ‘n’ non empty children.
All leaves nodes will be at the same level.
All the leaf nodes contain minimum n-1 keys.
Keys are arranged in a defined order with in the node. All keys in sub
tree to the left of the key are the procedure of the key and that on the
right are successors of the key.
When a new key is to be inserted in to a full node, then split the nodes
with the median value is inserted in the parent node. In case the
parent node is root, a new node is created.
10 20 40 50
35 65
100
130 180
70 80 90 110 120 140 160 190 240 260
Each node at same leaves.
All non- leaf nodes have no empty subtree.
Keys 1 less than no. of their children.
Q. Construc a B – tree of order 5 inserting the keys:-
10, 70, 60, 20, 110, 40, 80, 130, 100, 50, 190, 90, 180, 140, 280.
Soln
:-

Insert 10
10
Insert 70
10 70
Insert 60
10 60 70
Insert 20
10 20 60 70
Insert 110
10 20 70 110
60
Insert 40
10 20 40 70 110
60
80, 130, 100, 150, 90, 180, 240, 30, 120, 140, 160
Insert 80
Insert 130
60
10 20 40 70 80 110 130
Insert 100 10 20 40 70 80 110 130
60 110
Insert 50
60 100
10 20 40 50 70 80 100 130
60
10 20 30 70 80 110

Insert 190
60 100
10 20 40 50 70 80 100 130 190
Insert 90
60 100
10 20 40 50 70 80 90 100 130 190
Insert 180
60 100
10 20 40 50 70 80 90 100 130 180 190
Insert 240
60 100 180
10 20 40 50 70 80 90 100 130 190 240
Insert 30
30 60 100 180
10 20 40 50 70 80 90 100 130 190 240
Insert 120
30 60 100 180
10 20 40 50 70 80 90 110 120 130 190 240
Insert 140
30 60 100 180
10 20 40 50 70 80 90 110 120 130 140 190 240

Insert 160
30 60
10 20 40 50 70 80 90
100
130 180
110 120 140 160 190 240
Deletion in B- tree:-
node is leaf node.
node is non leaf
-if node has more than minimum no. of keys than it can be easily deleted.
-if it has only minimum no. of keys, than first we see the no. of keys in adjacent
leaf node. If it has more than minimum no. of keys then first key of the of the
adjacent node will go to the parent node & key in parent node will be combined
together in one node.
-If now parent has also less than minimum no. of keys then the same thing will be
repeated until it will gate the node which has more than the minimum no. of keys.
Node is non leaf:-
-In this case key will be deleted & it’s predessor or successor key will condition
it’s place.
-If both nodes of predessor or successor key have minimum no. of keys then the
rates of predessor & successor keys will be combine.
100
30 60 100
140 120 140 160 190 240 26070 80 9040 5010 20 25
Delete 190
100
30 60 100
110 120 140 160 240 26070 80 9040 5010 20 25

Delete 60
100
30 60 130 180
110 120 140 16070 80 9040 5010 20 25
Delete 40
100
25 70
10 20 30 50 80 90
130 180
110 120 130 160 240 260
Delete 140
240 260
25 70 100 180
80 9030 5010 20 110 120 130 160
Question:- Construct a B- tree of order 5 I/P the element when keys are 659, 767,
702, 157, 728, 102, 461, 899, 920, 44, 744, 264, 384, 344, 973, 905, 999
Perform delete operation fro 44, 344, 920

Chapter – 8
Game tree:- 659, 767, 702, 157, 728, 102, 461, 899, 920, 44, 744, 264, 344, 973,
905, 999
Given,
Maximum no. of key = n – 1 = 4 element
Minimum no. of element = n/2 = 2 element in each node.
Maximum no of children = 5
Insert 659
659
Insert 767
659 767
Insert 702 659 702 767
Insert 157 157 659 702 767
Insert 728
702
157 659 728 767
Insert 102
702
102 157 659 728 767
Insert 461
702
102 157 461 659 728 767
Insert 899
702
102 157 461 659 728 767 899
Insert 920
702
102 157 461 659 728 767 899 920
Insert 44
702
102 157 461 659 728 767 899 920
Insert 744

Insert 264 & 384
Insert 344
Insert 973 & 905 & 999
delete 44
delete 344 direct
Sorting:-
Sorting is storage of data in sorted order it can be in ascending or
descending order.
Type
Internal sort
External sort
In internal sorting data i.e. going to be sorted will be in many memory. In
external …….will be on auxiliary storage , tape floppy disk etc.
Sorting Technique:-
Insertion Sort:-
The insertion sort inserts each element in proper place if there are n-
element in array and we place each element of array at proper place in the
previously sorted element list.
Algorithm:-
Consider N elements in the array are
Pass 1 arr[0] is already sorted because of only one element.
Pass 2 arr[0] is inserted before or after arr[0]. So arr[0] & arr[i] are sorted.
Pass 3 arr[2] is inserted before arr[0], in between arr[0] & arr[1] or after
arr[0] . so arr[0] , arr[1] & arr [2] are sorted.
Pass 4 arr[3] is inserted in to it’s proper place in array arr[0], arr[1], arr[2],
arr[3] & are sorted.
…………………
……………………
Pass N arr[ N-1] is inserted in to it’s proper place in array arr[0], arr[1],
……. Arr[N-1]. So arr[0], arr[N-1] are sorted.

Q. Trace algo. With the given data using insertion sort.
82 42 49 8 92 25 59 52
Pass1 82 42 49 8 92 25 59 52
Pass 2 82 42 49 8 92 25 59 52
Pass 3 42 82 49 8 92 25 59 52
Pass 4 42 49 82 8 92 25 59 52
Pass 5 8 42 49 82 92 25 59 52
Pass 6 8 42 49 82 92 25 59 52
Pass 7 8 25 42 49 82 92 59 52
Pass 8 8 25 42 49 59 82 92 52
Pass 9 8 25 42 49 52 59 82 92
Selection Sort:-
Selection sort is the selection of an element & keepingit in sorted order. Let
us take an array arr[0]……..arr[N-1]. First find the position of smallest element
from arr[0] to arr [n-1]. Then interchange the smallest element from arr[1] to
arr[n-1], then interchanging the smallest element with arr[1]. Similarly, the
process will be for arr[0] to arr[n-1] & so on.
Algorithm:-
Pass 1:- search the smallest element for arr[0] ……..arr[N-1].
- Interchange arr[0] with smallest element
Result : arr[0] is sorted.
Pass 2:- search the smallest element from arr[1],……….arr[N-1]
- Interchange arr[1] with smallest element
Result: arr[0], arr[1] is sorted.
…………………
…………………
Pass N-1:-
- search the smallest element from arr[N-2] & arr[N-1]
- Interchange arr[N-1] with smallest element
Result: arr[0]…………. Arr[N-1] is sorted.

Q. Show all the passes using selecting sort.
75 35 42 13 87 27 64 57
Pass 1 75 35 42 13 87 27 64 57
Pass 2 13 35 42 75 87 27 64 57
Pass 3 13 27 42 75 87 35 64 57
Pass 4 13 27 35 75 87 42 64 57
Pass 5 13 27 35 42 87 75 64 57
Pass 6 13 27 35 42 57 75 64 87
Pass 7 13 27 35 42 57 64 75 87
Buibble sort:- (exchange sort)
In bubble sort each element is compared with I’t adjacent element. If the 1st
element is large than the second one then the position of the elements are
interchanged otherwise it is not changed.
Alg.
If N elements are given in memory then for sorting we do the following
steps:-
Pass:-
1. first compare the 1st
element and 2nd
element of array. If 1st
< 2nd
then compare
the 2nd
with 3rd
.
2. if 2nd
> 3rd
.
Then interchange the value of 2nd
& 3rd
.
now compare the value of 3rd
with 4th
.
similarly compare until N-1th
element is compared with nth
element.
Now, highest value element is reached at the Nth place.
Elements will be compared until N-1 elements.
Or
Repeat through step 1 to 5 for N-1 elements.

Q. Show all the passes using bubble sort.
13, 32, 20, 62, 68, 52, 38, 46
Pass 1. 13 32 20 62 68 52 38 46
Pass II 2nd
> 3rd
-interchange.
13 32 62 68 52 38 46
Pass III 32<62 - no change.
13 32 20 62 68 52 38 46
Pass IV 62<68 - no change.
13 32 20 62 68 52 38 46
Pass V 68< 52 - interchange.
13 32 20 62 52 68 38 46
Pass VI 68 <38 - interchange
13 32 20 62 68 38 68 46
Pass VII 68<46; interchange
13 32 20 62 52 38 46 68
Pass 1 (i) 13 20 32 62 52 38 46 68
13< 20 ; no change
(ii) 13 20 32 62 52 38 46 68
(iii) 13 20 32 62 52 38 46 68
(iv) 13 20 32 52 62 38 46 68
(v) 13 20 32 52 38 62 46 68
(vi) 13 20 32 52 38 46 62 68
(vii) 13 20 32 52 38 46 62 68
Pass:- 3
(i) 13 20 32 52 38 46 62 68 ;13<20 no change
(ii) 13 20 32 52 38 46 62 68 ;20<32 “
(iii)13 20 32 52 38 46 62 68 ;32<52
(iv)13 20 32 38 52 46 62 68 ;52< 38 change
(v)13 20 32 52 38 46 62 68
Pass :- 4
O/P 13 20 32 38 46 52 62 68

Q. Quick sort (partition exchange sort)
Quick sort is base on divide and conquer algorithm . In thi9s we divide the
original list in to two sub list . we choose the item from list called key or pivot
from which all the left side of elements are smaller & all the right side of elements
are greater than that element. Thus we can create two list, one list is an the left
side of pivot and 2nd
list is an right side of pivot. Similarly, we choose the pivot
for dividing the sub-list until there are two or more elements in the sub list.
Algorithm:-
(i) take the first element of list as pivot.
(ii) Place pivot at ht proper place in list.
For placing pivot at it’s proper place it follows following steps
→ Compare the pivot element one by one from right to left for getting the
element which has value less than pivot element.
→ Interchange the element with element.
→ Now, comparison will start from the interchange element position from
left to right for getting the element which has higher value than pivot.
→ Repeat the same process until pivot is at it’s proper position.
(iii) Create two subsists left & right side of pivot.
(iv) Repeat the same process until all elements of first are at proper position
in list.
Q. Show all the same process until all elements quick with the following.
48 29 8 59 72 88 42 65 95 19 82 68
Soln:- Pass(i)
48 29 8 59 72 88 42 65 95 19 82
68
(ii) interchange
19 29 8 59 72 88 42 65 95 48 82
68
(iii) large element interchange
19 29 8 48 72 88 42 65 95 59 82
68

(iv) interchange
19 29 8 42 72 88 48 65 95 59 82
68
(v) interchange
19 29 8 42 48 88 72 65 95 59 82
68
For sub list 1.
(i) 19 29 8 42 search small element
(ii) 8 29 19 42
Search large element
(iii) 8 19 29 42
Sublist sublist 22
Now, consquaring sublist 11 & sublist 12 with pivot (19) we get,
8, 19, 29, 42
For sublist 2
88 72 65 95 59 82 68
68 72 65 88 59 82 88
68 72 65 88 59 82 95
68 72 65 82 59 88 95
Sub list 21
For sub list 21.
Similarly, perform above process
Merge Sort:-
IN merge sort we take a pair of conjugative array elements then merge them
in sorted array & take adjacent pair of array elements & so on. Until all elements
of array are in single list.
e.g. 5 8 89 30 92 64 4 21 56
pass1 5 8 89 30 92 64 4 21 56
pass2 5 8 30 89 64 92 4 21 56
pass3 5 8 30 89 4 21 64 92 56
4 5 8 21 30 56 64 89 92
Algorithm :- at last:-

Shell sort (Diminishing increment sort)
Shell sort is an improvement an insertion sort. In this case we take on itme
at a particular distance (increments) than we compare items which are far apart
and then sort them. After- words we decrease the increments & repeate this
process again. At last we take the increment one & sort them with insertion sort.
Procedure:-
Let us take an array from arr[0], arr[1]……… arr[N-1] & take the distance 5 (say)
for grouping together the items then in terms will grouped as:
First: arr[0], arr[5], arr[10],……….
Second: arr[1], arr[6], arr[11],………
Third: arr[1], arr[7], arr[12],……..
Fourth: arr[9], arr[8], arr[13]…….
We can see that we have to make the list equal to the increments & it will
cover all the items from arr[0]………arr[N-1].
First sort this list with insertion sort then decrease the increments & repeat this
process again.
At ht end, list is maintained with increment 1 & sort them with insertion sort.
Question:- Show all the passes using shell sort with following list:-
75 35 42 13 87 27 64 57
Soln
:-
Consider increments 5
Pass1: 75 35 42 13 87 27 64 57
Pass 2 27 35 42 13 87 75 64 57
Pass 3 13 35 42 27 57 75 64 87
⇒ 13 27 35 42 57 64 75 87

Explanation:-
Binary Tree sort:-
Create binary search tree.
Find in order traversal of binary tree.
Question:-
19 35 10 12 46 40 6 90 3 8
19
10
6
3
8
12
35
46
40
90
[do step by step]
In order traversal
3 6 8 10 12 19 35 40 46 90
Heap Sort:-
Heaps:-
A heap is a binary tree that satisfied the following propertiers:-
→ shape property
→ Order property.
By the shape property we mean that heap must be a complete binary tree
where as by order property we mean that for every node in the heap the value store
in the heap node is greater than or equal to the value to the value an each of its’
children. A heap that satisfied this property is known as max heap.
However, If the order property is such that for every node. In the heap the
value stored in that node is less than or equal to the value in each of it’s children.
That heap is known as minimum heap.

e.g create the heap tree.
10 5 70 15 12 35 50
10
5
15
12
70
50
35
- Here root no must be greater than children here 5< 15 so, we inter change it’s
position.
10
5
15
12
70
50
35
- Now, in position 2, 5, & 6 the 70 is greater than 35 & 50. so we can’t
interchange.
- On the other hand the.
70
15
5
12
50
10
35
Heap sort:-
The elements of the heap tree are represented by an array . The root is the
larget element of heap tree. As it is maintained in the array. The largest element of
heap tree. As it is maintained in the array. The largest value should be the last
element of array. For heap sorting root is deleted till there is only one element in
the tree.
Steps:-
Replace the4 root with the last node of heap tree.
Keep the last node at proper position.
Repeat step 1 & 2 until there are only one root node in the tree.

72
64
56
32
65
54
46
29 48 arr[8]arr[7]
arr[9]
arr[4]
arr[1]
arr[0]
arr[ 2]
arr[6]
arr[ 5]
72 64 65 56 32 46 54 29 48
0 1 2 3 4 5 6 7 8
arr
Step:-1 Root node is delete and this root node is replaced by last node and
the previous value of root node is placed in proper place in array.
48
64
29
12
65
46
54
56
4 8 64 65 56 3 2 5 4 2 9 72
Here, root node is less than 65. so we interchange the position to make heap tree.
65
64 54
56 32 46 48
29
Again , 48< 54
65
64 54
56 32 46 48
29
Now, which is in Heap tree form.
6 5 6 4 5 4 5 6 3 2 4 6 4 8 29 72
Again , delete root node & put 29 in root node.
2 9 6 4 5 4 5 6 3 2 3 2 4 6 48 6 5 7 2
2 9 6 4 5 4 5 6 3 2 3 2 4 6 48 6 5 7 2
6 5 6 4 5 4 5 6 3 2 4 6 4 8 29 72

Radix sort:-
In radix sort, we sort the item in terms of it’s digits. If we have list of nos.
then there will be 10 parts from 0 to 9 because radix is 10.
Algo
consider the list n digit of nos, then there will be 10 parts from 0 to 9.
in the first pass take the nos in parts on the basis of unit digits.
In the second pass the base will be ten digit.
Repeat similarly for n passes for n digits.
Show all the passes using radix sort.
233 124 209 345 498 567 328 163
Pass 1
Numbers [0] [1] [2] [3] [4] [5] [6] [7] [8] [9]
233 233
124 124
209 209
345 345
498 498
567 567
328 328
163 163
Pass 2
233 163 124 345 567 498 328 209
Number [0] [1] [2] [3] [4] [5] [6] [7] [8] [9]
233 233
163 1 163
124 124
345 345
567 567
498 498
328 328
209 209
209, 124, 328, 233, 345, 163, 567, 498.

Taking hundred position:-
Number 0 1 2 3 4 5 6 7 8 9
209 209
124 124
328 328
233 233
345 345
163 163
567 567
498 498
Hence, required sequence is
124 163 209 233 328 345 498 567

Chapter – 9
Searching
Definition:-
Searching is used to find the location whether element is available or not.
There are various kinds of searching techniques.
sequential search:-
simplest technique for searching on unordered table for particular record is to
scan each entry in sequential manner until the desired record is found.
If search is successful then it will return the location of element otherwise it
will return failene notification.
Consider, sequential search in array.
Arr [0] [1] [2] [3] [4] [5] [6]
10 20 30 40 50
Algorithm:-
Put a unique value at the end of array. Then.
(i) index =0
(ii) scan each element of array one by one.
(iii) (a) If match occurs then return the index value.
(b) otherwise index = index +1
(iv) Repeat the same process until unique value comes in scanning.
(v) Return the failure notification.
Consider sequential search in linked list.
Algorithm:-
Take a pointer of node type and initialize it with sort
Ptr = start
Scan each node of the linked list by traversing the list with the help of ptr.
Ptr = ptr → link;
If match occur then return.
Repeat the same process until null comes in scanning.
Return the failure notification.
Performance analysis:-

Number of key comparison taken to find a particular record
Average case : (n+1)/2
Warpe case: n+1
Best case:- if desired record is present in the 1st
position of search table i.e. only
one comparison is made.
Binary search:-
The sequential search situation will be in worse case i.e. if the element is at
the end of the list. For eliminating this problem one efficient searching technique
called binary search is used in this case.
the entries are stored in sorted array.
The element to be searched is compared with middle element of the array.
a. If it is less than the middle element then we search. It in the left portion
of the array.
b. If it is greater than the middle element then search will be in the right
portion of the array.
The process will be in iteration till the element is searched or middle element
has no left or right portion to search.
Algorithm:-
Binary search (K.N, x)
→ entries in ascending order.
(i) [initialize]
Start ← 0, end ← N
(ii) Perform search
Repeat through step (iv) while low ≤ high
(iii) [obtain index of mid point of interval]
Middle ← [ (start + end)/2]
(iv) Compare
If X<K (middle)
Then end ← middle -1
Else
If X>K(middle);

Then start ← middle +1
Else
Write (“Successful search”)
Return (middle)
(v) [unsuccessful search]
Return (0)
Question:- Tress a Binary search algorithm I/P data:-
75, 151, 203, 275, 318, 489, 524, 591, 647, 727
Search Pop x = 275, 727, 725.
Soln
:- [0] [1] [2] [3] [4] [5] [6] [7] [8] [9]
75 151 203 275 318 489 524 591 647 727
Iteration:1
Start = 0 end = 9 middle =
4
2
90
=
+
Since middle element [4] = 318
Since, the middle element is greater than search element, so we assign,
Iteration:-2
1
2
30
=
+
[0] [1] [2] [3] [4] [5] [6] [7] [8] [9]
75 151 203 275 318 489 524 591 647 727
151 < 275
Now,
Start = middle + 1 = 2 end = 3
Iteration:- 3
[0] [1] [2] [3] [4] [5] [6] [7] [8] [9]
75 151 203 275 318 489 524 591 647 725
203 < 275 end = 3
Iteration :- 4

Middle =
3
2
33
=
+
275 = 275
Middle [3] = 275
For 725
[0] [1] [2] [3] [4] [5] [6] [7] [8] [9]
75 151 203 275 318 469 524 591 647 727
Iteration:-1
4
2
90
=
+
725> 318
Iteration:-2
7
2
95
=
+
[0] [1] [2] [3] [4] [5] [6] [7] [8] [9]
75 151 203 275 318 489 524 591 647 727
725 < 591
Iteration:-3
8
2
98
=
+
[0] [1] [2] [3] [4] [5] [6] [7] [8] [9]
75 151 203 275 318 489 524 591 647 727
725 < 647
Iteration:- 4
9
2
99
=
+
[0] [1] [2] [3] [4] [5] [6] [7] [8] [9]
75 151 203 275 318 489 524 591 647 727
725 < 727
Binary search tree:- (BST)

Algorithm:-
[initialize]
Read (no)
Root node content = 0
Right subtree = NULL
Left sub tree = NULL
while there is data
Do
Begin
Read(no)
compare no. with the content of root.
Repeat
If match then declare duplicates
Else
If no < root, then
Root = left sub tree root
Else
Root = right sub tree root.
Until
Duplicate found or (root = = NULL)
If(root = = NULL) then place it as root.
Application of BST:-
sorting a list
→ construct a BST &
→ Traverse in in order
For conversion prefix, infix & posfix expression.
→ construct a BST &
→ traverse in preorder, post order, inorder
Algorithm to build a binary search tree (BST) from post fix expression

→ Read symbol at a time.
→ If symbol is a operand.
* create one node tree push pointer to stack.
→ if symbol is operator
pop pointer to two trees T! and T2 & form a new tree whose root is children
point to T1 & T2 respectively.
Pointer to this new tree is then pushed to stack.
Ab+ cde + **
a b
+
1.
+
a b
c
d e
+
3.
+
a b
c +
d e
+
2.
+ *
a b
+
ed
c
*
4.
*
a b
+
+
5.
*
c
d e

Inorder:-
Preorder:-
Post order:-
“Hashing”
Sequential search, binary search and all the search trees are totally
dependent on no. of element and may key comparisons are involved. Now, our
need is to search the element in constant time and loss key comparisons should be
involved.
Suppose all the elements are in array of size ‘N’. Let us take all the keys are
unique and in the rage ‘o’ to N-1. Now we are sorting the record in array based on
key, where array index and keys are same. Then we can access the record in
constant time and no key comparisons are involved.
Consider 5 records where keys are :-
9, 4, 6, 7, 2
The keys can be stored in array up
Arr [0] [1] [2] [3] [4] [5] [6] [7] [8] [9]
2 4 6 7 9
Here, we can see the record which has key value can be directly accessed through
array index.
In hashing key is converted into array index and records are kept in array.
In the same way for searching the record, key are converted into array index and
get the records from array.
For storing records:-
Key
↓
Generate array index
Key
↓
Stored the record on that array index.
For accessing record:-
Key
↓
Generate array index
Key
↓
Get the records from the array index

The array which supports hashing for storing records or searching records
is called hash table. Each key is mapped on a particular array index through hash
function. If each key is mapped on a unique hash table then this situation is ideal
but there may be possibility that hash function generating some hash table address
for different keys & this situation is called collision. For missing collision we have
to perform:
Choose a good hash table which perform minimum collision.
Resolving the collision.
Technique for choosing hash function.
Trunction method:-
In this method a part of hey is considered as address, it can be some
rightmost digit or leftmost digit.
Q. apply the truncation method to get the hash index of table size of 100 for
following keys.
82394561, 87139465, 83567271, 85943228
Soln
:
Table size = 100
Then,
Take 2 rightmost digits for getting the hash table address.
Key value address
82394561 61
87139465 65
83567271 71
85943228 28
Example. h(key) = a
h(823994561) = 61
Mid square method:-
In this method we square the key , after getting number we take some
middle of that number as an address, suppose keys are 4 digits and maximum
address = 100.
Key value square address
1456 02119936 19 h(1456) = 19
1892 03579664 79 h(1892) = 79
3. folding method:-
In this method we break the key in to pieces & add them and get the
address. Suppose 10 digit key 1234567890 may be reduced to key 50, 000 storage

location by carrying out the steps.
1 2 3 4 5 6 7 8 9 0
Steps:-
1 2 3 4 5 6 7 8 9 0
3 4 5 6 7
+ 1 2 8 9 0
4 7 4 5 7
3. number > 50, 000?
4. if no; number = address
5. if yes, subtract 50,000 with the no until no<50,000.
H(1234567890) = 47457
Modulus method:-
Modular method is best way for getting address from key take the key, do
they modulus operation & get the remainder as address for hash table in order to
minimize the collision table size should a prime number consider keys.
82394561, 87139465, 83567271, 89943228
& Table size = 97
Then, address are
82394561 % 97 = 45
87139465 % 97= 0
83567271% 97=25
89943228% 97= 64
82394561 = 82 = 3945 + 61= 4088
87139465 = 87 + 1394 + 65=
83567271= 83 + 5672 +71=
89943228 = 89 + 9432 + 28=
Applying modulus method we get,
4088 % 97 = 14.
Hash function for floating point numbers:-
The operations can be performed as
i. Check the fractional part of key.
ii. Multiply the fractional part with the size of the hash value.
iii. Take the integer part of the multiplication result as a hash address of
key.
e.g. Question:- Consider the key as floating point number as
123.4321, 19.469, 2.0289, 8.9956
Hash size = 97
Soln
:-
0.4321 x 97 = 41. 9139

0.463 x 97 = 44.911
0.0289 x 97 = 2.8906
0.9956 x 97 = 96.5732
Here, hash address will be integer part of these numbers
i.e. H(123.4321) = 41
H(19.463) = 4
H(2.0289) = 2
H(8.9956) = 96
Hash function for string:-
iv. In this case we add the ASCII value of each character and then apply
modulus operation on this value.
e.g consider the table size = 97
PUSET
PUSET = P + U + S + E + T
= 112+ 117 + 115 + 101 + 116
= 561
Now,
Applying modulus operation, we get
H(PUSET) = 561% 97
= 76
Hence, puset lies on the 76th
address of hash table.
v. In second method ASCILL value of each character is multiplied by 127 i.e.
maximum value of ASCII character and then perform modulus operation.
e.g.
table size = 997
PUSET =
PUSET = P + U + S + E + T
= (112 + 117+ 115 + 101 + 116) x 127
= 71247
Now, applying modulus operation, we get
H(PUSET) = 71247 % 997
= 460
Q. 1. Apply the folding and modulus method to get the hash index of table size 11
for following keys.
8925, 23197,37565, 48693, 46453
Q. 2. :- Get the hash index in the table size 29 for the following floating point
number.
56.9281, 145.0092, 28.45, 28.45 , 89.3967, 2.877
Q 3. :- Get the hash index in the table size 19 for the following keys.
PUSET, eastern, khoppa, acme, novel
Q 4. :- apply the truncation method to get the hash index of the table size 997 for
following keys.
699934, 674352, 632433, 678433, 678668, 629871,

0
0
0
0
0
0
0
0
1
2
3
4
5
6
15
23
24
13 6
20
653420
Q 5:- apply folding method to get the hash index of table size 79 for the following
keys.
56497,79256,27143,49239,18942,77722
Q 6:- apply the mid- square method to get the index of table size 97 for the
following keys.
1123, 1234 1012, 1034, 1103, 1005
# collision Resolution technique:-
open chaining : separate chaining
closed chaining.
i. Linear probing
ii. Quadratic probing
iii. Double hashing
iv. Rehashing.
Open chaining (separate chaining):-
This method maintains the chain of elements which have same hash
address. We can take hash table as an array of pointers. In this method each
pointer will point to one linked list and the elements which have same hash
address will be maintained in the linked list basically it involves two
operations.
(i) creation of god hash function for getting hash key value in the
hash table.
(ii) Maintain the elements in the linked list which is pointed by the
pointer available in hash table.
e.g.
I/P keys:- 13, 15, 6, 24, 23 & 20
Table size :- 7
Hash function h(x) = x %7
h(13) = 13%7 = 6
h(15) = 15%7 = 1
h(6) = 6%7 = 6
h(24) = 24%7 = 3
h(23) = 23 % 7 = 2
h(20) = 20%7= 6
Closed chaining (open addressing)
(iii) Linear probing:-
This hashing technique finds the hash key value through hash function and
maps the key on the particular position in hash in hash, table. In case if key
has same hash address then it will find the next empty position in the hash
table we take the hash table as circular array. If table size in n then after n-1

0
1
2
3
4
5
6
7
8
9
10
10
46
54
43
18
29
0
1
2
3
4
5
6
7
8
9
10
position it will search form zeroth
position in the array.
e.g. consider table size 11 & elements are
29, 18, 23, 10, 36, 26, 46, 43
Soln
:-
H(29) = 29%11 = 7
H(18) = 18 %11= 7
H(23) = 23%11 = 1
H(10) = 10%11= 10
H(36) = 36%11= 3
H(25) = 25%11= 3
H(46) = 46%11= 2
H(43) = 43%11= 10
Disadvantage of linear probing:-
→ dustering problem.
→ searching is slow.
Quadratic probing:-
In case of collision
Rehash functions : (Hash value + 12
) % size
(Hash value + 22
) % size & so on.
This technique decrease the problem of dustering but can’t search all the
locations, if hash table is prime then it will search at least half of the locations of
hash table.
e. g.
table size = 11
given elements = 29, 18, 43, 10, 46, 54
H(29) = 29% 11= 7
H(18) = 18%11 = 7
(hash value + 12
) %11
(7 + 1 ) % 11 = 8
H(43) = 43%11= 10
H(10) = 10 % 11 = 10
(Hash value + 12
) %11
H(46) = 46%11 = 2
H(54) = 54%11 = 10
(10 + 12
) %11 = 0
(10 + 22
) % 11= 3
Double hashing :-
This technique requires hashing second time in case of collision. Suppose h
is a hash key then in case of collision we will again do the hashing of this hash
key.

55
68
8
0
1
2
3
4
5
6
7
8
9
10
11
12
48
0
1
2
3
4
5
6
7
8
9
10
10
36
25
7
18
43
i.e. Hash( h) = h’
now,
we will search the hash key location as h, h+h’, h+2h’ & h+3h’ & so on.
Consider, the table size = 13 then two hash functions are
H = key % 13
& h’ = 11(key % 11)
At the time of collision, hash address for next probability is (h+h’) % 13
((key % 13) + (11 – (key %11) ))) % 13
Or, h+ 2h’
Or, (key % 13) + 2(1) – key % 11))
Question:- consider elements 8, 55, 48, 68 table size = 13
h = key % 13
h’ = 11 – (key % 11)
now, applying modulus operation,
h(8) = 8%13= 8
h(55) = 55%13= 3
h(48) = 48%13 = 9
h(68) = 68 % 13= 3
hence, collision occurs at table location 3.
so, applying double hashing, we get
(key % 13) + (11- key %11) ) % 13
(3 + 11 – 2) % 13 = 12 % 13 = 12
Advantages:-
→ we do operation faster.
Draw backs:-
Requires two times calculation of hash function, which creates if complex
& searching will be slower than linear & quadratic probing.
Rehashing:-
There are chances of insertion failure when hash table is full, so the
solution for this particular case is to create a new hash table with the double size of
previous hash table. Here, we will use new hash function and we
will insert all the elements of the pervious hash table. So, we will
scan the elements of previous hash table one by open & calculate
the hash key with new hash function & insert them into new hash
table.
Consider the table size 11 & elements are
7, 18, 43, 10, 36, 25………..

25
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
43
18
36
10
Soln
:- Size = 11
Applying linear probing we get
h(7) = 7%11= 7
h( 18) = 18%11= 7
h(43) = 43%11= 10
h(10) = 10%11= 10
h(36) = 36%11= 3
h(25) = 25%11= 3
now, if we want to insert six more element then size will not be sufficient .
In order to fit all the elements or key with in a table we take new table of
size more than double with prime number. Thus total size is 23.
Applying linear probing, we get,
h(7) = 7%23= 7
h(18) = 18%23 = 18
h(43) = 43%23 = 20
h(10) = 10%23 = 10
h(36) = 36%23= 13
h(25) = 25%23 = 2
.. ….
.. ..
.. ..
Disadvantage:-
This method is more expression.

1 2
4
3
1 2
4
3
1 2
4
3
3
6
CHAPTER :- 10
Graph:-
A graph B is a collection of two sets V & E where V is the collection of
vertices . V0, V1……….Vn-1 also called nodes & E is the collection of edges e1, e2,
……….en .where an edges is an arc which connects two nodes. This can be
represented as
G = (V,E)
V(G) = (V0, V1……….Vn-1 ) or set of vertices.
E(G) = (e1, e2, ……….en) or set of edges.
Types:-
undirected
directed.
Undirected:-
A graph which has unordered pair of vertices is called undirected graph.
Suppose there is an edge between V0 & V1 then it can be represented as
(V0, V1) or (V1, V0).
Vo V1
V(G) = { 1, 2, 3, 4}
E(G) = {(1, 2), (1, 4), (1, 3), (2, 3), (2, 4), (3, 4)}
Directed:-
A directed graph or digraph is a graph which has ordered pair of vertices
(V1, V2) where V1 is the tail & V2 is the head of the edge.
Vo V1
(V1, V2)
V(G) = { 1, 2, 3, 4}
E(G) = {(1, 2), (1, 4), (1, 3), (2, 3), (2, 4), (3, 4)}
Weighted graph:-
A graph is said to be weighted if it’s edges have been assigned with some
non negative as weight.
Adjacent nodes:-
A node u is adjacent to another node or is a neighbor of another
node V if there is an edge from u to node V. In undirected graph if(Vo,
V1) is an edge then Vo is adjacent to V1 & V1 is adjacent to Vo. In a
digraph if (Vo, V1) is an edge then Vo is adjacent to V1 & V1 is adjacent
from Vo.

1 2
4
3
A
C B
A
C B
A
C D
B
Path:-
A path from node u0 to node un is a sequence of nodes u0, u1 , u2……… un-1 ,
un such that u0 is adjacent to u1, u1 is adjacent to u2 …………….. un-1. is adjacent
of un
Length of Path:-
Length of a path is the total number of edges included in the path.
Closed path:-
A path is said to be closed if first & last node of the path are same.
1 2
3 4
Simple Path:-
It is a path in which all the nodes are distinct with an exception
that the first & the last nodes of the path can be same.
Cycle:-
Cycle is a simple path in which first & last nodes are the same.
A
Cyclic graph:-
A graph that has no cycles is called Acyclic graph.
Degree:-
IN an undirected graph the number of edges connected to a node is called
the degree of that node or degree of a node is the no. of edges incident on it.

BA
C D E
GF
Indegree:-
The indegree of a node is the number of edges coming to that node.
Here, in degree of nodes:-
A = 0
B = 2
C = 2
D = 6
E = 1
F = 0
G = 1
Outdegree:-
The out degree of a node is the number of edges going outside form the
node.
Out degree of A = 3
B = 1
C = 1
D = 0
E = 2
F = 3
G = 2
Isolated node:-
If a node as no edges connected with any other node then it’s degree will be
zero & is called as isolated node
A C
B D
C is isolated node.
Successor or predessor:-
In digraph if a node Vo is adjacent to node V1 then Vo is the predessor of
V1 & V1 is the successor of Vo,
V1Vo
Connected graph:-
An undirected graph is connected if there is a path from any node of a
graph to any other node.

A
C B
A
C B
A B
C
D
A B
C
D
Here, G1 is connected Graph while G2 is unconnected graph.
Strongly Connected:-
A digraph is strongly connected if there is a directed path from
any node of graph to any other node.
Weakly connected:-
A digraph is called weakly connected for any pair of nodes u & v,
there is a path form u to v r a path from v to u but not both.
Maximum, edges in graph:-
In undirected graph maximum edge = 2
)1( −nn
& in digraph.
Maximum edges = n(n -1)
Where, n is the total no. of nodes in the graph. A
n(n-1)/2
= 3(8 -1)/2
= 3 C B
n(n -1) A
3(3 -1) = 6
C B
Representation of Graph
Two ways:-
→ Sequential representation (adjacency matrix)
→ linked list representation (adjacency list)
Adjancy matrix:-
Adjancy matrix is the matrix which keeps information of adjacent nodes i.e.
keeps the information weather node is adjacent to any other node or not. Suppose
there are four nodes in a graph then row one reprew3ents the node 1, row 2

A B
D
C
A B
C
D
A B
D
C
9
2
3
7
5
6
8
represents the node 2 & so on . similarly, column 1 represents node 1 & column 2
represents node 2 & so on. The entry of the matrix will be.
Arr[i][j] = 1 if there is an edge from node I to node j
= 0 if there is no edge from node I to node j.
A B C D outdegree
A 0 1 0 1 2
B 1 0 1 1 3
C 0 0 0 1 1
D 1 0 1 0 2
indegree 2 1 2 3
again,
For undirected graph:-
A B C D
A 0 1 1 1
B 1 0 1 1
C 1 1 0 1
D 1 1 1 0
It has no in degree & out degree because it has no direction of node.
Representation of weighted graph in matrix form:-
If graph has some weight on it’s edge then,
Arr[i][j] = weight on edge (if there is an edge from I to node j
= 0 (otherwise)
Weighted adjacency matrix
A B C D` out degree
A 0 2 0 8 10
B 3 0 4 7
C 0 0 0 5
D 9 0 6 0
Indegree =
Warshall Algorithm:-
Used for finding path matrix of a graph
Algorithm:-
Initialize
P ← A

A B
D
C
[perform a pass]
Repeat through step 4 fro K = 1, 2, ……….n
process row]
Repeat step 4 for I = 1, 2, ………..n
process columns]
Repeat for j= 1, 2, ……….n
Pij U (Pik ∩Pkj)
[finish]
Return
Q:- From the given graph find out the path matrix by warshal algorithm.
A B C D`
A 0 1 0 1
P0 = B 1 0 1 1
C 1 1 0 1
D 1 1 1 0
Pij ∪ (Pik ∩ Pkj)
For k = 1
Pij ∪ (Pi1 ∩ P1j)
A B C D
A 1 0 1 0
P1 = B 1 1 1 1
C 0 0 0 1
D 1 1 1 1
Similarly taking k = 2 Pij = Pij ∪ (Pik ∩ Pkj)
Pij = Pij ∪ (Pi2 ∩ P2j)
A B C D`
A 1 1 1 1
P2 = B 1 1 1 1
C 0 0 0 1
D 1 1 1 1
Now taking k = 3,
A B C D`
A 1 1 1 1
P3 = B 1 1 1 1
C 0 0 0 1
D 1 1 1 1
Again taking k = 4

A B C D`
A 1 1 1 1
P4 = B 1 1 1 1
C 1 1 1 1
D 1 1 1 1
Here, P0 is the adjency matrix & Pu is the path matrix of the graph.
Q:- Modified warshal’s algorithm:-
Warshall’s algoritham give the path matrix of graph. By modifying this
algorithm, we will find out the shortest path matrix Q. Qij represent the length of
shortest path from Vi to Vj. Here, we consider the matrices q0, q1, q2, ………. qn.
Thus, length of shortest path from Vi to Vj using nodes Vj,V2, ….Vn
Qk[i][j] =
∞ [i] there is no path from Vj to Vj using nodes V1, V2…………….
Vn
Procedures:-
→ In this algorithm length of 1st
path will be Qk-1 [i][j]
→ length of 2nd
path will be Qk-1 [i][j] + Qk-1 [i][j].
Now, select the smaller one from these two path length so value of
Qk [i][j] = Minimum [Qk-1 [i][j], Qk-1 [i][j] + Qk-1 [i][j] ]
Algorithm:-
→→→→ Q ←←←← A
- adjacency matrix with 0 replaced by ∞
→→→→ [Perform a pass]
- Repeat through step 4 for k = 1,2, ………..0.
→→→→ [process rows]
- Repeat step 4 fro j = 1,2, …………n
→→→→ [process column]
- Repeat for j = 1,2, ……….n
Qk [i][j] ← Min [Qk-1 [i][j], Qk-1 [i][j] + Qk-1 [i][j] ]
→→→→ [finish]
- Return
Case :-1
Qk [i][j] = ∞ & Qk-1 [i][j] + Qk-1 [i][j] = ∞
Then, Qk [i][j] =min (∞,∞ ) = ∞
Case :-2
Qk [i][j] = ∞ & Qk-1 [i][j] + Qk-1 [i][j] = b

Then, Qk [i][j] =min (∞,b ) = b
Case :-3
Qk [i][j] = a & Qk-1 [i][j] + Qk-1 [i][j] = ∞
Then, Qk [i][j] =min (a,∞ ) = a
Case :-4
Qk [i][j] = a & Qk-1 [i][j] + Qk-1 [i][j] = b
Then, Qk [i][j] =min (a,b)
Traversal in graph:-
There are two efficient techniques for traversing the graph.
→ 1) depth first search (DFS)
→ 2) Breadth first search (BFS)
Difference between traversal in graph & traversal in tree or us
There is no 1st
node or root node in graph . Hence the traversal can start from
any node
In tree or list when we start traversing from the 1st
node, all the nodes are
traversed which are reachable from the starting node. If we want to
traverse al the reachable nodes we again have to select another starting
node for traversing the remaining nodes.
In tree or list while traversing we never encounter a node more then once but
while traversing graph,. There may be a possibility that we reach a node
more than once.
In tree traversal, there is only one sequence in which nodes are visited but in
graph for the same technique of traversal there can be different sequences
in which node can be visited.
Breadth first search:-
This technique uses queue for traversing all the nodes of the graph. In this
we take any node as a starting node than we take all the nodes adjacent to that
starting node. Similar approach we take for al other adjacent nodes which are
adjacent to the starting node & so on. We maintain the start up of the visited node
in one array so that no node can be traversed again.
Algorithm:-
[initialize]
Mark all vertex unvisited
begin with any node.
Insert it into queue (initially queue empty)
Remove node from queue.
append it to traversal list.
2 Mark it visited.

4. insert al the unvisited or node snot an queue in to the queue.
2. Repeat step 3 to 5 until queue is empty.
3. [Finish]
Return.
1
2
8
3
5
6
7
8
3
3
5
5
5
6
6
6
7
7
7
7
Thus the traversal list is
1→ 2 → 8 → 3 → 4 → 5 → 6 → 7
Depth first Search:-
This technique uses stack for traversing all the nodes of the graph in this
we take one as starting node then go to the path which is from starting node &
visit all the nodes which are in that path. When we reach at the last node then we
traverse another path starting from that node. If there is no path in the graph from
the last node then it returns to the previous node in the path & traverse another &
so on.
Algorithm:-
1
*
2 8 3
2
*
1 4 5
3 1
*
6
*
7
4
*
2
*
8
5 2 8
6
*
3 8
7
*
3
*
8
8
*
4
*
5
*
1
*
6 7

Traversal:-
1→ 2 → 8 → 3 → 4 → 5 → 6 → 7
Question :-
1
4
2 3
6
9
5
87
Breadth first search:-
4
1
2
4
5
6
7
6
8
9
5
5 3
3
7
6
8
7
8
86
6
Thus traversal list is
1→ 2 → 4 → 5 → 3 → 7 → 6 → 8 → 9
Depth first search:-
→1
*
4 5 2
2
*
5 3
3 1 6
→4
*
7
5
*
8
*
6
6
7
*
5 8
8
*
9
*
9
Traversal is
1→ 4 →7 → 5 → 8 → 9 → 6 → 2 → 3

Shortest path algorithm:-
→→→→ used to find the shortest path form one node to another node.
→ A single source vertex & seek shortest path to all other vertices.
→→→→ shortest path is that path in which the sum of weight of included edges is
minimum.
A
B
D
F
E
C
G
H
1
2
3
3
2
2
1
4
3
7
5
2
From A
Shortest path to B = 2
C = 2 + 2 = 4
D = ( 2 + 2) or (1 + 3) = 4
E = 4
F = 1
G = 6
H = 5
A
B
F
D
V
H
G
E
1
2
1
2
2
3
5
Dis Kstra algorithm:-
In this technique each node is labeled with distance (dist) predecessor &
status, distance of node represents, the shortest distance of that node from the
source node & predecessor of a node represents the node which precedes the
given node in shortest path form source. Status of a node can be permanent or
temporary.
→ shaded circle represent permanent nodes which indicates that it has been
included in the shortest path.
→ Temporary nodes can be relabeled if required but once node is made
permanent, it can’t be rebelled.

Procedure:-
→ Initially make source node permanent and make it the current working node
. al other nodes are node temporary.
→ Examine all the temporary neighbors of the current working nodes & after
checking the condition for minimum weight reliable the require node.
→ From all the temporary nodes find out the node which ahs minimum value
of distance, make the node permanent & now this is the current working
node.
→ Repeat step 2 & 3 until destination node is made permanent.
V3
V1
V4
V7
V8
V5 V6
V2
5
2
5
4
7
3
36
3
5
4 2
5
16
Let , V1 = source node
Node dist pred status
V1 0 0 permanent
V2 ∞ 0 temp
V3 ∞ 0 temp
V4 ∞ 0 temp
V5 ∞ 0 temp
V6 ∞ 0 temp
V7 ∞ 0 temp
V8 ∞ 0 temp
Check adjacent node of V3
V4 Dis > V3 dis + distance (V3,V4) 7<2+ 4 relable.
V7 Dis > V3sis + sistance (V3, V7) ∞> 2 + 3 relable
V1 0 0 permanent
V2 ∞ 0 temp
V3 ∞ 0 temp
V4 ∞ 0 temp
V5 ∞ 0 temp
V6] ∞ 0 temp
V7 ∞ 0 temp
V8 ∞ 0 temp

V4 Dis > V37 dis + distance (V7,V4) 7<2+ 4 relable.
V5 Dis > V7 sis + sistance (V7, V5) ∞> 2 + 3 relable V5
V1 0 0 permanent
V2 8 V1 temp
V3 2 V1 permanent
V4 6 V3 permanent
V5 9 V7 temp
V6] ∞ 0 temp
V7 5 V3 permanent
V8 ∞ 0 temp
V5 Dis > V4 dis + distance (V4,V5) 9<6+9 leave.
V1 0 0 permanent
V2 0 V1 permanent
V3 2 V1 permanent
V4 6 V3 permanent
V5 9 V7 temp
V6] ∞ 0 temp
V7 5 V3 permanent
V8 ∞ 0 temp
V6 Dis > V2 dis + distance (V2,V6) ∞>8 + 16 relable.
V1 0 0 permanent
V2 8 V1 permanent
V3 2 V1 permanent
V4 6 V3 permanent
V5 9 V7 permanent
V6] 24 0 temp
V7 55 V3 permanent
V8 ∞ 0 temp

V6 Dis > V5 dis + distance (V5,V8) ∞ >9 + 8
V6 Dis > V5 dis + distance (V6, V5) 24> 9+5
V1 0 0 permanent
V2 8 V1 permanent
V3 2 V1 permanent
V4 6 V3 permanent
V5 9 V7 permanent
V6 14 V5 permanent
V7 5 V3 permanent
V8 17 0 temp
Here, V6 is smallest & make it permanent since. V6 is the destination node
make it permanent & stop.
Now, start from destination node V6 & keep on seeing it’s perdecessors
until we get source node as predecessor.
Predecessor of V6 is V5
“ : “ V5 is V7
“ : “ V7 is V3
“ : “ V3 is V1
V1 →V3 →V7 →V5 →V6
Spanning tree:-
A spanning tree of a connected graph G contains all the nodes and has the
edges which connects all the nodes so that number of edges will be one less than
the no. of nodes.
A
C
B
D
A
C
B
D
A
C
B
D
A
C
B
D
A
C
B
D
Fig a Fig a
Fig b Fig c
Fig d

Here, A B C D is a spanning tree & D is the minimum spacing tree.
Node:- 1 2 3 4 5 6 7 8 9
Father:- 0 0 0 0 1 0 0 0 0
Step :2 = selected is 4 -5 wet = 3.
n1 = 4 root n1 = 4 n2 = 5 root n2 = 1
Roots are different, so edge is inserted.
Father [1] = 4
Node:- 1 2 3 4 5 6 7 8 9
Father:- 4 0 0 0 1 0 0 0 0
Step:- 4 edge selected is 3 – 6 wt = 5
n1 = 3 root – n1 = 3 n2= 6 root – n2= 6
father [6] = 3
roots are different so edge is inserted.
Node:- 1 2 3 4 5 6 7 8 9
Father:- 4 0 0 0 1 3 0 0 0
Step: 5 edge selected is 5-6 wt = 6
n1 = 5 root – n1 = 4 n2= 6 root – n2= 3
Roots are different so edge is inserted in spanning tree. Father of [3] = 4
Node:- 1 2 3 4 5 6 7 8 9
Father:- 4 0 0 0 1 3 0 0 0
Step: 6 edge selected is 3 -5 wt = 7
n1 = 3 root – n1 = 4 n2= 5 root – n2= 4
Roots are same so edge is not inserted i
Node:- 1 2 3 4 5 6 7 8 9
Father:- 4 0 4 0 1 3 0 0 0
n1 = 2 root – n1 = 2 n2= 5 root – n2= 4
Roots are different so edge is inserted Father of [4] = 2
Node:- 1 2 3 4 5 6 7 8 9
Father:- 4 0 4 2 1 3 0 0 0
n1 = 1 root – n1 = 2 n2= 2 root – n2= 32
Roots are same so edge is not inserted
Node:- 1 2 3 4 5 6 7 8 9
Father:- 4 0 4 2 1 3 0 0 0

n1 = 2 root – n1 = 2 n2= 3 root – n2= 2
Node:- 1 2 3 4 5 6 7 8 9
Father:- 4 0 4 2 1 3 0 0 0
n1 = 5 root – n1 = 2 n2= 7 root – n2= 7
Node:- 1 2 3 4 5 6 7 8 9
Father:- 4 0 4 2 1 3 2 0 0
n1 = 5 root – n1 = 2 n2= 8 root – n2= 8
Node:- 1 2 3 4 5 6 7 8 9
Father:- 4 0 4 2 1 3 2 2 0
n1 = 7 root – n1 = 2 n2= 8 root – n2= 2
Node:- 1 2 3 4 5 6 7 8 9
Father:- 4 0 4 2 1 3 2 2 0
Step: 13 edge selected is 5- 9 wt = 15
n1 = 5 root – n1 = 2 n2= 9 root – n2= 9
Node:- 1 2 3 4 5 6 7 8 9
Father:- 4 0 4 2 1 3 2 2 2
0 3
7
4 0 6
9
2
8
Since, minimum spacing tree should contain n-1 edges where n is the no. of nodes
in the graph. This graph contains nine nodes so after inserting 8 edges in the
spanning tree we will not examine other edges & stop the process.
Here, edge included in the spacing tree are (1, 5) (4, 5) (3, 6), (5, 6), (2, 5)
,(5, 7), (5, 8) & (5, 9) & weight of spanning tree = 2 + 3+ 5+ *+ 11+12 + 15
` = 62

Round- Robin algorithm:-
Initialize the spanning forest to contain the node but not edges.
1. the node but not edges.
a. Each vertex is in It’s own set called partial tree & maintained in aqueue
arbitrarily.
2. maintain each edges associated with each node or partial tree in priority
queure ordered by the weight of the edges.
3. Select a partial tree from queue & find the minimum weight edge incident
to the partial tree from priority queue.
4. find the practical tree that is connected by minimum path edge. Remove
two sub tree joining by the edge & combine in to a single new tree & add to
rear of the queue.
b. Combine two priority queue of the node A & B (partial tree ) &
delete the edge connecting them from queue.
5. Repeat the algorithm unit a queue contains a single tree which is called
minimum cost spanning tree.
0
1
6
5
4
0
2
25
10
28
16
12
22
24
14
18
0
1
6
5
4
0
2
25
10
16
12
22
14
Soln:-
Priority queue. Partial tree weight of edges.
{4} 22, 24, 25
{3} 12, 18 22
{5} 10, 25
{6} 14, 18, 24
{0} 10, 28
{2} 12, 16
{1} 14, 16, 28
⇒
{3, 4} 12, 18, 24, 25
{0, 5} 25, 28
{1, 6} 16, 18, 24, 28
{2, 3, 4} 16, 18, 24, 25
{0, 2,3, 4, 5} 16, 18, 24, 28
{0, 1, 2, 3, 4, 5, 6} all cancel
Here, weight of spanning tree = 10+25+22+12+16+14 =99

Greedy algorithm:-
Consider the problem of making changes Assume cosines of value 25φ
(quarter), 10φ & (dine), 5φ (nick) , & 1φ (penny) & suppose we want to return 63
φ in change, almost without thinking , we convert this amount to two quarters, one
dine & 3 pennies. Not only where we able to determine quickly lost of coins & the
correct value but we produce certain list of value with that coin.
The algorithm probably used to select the largest will whose value was not
greater than 63, & add it to the list of subtract it’s value from 63 getting 38[(63-
25) = 38]
We then select the largest coi9n whose value is not greater than 38 and add
it to the list & so one. According to this
63-25 = 38
38-25 = 13
13-10 = 3
3-1 =2
2-1 =1
1-1 = 0
This method of making charge is a greedy algorithm.
At any individual stage a greedy algorithm selects that option which is
locally optimum in same some particular sence. not that the greedy algorithm for
making change produces on over al optimum solution only because of special
properties of the coins. If the coins had value 1, 9 & 11 & we first select an 11
coin & then four 1 coins total of 5 coins.
We have seen several greedy algorithm such as Dijkstrals shortes path
algorithm & kruskal’s minimum cost spanning tree algorithm.
Kruskal’s algorithm is also greedy as it picks from remaining edges the
shortest among these that do not create a cycle.
/*Program of stack using Array */
# define max 5
Int top =-1;
Int stack_arr[Max];
Main()
{
Int choice;
While(1)
{
Printf(“1.psh”n”);
Printf(“2.pop”);
Printf(3. display”);
Printf( 4. quit”);
Printf(“Enter your choice”)

Scanf(“%d”, choice);
Switch(choice)
{
Case 1:- psh();
Break;
Case 2: pop();
Break;
Case 3:- display()l;
Break;
Case 4: exit(4);
Default:
Printf(“wrong choice”);
}
}
}
Void push()
{
Int pushed_item;
If(top== (max -1)
Printf(“stack overflow”);
Else
{
Printf(“Enter that item to be pushed in stack”);
Scanf(“%d”, & pushed_item);
Top =top +1;
Stack_arr[top] = pushed_item;
}
}
Void pop()
{
If (top = =-1)
Printf(“stack uinderflow”);
Else
{
Printf(“popped element is %d”, stack_arr[top];
Top = top -1;
}
}
Void display()
{
Int I;
If(top == -1)
Printf(“Stack empty”);

Else
{
Printf(“Stack element”);
For(i=top; I > = 0; i- -)
Printf(“%d”, stack_arr[i]);
}
}
/*Program of circular queue */
# define Max 5
Int cqueue_arr[Max];
Int front =-1;
Int rear =-1;
Main()
{
Int choice;
While (1)
{
Printf(“1. insert”);
Printf(2.delete”);
Printf(3.dispalay”);
Printf(4.quit”);
Printf(“Enter your choice”);
Scanf(“%d”, & choice);
Switch(choice)
{
Case 1: insert();
Break;
Case 2: del();
Break;
Case 3: display();
Break;
Case 4: exit(1);
Default:
Printf(“Wrong choice”);
}}}
Insert()
{
Int added_item;
If((front = = 0 & & rear = = Max -1)) || (front = = rear +1))
{
Return;

}
If (front = = -1)
{
Front = 0;
Rear = 0;
}
Else
If (rear = = max -1)
Return =0;
}
else
Reae = rear +1;
Printf(“I/P element for insertion in queue”);
Scanf(“%d”, & added_item);
Cqueue_arr[rear] = added item;
Dle()
{
If(front ==-1)
{
Printf(“Queue underflow”);
Return;
}
Printf(“Element deleted from queue is %d”, cqueue_arr[front]);
If (front == rear)
{
Front =-1;
Rear =-1;
}
Else
If(front = max-1)
Font =0;
Else
Front = front +1;
}
Display()
{
Int fornt _pos = front; rear_pos = rear;
If (front == -1)
{
Printf(“queue is empty”);
Return;
}
Printf(“Queue elements”);

If(front_pos<=rear_pos)
While(front_pos<=rear_pos)
{
Printf(%d”, cqueue_arr[front_pos]);
Front_pos + +;
}
Font_pos= 0;
While (front_pos<= rear_pos]);
Front_pos =0;
Whiel(frot_pos<=rear_pos)
{
Printf(%d”, cqueue arr[front_pos]);
Front_pos ++;
}}}
Output:-
1. insert
delete
display.
quit
enter Your choice:-1
input the element for insertion in queue 7
insert. 2. delete. 3. delete. 4. quit
enter your choice 1. input = 8
“ input = 9
+ input = 10
“ input = 11
# Define Max 5
Int deque_arr[max];
Int left =-1;
Int right =-1;
Main()
{
Int choice;
Printf(“1. I/P restricted dequeue”);
Printf(“2. O/P restricted dequeue”);
Scanf(“%D”, & choice);
Switch(choice)
{
Case 1: input_que();

Break;
Default: printf(“Wrong choice”);
}}
Input_que()
{
Input_que()
{
Int choice;
While(1)
{
Printf(1. insert at right”);
Printf(2. delete from left”);
Printf(3. delete from right”);
Printf(4. display”);
Printf(5. quite”);
Switch(choice)
{
Case 1: insert_right();
Break;
Case 2: delete_left();
Break;
Case 3: delete_right();
Break;
Case 4: display_queue();
Break;
Case 5: exit();
Default: printf(“Enter wrong choice”);
}}}
Insert_right()
{
Int added_item;
If((left == 0 && right == max -1) || (lseft == right +1))
{
Return;
}
If(left ==-1)
{
Left =0;
Right =0;
}
Else
(f(right = max-1)

Right =0;
Else
Right = right +1;
Printf(“I/P the element for adding in queue”);
Deque_arr[right] = added_item;
}
Insert_lfet()
{
Int added_item;
If(left== 0&& right = max -1) || (left = right +1));
{
Return;
}
If(left ==-1
{
Left = 0;
Right =0;
}
Else
If(left = 0)
Left = max -1;
Else
Left = left-1;
Printf(“I/P thelement for adding”);
Deque_arr(left) = added_item;
}
Delete_left()
{
If(left ==-1)
{
Printf(“Queue underflow”):
Return;
}
Printf(“Element deleted from queue is %d”, deque_arr[left]);
If(left == right)
{
Left =-1;
Right =-1;
}
Else

If()left== max-1)
Left =0;
Eles
Left = left +1;
}
Delete_right()
{
If(left ==-1)
{
Printf(“Queue under flow”);
Return;
}
Printf(“Element deleted from queue is %d”, deque_arr[right]);
If(left = = right)
{
Left = -1;
Right =-1;
}
Else
If(right = =0)
Right = max -1;
Else
Right = rightg -1;
}
Display_queue()
}
Program of list using array
#defince Max 10
Int arr[max];
Int n;
Main()
{
Int choice, item_pos;
While(1)
{
Print(“1. input list”);
Printf(“2. insert”);
Printf(“3. search”);
Printf(“4. display”);
Printf(“5. quit”);

Scanf(“%d”, & choce);
Switch(choice)
{
Case 1:
Printf(“Enter the no. of element to be inserted”);
Scanf(“%d”, &n)
Input(n);
Break;
Case 2:
Break;
Case 3. insert();
Break;
Case 3: printf(“Enter elements to be serched”);
Scanf(“%d”, & item);
Pos = search(item);
If(p0os>=1)
Printf(“%d found at postion %d”, item, pos);
Else
Printf(“element not found”);
Break;
Case 4:
Del();
Break;
Case 5: display();
Break;
Case 6: exit();
Break;
Default:
Print(“Wrong choice”);
}}}
Input()
{
Int I;
For (I =0; i<n; i++)
{
Printf(“I/P value for element %d”,j+1);
Scanf(%d”, &arr[i]);
}}
Int search *(int item)
{
Int I;
For (I =0; i<n;i++)
{

If (item == arr[i])
Return(i+1);
}
Return(0) /*if element not found */
}
Insert ()
{
Int temp, item, position;
If(n == max)
{
Printf(“list overflow”);
Return;
}
Printf(“enter positionfor insertion”)
Scanf(“%d”, & position);
Printf(“Enter the value”);
Scanf(“%d”, & item);
If position > n+1)
{
Printf(“Enter position less than or equal to n+1);
Return;
}
If position = n+1;
}
Arr[n] = item
n = n+1
return;
}
/*insertion in between */
Temp = n-1;
While (tem> = posiotn-1)
{
Arr[tem +1] = arr[temp];
Temp --;
}
Arr[position -1] = item;
n = n+1;
}
Del()
{
Int tem , position, item;
If(n==0)
{

Printf(“list underflow”);
Return;
}
Printf(“Enter the element to be deleted”);
Scanf(“%d”, & item)
If (item = arr[n-1])
{
N = n-1;
Retun;
}
Position = search (item);
If (position == 0)
{
Printf(“Element not present in array”);
Return;
}
//Deletion in between
Temp = position -1;
While (tem<=n-1)
{
Arr[temp] = arr[temp +1];
Temp ++
{
N = n-1;
}
Display()
{
Int I;
If (n = = 0)
{
Printf(“List is empty”);
Return;
}
For (I =0; i<n; i++)
Printf(“value at position % %d”, i+1, arr[i]);
}
/* program of single linked list */
#include <stdio.h>
#include<malloc.h>
Struct node
Int info;

Struct node * link;
}* start;
Main()
{
{
Int choice n, m, position, I;
Start = NULL;
While(1)
{
Printf(“1. create list n”);
Printf(“2. ad at beginning n”);
Printf(“3. add after n”);
Printf(“4. delete n”);
Printf(“5. display n”);
Printf(“6. count n”);
Printf(“7. reverse n”);
Printf(“8. search n”);
Printf(“ 9. quit n”);
Scanf(%d”, & choice);
Switch (choice)
{
Case 1: printf(“How many nodes you want”);
Scanf( “%d”, & n);
For(i=0; i<n;i++)
{
Printf(“Enter the element :”);
Scanf(“%d”<&m);
Create – list (m);
}
Break
Case 2:
Printf(“Enter the element:-“);
Scanf(“%d”, & m);
Addetheg(m);
Break;
Case 3:
Printf(“Enter the element:-“);
Scanf(“%d”, &m);
Printf(“Enter the position after which this element is );
Scanf(“%d”, & position);
Add after (m, position);
Break;

Case 4:
If(start = = NULL)
{
Printf(“List is empty n”);
Continue;
}
Printf(“Enter the element for deletion “);
Scanf(‘%d”, &m)
Del(m);
Break;
Case 5: display();
Break;
Case 6: count();
Break;
Case 7: rev();
Break;
Case 8:
Printf(“Enter the element to be searched”);
Scanf(“%d”, &m);
Search(m);
Break;
Case 9:
Exit()
Default: printf(“Wrong choice n”);
} /* end of switch */
} /* end of main() * /
Create – list (int data)
{
Struct node *q, * temp;
Temp = malloc (size of (struct node));
Temp –→info = data;
Temp –→link = NULL;
If(start == NULL) /*if list is empty*/
q = start;
While(q → link ! = NULL)
q = q → link;
q → link = temp
}
Return;
} /* end of create – list is */
Addatbeg (int data)
{
Strct node * temp;

Temp = malloc (size of (struct node));
temp→ infor = data;
temp → link = start;
start = temp ;
return;
} /* End of addatbeg() */
Add after (int data, int pos)
{
Struct node * temp *q;
Int I;
Q start;
For (i=0; i<pos-1; i++)
{
Q = q→ link;
If(q == NULL)
{
Printf(“There are less than %d element”, pos);
Return;
}} /* end of for */
Temp = malloc (size of (struct node)
Temp → link = q → link;
Temp →info = data;
Q → link = temp;
Return;
} /*end of add after () */
Del (int data)
{
Struct node * temp, * q;
If(start → info = = data)
{
Temp = start;
Start start → link; /* first element deleted */
Free (temp);
Return;
}
Q = start;
While (q→ link→ link ! NULL)
{
If (q→ link → info = = data)
{
Temp = q → link;
Q →link = temp → link;

Free (temp);
Return;
}
Q = q→ link ;
/*end of while */
If (q→link →info = data) /* last element deleted */
{
Temp = q → link ;
Free (temp);
Q → link = NULL
Return;
}
Printf(“element %d not found n”, data);
Return;
} /* End of del() */
Display()
{
Strcut node * q;
If (start == NULL)
{
Printf(“List is empty n”);
Return;
}
q = start
printf(“List is: n”)
while (q ! = NULL)
{
Printf(“%d”, a →info);
Q = q→ link;
}
Printf(“n”);
Return;
} /*End of display */
Count()
{
Struct node * q = start;
Int cnt = 0;
While (q ! = NULL)
{
Q = q →link;
Cnt ++ ;
}

Printf(“No. of element are %d n”, cnt);
} /* End of count () */
Rev()
{
Struct node * P1, *P2, *P3;
If (start → link = = NULL) / *only one element
*/
Return;
P1 = start;
P2 = P1 → link
P3 = P2 → Link;
P1 → Link = NULL
P2 →link = P1;
While (P3 ! = NULL)
{
P1 = P2;
P2 = P3;
P3 = P3 → link
P2 → link = P1;
}
Start = P2;
} /*End of rev() */
Search(int data)
{
Struct node * ptr = start;
Int pas = 1;
While (ptr ! = NULL)
{
If (ptr → infor == data)
{
Printf(“Item %d found at position %d”, data, post);
Return;
}
Ptr = ptr → link;
Pos + + ;
}
If (ptr = = NULL)
Printf(“Item %d not found in list n”, data);
} /* end of search() */

Data structure and algorithm.(dsa)

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