Sort array containing only 0, 1 and 2 as elements

Concepts Used:

Sorting

Difficulty Level:

Easy

Problem Statement (Simplified):

We have to print the array containing 0_s, 1_s and 2_s in non-decreasing order.

Test Case:

Input:
1
10
0 1 0 0 1 0 2 2 0 1 

Output:
0 0 0 0 0 1 1 1 2 2 

Explanation:
We can get this answer after sorting the whole array.

Solving Approach :

1) Here we can sort the array too but that would cost us at least O(N x log(n)) if we use Merge Sort.

2) We know array contains only three elements i.e. 0_s, 1_s and 2_s, we can count their total number of occurrences in the array.

3) We later can print 0_s, 1_s and 2_s number of times they appear in array.

4) For example if 0 appears a times,1 appears b times,and 2 appears c times, then we’ll print 0 for a times first then 1 for b times and print 2 for c times aferwards.

Approach-1 : Using Sorting (Merge Sort)

We can sort array and print it, it takes O(N x log(N))

Let, array A be,

Sorting array A produces our answer that is,

Approach-2 : Using counters

We know, array contains only 0, 1, and 2 as elements, so we find total number of appearances of 0,1, and 2 and print them in order. This take O(N) time complexity.

Let, the array A be,

We maintain hash table H for counting 0, 1, and 2, hash table after counting is,

So our final answer is,

Solutions:

#include <stdio.h>

int main()
{

  int test;
  scanf("%d",&test);

  while(test--){

    int n;
    scanf("%d",&n);

    int arr[n];
    for(int i=0; i<n; i++)
      scanf("%d",&arr[i]);

    int count[3] = {0};
    for(int i=0; i<n; i++)
      count[arr[i]]++;

    for(int i=0; i<=2;i++)
      for(int j=0; j<count[i]; j++)
        printf("%d ",i);

    printf("\n"); 
  }
}

#include <bits/stdc++.h>
using namespace std;

int main()
{

  int test;
  cin>>test;

  while(test--){

    int n;
    cin>>n;

    int arr[n];
    for(int i=0; i<n; i++)
      cin>>arr[i];

    int count[3] = {0};
    for(int i=0; i<n; i++)
      count[arr[i]]++;

    for(int i=0; i<=2;i++)
      for(int j=0; j<count[i]; j++)
        cout<<i<<" ";

    cout<<endl;
  }
}

import java.util.*;
import java.io.*;

public class Main {
  public static void main(String args[]) throws IOException {

    Scanner sc = new Scanner(System.in);
    int test = sc.nextInt();

    while(test--!=0){

      int n = sc.nextInt();

      int arr[] = new int[n];
      for(int i=0; i<n; i++)
        arr[i] = sc.nextInt();

      int count[] = new int[3];
      for(int i=0; i<n; i++)
        count[arr[i]]++;

      for(int i=0; i<=2;i++)
        for(int j=0; j<count[i]; j++)
          System.out.print(i+" ");

      System.out.println(); 
    }

  }
}

for _ in range(int(input())):
	
	n = int(input())
	arr = list(map(int,input().split()))
	count = [0] * 3

	for i in arr:
		count[i] += 1

	for i in range(3):
		for j in range(count[i]):
			print(i, end = " ")

	print()

[forminator_quiz id="1136"]

This article tried to discuss the concept of sorting. Hope this blog helps you understand and solve the problem. To practice more problems on sorting you can check out .