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Sort an array of 0's, 1's and 2's in linear time complexity

Here, we are going to learn how to sort an array of 0's, 1's and 2's linear time complexity using C++ programming code?
Submitted by Radib Kar, on November 17, 2018

Description

Solution to the coding problem of sorting an array of 0's, 1's and 2's in linear time complexity.

Problem statement

Given an array consisting only 0's, 1's and 2's. Give an algorithm for sorting the array in O(n) time complexity ( in the sorted array, 0's will be at starting ,then the 1's & then the 2's).

Solution

Algorithm

1. Set three variables low=0,mid=0, high=n-1 where n=length of input array
2. While (mid<=high), consider three cases on basis of the value of array [mid].
3. If array[mid] is 0
   Swap (array [low], array [mid]) since 0's should be at starting
   Increment low
   Increment mid
   Break statement
4. If array[mid] is 1
   Keep as it is since 1's should be at middle after being sorted
   Increment mid
   Break statement
5. If array[mid] is 2
   Swap (array [mid], array [high])
   Decrement high
   Break statement
6. Resulting array is sorted.

C++ program to sort an array of 0's, 1's and 2's in linear time complexity

#include<stdio.h>
#include<stdlib.h>

//function to swap by reference
void swap(int*a,int*b){
	int temp;
	temp=*b;
	*b=*a;
	*a=temp;
	return;
}
int* asort(int *a,int n){
	int low=0,mid=0,high=n-1;   //variables are set
	
	while(mid<=high){
		switch(a[mid]){
			case 0:     //if a[mid]==0
				//swap a[low] & a[mid], swapping by reference
				swap(&a[low],&a[mid]);  
				low++;      //increment low
				mid++;      //increment mid
				break;
			case 1:     //if a[mid]==1
				mid++;      //increment mid
				break;
			case 2:     //if a[mid]==2
				//swap a[mid] & a[high], swapping by reference
				swap(&a[mid],&a[high]);  
				high--;     //decrement high
				break;
		}
	}
	//returning adress of array(sorted)
	return a;   
}

int main() {
	int n;

	printf("enter no of array elements\n");
	//input array length
	scanf("%d",&n);

	int* a=(int*)malloc(sizeof(int)*n);

	printf("enter array elements\n");
	//input array elements
	for(int j=0;j<n;j++)      
		scanf("%d",&a[j]);
	//array is modified
	a=asort(a,n);   
	printf("after being sorted..............\n");
	//printing the sorted array
	for(int j=0;j<n-1;j++)   
		printf("%d ",a[j]);
	
	printf("%d\n",a[n-1]);

	return 0;
}

Output

First run:
enter no of array elements 
10 
enter array elements 
0
2
2
1
1
1
2
1
2
1
after being sorted.............. 
0 1 1 1 1 1 2 2 2 2 


Second run:
enter no of array elements 
10 
enter array elements 
2
2
2
2
0
0
0
0
1
1
after being sorted.............. 
0 0 0 0 1 1 2 2 2 2