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  • Comb Sort Algorithm Assignment Help

Comb Sort Algorithm Assignment Help


Are you stressed about the allocated assignment on the subject of Comb Sort Algorithm Assignment Help? Are you thinking about will you be able to deliver the assignment on time? Algorithm Assignment  is a huge subject and it exclusively requires time and knowledge to prepare an assignment. 


ABC Assignment Help is a reliable Comb Sort Algorithm Assignment Help service provider helping a number of scholars who are facing difficulties in preparing an assignment on Algorithm.


What is Comb Sort Algorithm?


1) Comb sort improves on bubble sort by using gap of size more than 1.

2) The gap starts with a large value and shrinks by a factor of 1.3 in every iteration until it reaches the value 1.


EXAMPLE: To start with we have the following array: 4 6 2 3 7 1 8 0 9 5


Iteration 1: Gap : 8

Input
4
6
2
3
7
1
8
0
9
5

Index 0 with index8- we find the elements in order, hence no swap needed.



4
6
2
3
7
1
8
0
9
5

Index 1 with index 9 – we find the elements not in order, hence swap needed. 


Iteration 2:- Gap : 6

Input
4
5
2
3
7
1
8
0
9
6

Index 0 with index 6 – we find the elements in order, hence no swap needed.



4
5
2
3
7
1
8
0
9
6

Index 1 with index 7 – we find the elements not in order, hence swap needed. 



4
0
2
3
7
1
8
0
9
6

Index 2 with index 8 – we find the elements in order, hence no swap needed.



4
0
2
3
7
1
8
5
9
6

Index 3 with index 9 – we find the elements in order, hence no swap needed.


RESULT
4
0
2
3
7
1
8
5
9
6


Iteration 3:- Gap : 4

Input
4
0
2
3
7
1
8
5
9
6

Index 0 with index 4 – we find the elements in order, hence no swap needed.



4
0
2
3
7
1
8
5
9
6

Index 1 with index 5 – we find the elements in order, hence no swap needed.



4
0
2
3
7
1
8
5
9
6

Index 2 with index 6 – we find the elements in order, hence no swap needed.



4
0
2
3
7
1
8
5
9
6

Index 3 with index 7 – we find the elements in order, hence no swap needed.



4
0
2
3
7
1
8
5
9
6

Index 4 with index 8 – we find the elements in order, hence no swap needed.



4
0
2
3
7
1
8
5
9
6

Index 5 with index 9 – we find the elements in order, hence no swap needed.


RESULT
4
0
2
3
7
1
8
5
9
6


Iteration 4:- Gap : 3

Input
4
0
2
3
7
1
8
5
9
6

Index 0 with index 3- we find the elements not in order, hence swap needed. 



3
0
2
4
7
1
8
5
9
6

Index 1 with index 4 – we find the elements in order, hence no swap needed.



3
0
2
4
7
1
8
5
9
6

Index 2 with index 5 – we find the elements not in order, hence swap needed. 



3
0
1
4
7
2
8
5
9
6

Index 3 with index 6 – we find the elements in order, hence no swap needed.



3
0
1
4
7
2
8
5
9
6

Index 4 with index 7- we find the elements not in order, hence swap needed. 



3
0
1
4
5
2
8
7
9
6

Index 5 with index 8 – we find the elements in order, hence no swap needed.



3
0
1
4
5
2
8
7
9
6

Index 6 with index 9 – we find the elements not in order, hence swap needed.

RESULT
3
0
1
4
5
2
6
7
9
8



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