Home » C++ programming language

# Counting Sort with C++ Example

Submitted by Shubham Singh Rajawat, on June 13, 2017

Learn: **Counting Sort in Data Structure using C++ Example**: Counting sort is used for small integers it is an algorithm with a complexity of O(n+k) as worst case.

**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 .

**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

TOP Interview Coding Problems/Challenges

- Run-length encoding (find/print frequency of letters in a string)
- Sort an array of 0's, 1's and 2's in linear time complexity
- Checking Anagrams (check whether two string is anagrams or not)
- Relative sorting algorithm
- Finding subarray with given sum
- Find the level in a binary tree with given sum K
- Check whether a Binary Tree is BST (Binary Search Tree) or not
- 1[0]1 Pattern Count
- Capitalize first and last letter of each word in a line
- Print vertical sum of a binary tree
- Print Boundary Sum of a Binary Tree
- Reverse a single linked list
- Greedy Strategy to solve major algorithm problems
- Job sequencing problem
- Root to leaf Path Sum
- Exit Point in a Matrix
- Find length of loop in a linked list
- Toppers of Class
- Print All Nodes that don't have Sibling
- Transform to Sum Tree
- Shortest Source to Destination Path

Comments and Discussions

**Ad:**
Are you a blogger? Join our Blogging forum.