05 DS and Algorithm Session 07

8/2/2019 05 DS and Algorithm Session 07

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Ver. 1.0 Session 7

Data Structures and Algorithms

Objectives

In this session, you will learn to:

Identify the features of linked lists

Implement a singly-linked list


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Suppose you have to write an algorithm to generate andstore all prime numbers between 1 and 10,00,000 anddisplay them.

How will you solve this problem?

Linked List


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Consider the following algorithm, which uses an array tosolve this problem:

1. Set I = 0

2. Repeat step 3 varying N from 2 to 1000000

3. If N is a prime numbera. Set A[I] = N // If N is prime store it in an array

b. I = I + 1

4. Repeat step 5 varying J from 0 to I-1

5. Display A[J] // Display the prime numbers// stored in the array

Linked List (Contd.)


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What is the problem in this algorithm?

The number of prime numbers between 1 and 10,00,000 is notknown. Since you are using an array to store the primenumbers, you need to declare an array of arbitrarily large sizeto store the prime numbers.

Disadvantages of this approach, suppose you declare an arrayof size N:

If the number of prime numbers between 1 and 10,00,000is more than N then all the prime numbers cannot bestored.

If the number of prime numbers is much less than N, a lotof memory space is wasted.



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Thus, you cannot use an array to store a set of elements ifyou do not know the total number of elements in advance.

How do you solve this problem?

By having some way in which you can allocate memory as and

when it is required.



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When you declare an array, a contiguous block of memoryis allocated.

Let us suppose you declare an array of size 10 to store first10 prime numbers.

0

1

2

3

4

5

6

7

8

9

1000

1002

1004

1018

1008

1010

1012

1014

1016

1006One contiguous block ofmemory allocated forthe array to store 10

elements.

Memory representation

If you know the address of the first element in the array youcan calculate the address of any other elements as shown:

Address of the first element + (size of the element index ofthe element)

Dynamic Memory Allocation


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When memory is allocated dynamically, a block of memoryis assigned arbitrarily from any location in the memory.

Therefore, unlike arrays, these blocks may not becontiguous and may be spread randomly in the memory.

0

1

2

3

4

5

6

7

8

9

1000

1002

1004

1018

1008

1010

1012

1014

1016

1006One contiguous block ofmemory allocated forthe array to store 10

elements.


Dynamic Memory Allocation (Contd.)




Let us see how this happens by allocating memorydynamically for the prime numbers.

Allocate memory for 2 Memory allocated for 221002






Let us see how this happens.


3

1002

1036







Allocate memory for5 Memory allocated for 52

3

51008

1002

1036








3

5

71020

1008

1002

1036








3

5

7

111030

1020

1008

1002

1036






Now, if you know the address of the first element, youcannot calculate the address of the rest of the elements.

This is because, all the elements are spread at randomlocations in the memory.

2

3

5

7

111030

1020

1008

1002

1036


To access any element, you need to know its address.





Therefore, it would be good if every allocated block ofmemory contains the address of the next block in sequence.

This gives the list a linked structure where each block islinked to the next block in sequence.

2

3

5

7

11

1036

1008

1020

1030

1030

1020

1008

1036

1002






You can declare a variable, START, that stores the addressof the first block.

You can now begin at START and move through the list byfollowing the links.

START

1030

1020

1008

1036

1002 2

3

5

7

11

1036

1008

1020

1030


An example of a data structure that implements this conceptis a linked list.





Linked list:

Is a dynamic data structure.

Allows memory to be allocated as and when it is required.

Consists of a chain of elements, in which each element isreferred to as a node.

Defining Linked Lists




Defining Linked Lists (Contd.)

Data Link

Node

A node is the basic building block of a linked list.

A node consists of two parts:

Data: Refers to the information held by the node

Link: Holds the address of the next node in the list




10Data 3352 10Data 5689 10Data 1012 10Data

2403 3352 5689 1012

All the nodes in a linked list are present at arbitrary memorylocations.

Therefore, every node in a linked list has link field thatstores the address of the next node in sequence.

The last node in a linked list does not point to any othernode. Therefore, it points to NULL.

NULL





To keep track of the first node, declare a variable/pointer,START, which always points to the first node.

When the list is empty, START contains null.

START

10Data 3352 10Data 5689 10Data 1012 10Data

2403 3352 5689 1012

NULL





Let us solve the given problem of storing prime numbersusing a linked list.

1. Repeat step 2 varying N from 2 to 1000000.

2. If N is a prime number, insert it in the linked list:

a. Allocate memory for a new node.

b. Store the prime number in the new node.

c. Attach the new node in the linked list.

3. Display the prime numbers stored in the linked list.

Implementing a Singly-Linked List




Inserting a Node in a Singly-Linked List

START

2 3 5 7

Consider the following linked list that stores primenumbers.

When a new prime number is generated, it should beinserted at the end of the list.

Initially, when the list is empty, START contains NULL.


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Consider the given algorithm toinsert a node in a linked list.

1. Allocate memory for the new node.

2. Assign value to the data field of thenew node.

3. If START is NULL, then:a. Make START point to the new

node.b. Go to step 6.

4. Locate the last node in the list, andmark it as currentNode. To locate thelast node in the list, execute thefollowing steps:a. Mark the first node as

currentNode.b. Repeat step c until the

successor of currentNodebecomes NULL.

c. Make currentNode point tothe next node in sequence.

5. Make the next field of currentNodepoint to the new node.

6. Make the next field of the new nodepoint to NULL.

Inserting a Node in a Singly-Linked List (Contd.)


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Consider that the list is initiallyempty.START = NULL

Insert a prime number 2.













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START = NULL














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START = NULL

102













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START = NULL











102

START



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102

START













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102

START

Insertion complete













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102

START













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102

START













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102

START

103













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102

START

103













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102

START

103












S d l i h


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102

START

103

currentNode












D S d Al i h


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102

START

103

currentNode












D t St t d Al ith


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START

103

currentNode












D t St t d Al ith


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Insertion complete

102

START

103

currentNode












D t St t d Al ith


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102

START

103

currentNode












D t St t d Al ith


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102

START

103












Data St t e a d Al o ith


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102

START

103 105














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102

START

103 105














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102

START

103 105














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102

START

103 105

currentNode














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102

START

103 105

currentNode














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102

START

103 105

currentNode currentNode














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102

START

103 105

currentNode














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Insertion complete

102

START

103 105

currentNode














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What is the problem with thisalgorithm?

Consider a list of 10000 numbers.

To insert a number at the end ofthe list, you first need to locate the

last node in the list.To reach the last node, you haveto start from the first node andvisit all the preceding nodesbefore you reach the last node.

This approach is very timeconsuming for lengthy lists.

Therefore, a better approach is todeclare a variable, LAST, whichwill store the address of the lastnode in the list.

Hence, you need to maintain two

variables, START and LAST, tokeep track of the first and lastnodes in the list respectively.

If the list is empty, START andLAST point to NULL.














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Refer to the given algorithm toinsert a node at the end of thelinked list.



3. If START is NULL, then (If the list isempty):a. Make START point to the new

node.b. Make LAST point to the new

node.c. Go to step 6.

4. Make the next field of LAST point tothe new node.

5. Mark the new node as LAST.

6. Make the next field of the new node

point to NULL.




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Consider that the list is initiallyempty.









point to NULL.

START = NULL

LAST = NULL





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point to NULL.

START = NULL

LAST = NULL





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102

START = NULL

LAST = NULL









point to NULL.




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102

START = NULL

LAST = NULL









point to NULL.




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102

START = NULL

LAST = NULL

START









point to NULL.




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102

START LAST

LAST = NULL









point to NULL.




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102

START LAST









point to NULL.




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Insertion complete

102

START LAST









point to NULL.




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102

START LAST









point to NULL.




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102

START LAST










point to NULL.




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ata St uctu es a d go t s

102

START

103

LAST









point to NULL.




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g

102

START

103

LAST









point to NULL.




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g

102

START

103

LAST









point to NULL.




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g

102

START

103

LASTLAST









point to NULL.




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g

Insertion complete

102

START

103

LAST









point to NULL.




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g


102

START

103

LAST









point to NULL.




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g

102

START

103

LAST










point to NULL.




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g

102

START

103 105

LAST









point to NULL.




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102

START

103 105

LAST









point to NULL.




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START

103 105

LAST









point to NULL.




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102

START

103 105

LASTLAST









point to NULL.




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Insertion complete

102

START

103 105

LAST









point to NULL.




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Now consider another problem.Suppose you have to store the marks of 20 students in theascending order.

How will you solve this problem?




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Consider the following algorithm, which uses an array tosolve this problem.

1. Set N = 0 // Number of marks entered

2. Repeat until N = 20

a. Accept marks.

b. Locate position I where the marks must be inserted.

c. For J = N 1 down to I

Move A[J] to A[J+1] // Move elements to make

// place for the new element

d. Set A[I] = marks

e. N = N + 1




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1. Set N = 0

2. Repeat until N = 20a. Accept marks.

b. Locate position I wherethe marks must beinserted

c. For J = N-1 down to I

Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1

Let us perform some insertoperations in the array by placing theelements at their appropriatepositions to get a sorted list.

0 1 2 3 4 19




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1. Set N = 0

2. Repeat until N = 20a. Accept marks



Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1

10

0 1 2 3 4 19

Let us perform some insert operationsin the array by placing the elements attheir appropriate positions to get asorted list.

N = 0




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1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1

10

0 1 2 3 4 19

N = 0




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0 1 2 3 4 19

N = 0

marks = 10

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1




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0 1 2 3 4 19

N = 0

marks = 10

I

I = 0

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1




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0 1 2 3 4 19

N = 0

marks = 10

I = 0

I

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1




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0 1 2 3 4 19

N = 0

marks = 10

I = 0

10

I

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1




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0 1 2 3 4 19

N = 0

I = 0

10

I

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1




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0 1 2 3 4 19

N = 1

I = 0

10

I

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1




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0 1 2 3 4 19

N = 1

10

I

I = 0

marks = 20

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1




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0 1 2 3 4 19

N = 1

marks = 20

I = 1

10

II

I = 0

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1




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0 1 2 3 4 19

N = 1

marks = 20

I = 1

10

I

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1




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0 1 2 3 4 19

N = 1

marks = 20

I = 1

10 20

I

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1




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0 1 2 3 4 19

N = 1

I = 1

10 20

I

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1




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0 1 2 3 4 19

N = 2

I = 1

10 20

I

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1




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0 1 2 3 4 19

N = 2

marks = 17

10 20

I

I = 1

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1




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0 1 2 3 4 19

N = 2

marks = 17

I = 1

10 20

I

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1




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N = 2

marks = 17

I = 1

10 20

J = 1

I

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1




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N = 2

marks = 17

I = 1

10 20

J = 1

I

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1




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0 1 2 3 4 19

N = 2

marks = 17

I = 1

10 20

J = 1

20

I

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1




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0 1 2 3 4 19

N = 2

marks = 17

I = 1

10

J = 1

2017

I

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1




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0 1 2 3 4 19

N = 3

I = 1

10 2017

I

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1



i d i i l i k d i ( d )


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0 1 2 3 4 19

N = 3

marks = 15

10 2017

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1



I i N d i Si l Li k d Li (C d )


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0 1 2 3 4 19

N = 3

marks = 15

I = 1

10 2017

I

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1



I ti N d i Si l Li k d Li t (C td )


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0 1 2 3 4 19

N = 3

marks = 15

I = 1

10

J = 2

2017

I

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1





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0 1 2 3 4 19

N = 3

marks = 15

I = 1

10

J = 2

2017

I

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1





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0 1 2 3 4 19

N = 3

marks = 15

I = 1

10

J = 2

2017 20

I

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1





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0 1 2 3 4 19

N = 3

marks = 15

I = 1

10

J = 1

17 20

I

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1



Inse ting a Node in a Singl Linked List (Contd )


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0 1 2 3 4 19

N = 3

marks = 15

I = 1

10

J = 1

17 20

I

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1



Inserting a Node in a Singly Linked List (Contd )


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0 1 2 3 4 19

N = 3

marks = 15

I = 1

10

J = 1

17 2017

I

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1





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0 1 2 3 4 19

N = 3

marks = 15

I = 1

10

J = 1

201715

I

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1





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0 1 2 3 4 19

N = 4

I = 1

10

J = 1

201715

I

1. Set N = 0




Move A[J] to A[J+1]

d. Set A[I] = marks

e. N = N + 1





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What is the problem in this algorithm?To insert an element at any position other than the end of thelist, there is an additional overhead of shifting all succeedingelements one position forward.

Similarly, to delete an element at any position other than the

end of the list, you need to shift the succeeding elements oneposition backwards.

When the list is very large, this would be very time consuming.

From the preceding example we conclude that insertion anddeletion at any position other than the end of an array iscomplex and inefficient.

How can you overcome this limitation?

By using a data structure that does not require shifting of data

elements after every insert / delete operation.A linked list offers this flexibility.



Inserting a Node in a Singly-Linked List (Contd )


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Consider a linked list that stores the marks of students in anascending order.

Insert marks (17).

17 should be inserted after 15.

START

20

10

25

15


1002

1020

2496

1011

15 1002 20 102010 2496 25 NULL





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Allocate memory for 17.

10

START

10 2496 1015 1002 1020 1020 1025

20

10

25

15


17Memory allocated for 17

1002

1020

2496

1011

1008

NULL

17





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In the given list, node 15 contains the address of node 20.To add node 17 after node 15, you need to update theaddress field of node 15 so that it now contains the addressof node 17.

20

10

25

15


17

17

10

START

10 2496 1015 1002 1020 1020 1025 NULL

1002

1020

2496

1011

1008Memory allocated for 17





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Update the address field of node 15 to store the address ofnode 17.

20

10

25

15


17

1011

1020

2496

1008

1002

17

10

START

10 2496 1015 1002 1020 1020 1025 NULL17





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Update the address field of node 15 to store the address ofnode 17.

20

10

25

15


17

1011

1020

2496

1008

1002

10

START

10 2496 1015 1002 1020 1020 1025 NULL17

Updating the address field

1008

Inserting a Node in a Singly Linked List (Contd.)




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20

10

25

15


17

1011

1020

2496

1008

1002

10

START

10 2496 15 1020 1020 1025 NULL17

1008

Store the address of node 20 in the address field of node 17to make a complete list.

Address updated

1002

Node Inserted





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Now, let us solve the given problem using a linked list.Suppose the linked list currently contains the followingelements.

The linked list needs to be created in the ascending order of

values.Therefore, the position of a new node in the list will bedetermined by the value contained in the new node.

START

10 15 17 20





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Write an algorithm to locate the position of the new node tobe inserted in a linked list, where the nodes need to bestored in the increasing order of the values contained inthem.





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Algorithm for locating the positionof a new node in the linked list.

After executing this algorithm thevariables/pointers previous andcurrent will be placed on the

nodes between which the newnode is to be inserted.

1. Make current point to the first node.

2. Make previous point to NULL.

3. Repeat step 4 and step 5 untilcurrent.info becomes greater than thenewnode.info or current becomesequal to NULL.

4. Make previous point to current.

5. Make current point to the next node in

sequence.





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START

15 17 20

Insert 16 in the given list.







sequence.

10




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START

10 15 17 20

Insert 16 in the given list.







sequence.

current




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START

10 15 17 20

previous = NULL







sequence.

current




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START

10 15 17 20

previous = NULL

g g y ( )






sequence.

current




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currentprevious

g g y ( )






sequence.

previous = NULL




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current

g g y ( )

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current






sequence.

previous




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g g y ( )






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g g y ( )






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currentprevious




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Node located






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previous




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Once the position of the new node has been determined,the new node can be inserted in the linked list.

A new node can be inserted at any of the following positionsin the list:

Beginning of the list

End of the listBetween two nodes in the list




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Write an algorithm to insert a node in the beginning of alinked list.




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1. Allocate memory for the newnode.

2. Assign value to the data field ofthe new node.

3. Make the next field of the newnode point to the first node in thelist.

4. Make START, point to the new

node.

Algorithm to insert a node in thebeginning of a linked list

10




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newnode





node.

Insert 7




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newnode

7





node.

17

Insert 7




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newnode

7

newnode. next = START





node.




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node.

START = newnode

newnode. next = START

Insertion complete

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newnode

7




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Write an algorithm to insert a node between two nodes in alinked list.




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Algorithm to insert a node betweentwo nodes in a linked list.

1. Identify the nodes between which thenew node is to be inserted. Mark them

as previous and current. To locateprevious and current, execute thefollowing steps:a. Make current point to the first

node.b. Make previous point to NULL.c. Repeat step d and step e until

current.info becomes greaterthan newnode.info or current

becomes equal to NULL.d. Make previous point to current.e. Make current point to the next

node in sequence.


3. Assign value to the data field of the newnode.

4. Make the next field of the new nodepoint to current.

5. Make the next field of previous point tothe new node.

Insert 16

START

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Insert 16






node in sequence.








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current

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node in sequence.






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current

previous = NULL

Insert 16






node in sequence.








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current

previous = NULL

previous

Insert 16






node in sequence.








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current current

Insert 16






node in sequence.





previous




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current

Insert 16


as prev

Documents

05 DS and Algorithm Session 07