Counting Sort with C++ Example

belongs to collection: Data Structure programs using C and C++ (Sorting Programs)

mahmoud2506

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2022-05-08

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Counting sort is used for small integers it is an algorithm with a complexity of O(n+k) as worst case where 'n' is the number of elements and k is the greatest number among all the elements .

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mahmoud2506

2022-05-08

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Algorithm:

Method Counting_Sort() contains three arguments A contains the elements entered by user, B array in which sorted elements are stored , n is the size of array A.

First of all we will have to initialize the array C with zero then we will store the frequency of elements in another array C i.e. if the value of an input element is x , we increment C[i](to store the occurrence of each number) then we will modify C[x] so that it will contain the last occurrence of the element x this can be done by storing the sum of C[x] and C[x-1] in C[x].
Now we will traverse the array A from last and find its position from C and that element in B at that address and at last we will modify C so that duplicate element will not end up in the same position in B.

Let us understand this with the help of an example:

Here , n=7 and A[]={0,1,5,7,8,6,3}

At first C[]={0,0,0,0,0,0,0,0,0}

Now we will modify C[]
C[]={1,1,0,1,0,1,1,1,1}

Now we will modify C[] so that it contains the last occurrence of any element 
x at C[x].

C[1]=C[0]+c[1] , C[2]=C[1]+C[2]… and so on
C[]={1,2,2,3,3,4,5,6,7}    /*Index of will start from zero*/

  Now we will store the sorted array in B[] by traversing A[] 
  and checking the position of that element from C[]

/*Index of A[] and B[] will start from one*/
So first A[7]=3  So the position of 3 in B[] is 3 and then we will update C[3]=2
Next A[6]=6 so the position of 6 in B[] is 5 and then we will update C[6]=4
And this will keep on going until we reach the first element then we will get 
our sorted array B[].
B[]={0,1,3,5,6,7,8}

Counting sort program using C++

#include<iostream>
using namespace std;
int k=0;

/*Method to sort the array*/
void Counting_Sort(int A[],int B[],int n)    
{
	int C[k];
	for(int i=0;i<k+1;i++)
	{
		/*It will initialize the C with zero*/
		C[i]=0;
	}
	for(int j=1;j<=n;j++)
	{
		/*It will count the occurence of every element x in A 
		and increment it at position x in C*/
		C[A[j]]++;			    
	}
	for(int i=1;i<=k;i++)
	{
		/*It will store the last 
		occurence of the element i */
		C[i]+=C[i-1];            
	}
	for(int j=n;j>=1;j--)
	{
		/*It will place the elements at their 
		respective index*/
		B[C[A[j]]]=A[j];          
		/*It will help if an element occurs 
		more than one time*/
		C[A[j]]=C[A[j]]-1;		  
	}
}
int main()
{
	int n;
	cout<<"Enter the size of the array :";
	cin>>n;
	
	/*A stores the elements input by user */
	/*B stores the sorted sequence of elements*/  	
	int A[n],B[n]; 
	
	for(int i=1;i<=n;i++)        
	{
		cin>>A[i];
		if(A[i]>k)
		{
			/*It will modify k if an element 
			occurs whose value is greater than k*/
			k=A[i];              
		}
	}
	Counting_Sort(A,B,n);        
	/*It will print the sorted sequence on the 
	console*/ 
	for(int i=1;i<=n;i++)       
	{
		cout<<B[i]<<" ";
	}
	
	cout<<endl;
	return 0;
}

Output

First Run:
Enter the size of the array :10
12 345 89 100 23 0 18 44 111 1
0 1 12 18 23 44 89 100 111 345


Second Run:
Enter the size of the array :5
999 87 12 90 567
12 87 99 567 999

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