Sort an array of 0s, 1s and 2s in C

Given an array of elements consisting of 0, 1, 2. Sort the array.

{1, 2, 0, 0, 1, 2, 1}

The solution is very simple, we count the number of 0’s, 1’s and 2’s. Then we put those many 0’s, 1’s and 2’s. Here if you carefully examine, we are not sorting, instead we are counting and directly adding those many elements in the array.

#include<stdio.h>

void sortArray(int arr[], int length)
{

	int count_0 = 0;
	int count_1 = 0;
	int count_2 = 0;

	int itr = 0;

	for(itr = 0; itr < length; itr ++)
	{
		if(arr[itr] == 0)
			count_0 ++;

		if(arr[itr] == 1)
			count_1 ++;

		if(arr[itr] == 2)
			count_2 ++;
	}

	for (itr = 0; itr < count_0; itr++)
	{
		arr[itr] = 0;
	}

	for (itr = count_0 ; itr < (count_0+ count_1) ; itr++)
	{
		arr[itr] = 1;
	}

	for (itr = (count_0+count_1) ; itr < length; itr++)
	{
		arr[itr] = 2;
	}

}

void printArray(int arr[], int length)
{
	int itr = 0;

	for (itr = 0; itr < length; ++itr)
	{
		printf("%d ", arr[itr] );
	}
	printf("\n");
}


int main(int argc, char const *argv[])
{
	int arr[100] = {1, 2, 0, 0, 1, 2, 1};
	int lenght = 7;

	sortArray(arr, lenght);

	printArray(arr, lenght);

	return 0;
}

0 0 1 1 1 2 2