# Given an array which consists of only 0, 1 and 2 sort the array without using any sorting algorithm in C++

## Sort the array with elements 0, 1 and 2 in C++

Here, in this page we will discuss the program to sort the array with elements 0, 1 and 2 in C++ . We use the concept of counting the frequency of 0, 1 and 2 . We are giving with the size of the array along with array elements .We have to print sorted array.

## Algorithm :

• Take the size of the array from the user and store it in variable say n.
• Now, declare a vector of size n and take the n elements of the vector say arr from the user.
• Declare three variables say count_0, count_1 and count_2 and initialize them with 0. (These variable will hold the count of the frequency of 0, 1 and 2).
• Now, iterate over the array and increase the value of variable count_0 if arr[i] is 0 , count_1 if arr[i] is 1 and count_2 if arr[i] is 2 by value 1.
• Now, clear the vector.
• Declare variable say i and initialized it with 0.
• Run a while loop till count_0 is not zero and inside while loop set arr[i++] =0.
• Again Run a while loop till count_1 is not zero and inside while loop set arr[i++] =1.
• At last again run while loop till count_2 is not zero and inside while loop set arr[i++] =2.
• Finally the array get sorted without using any sorting technique and print the array.

## Code in C++

#include<bits/stdc++.h>
using namespace std;
int main ()
{
int n;
cin >> n;
int arr[n];
for (int i = 0; i < n; i++) cin >> arr[i];
int count_0 = 0, count_1 = 0, count_2 = 0;
for (int i = 0; i < n; i++)
{
if (arr[i] == 0)
count_0++;
else if (arr[i] == 1)
count_1++;
else
count_2++;
}
int i = 0;
while(count_0--)
arr[i++] = 0;
while (count_1--)
arr[i++] = 1;
while (count_2--)
arr[i++] = 2;
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
return 0;
}

Output :
6

1 0 0 1 0 0

0 0 0 0 1 1

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