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Sort an array of 0s, 1s and 2s | Dutch National Flag problem

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Given an array A[] consisting of only 0s, 1s, and 2s. The task is to write a function that sorts the given array. The functions should put all 0s first, then all 1s and all 2s in last.

This problem is also the same as the famous “Dutch National Flag problem”. The problem was proposed by Edsger Dijkstra. The problem is as follows:

Given N balls of colour red, white or blue arranged in a line in random order. You have to arrange all the balls such that the balls with the same colours are adjacent with the order of the balls, with the order of the colours being red, white and blue (i.e., all red coloured balls come first then the white coloured balls and then the blue coloured balls). 

Examples:

Input: {0, 1, 2, 0, 1, 2}
Output: {0, 0, 1, 1, 2, 2}

Input: {0, 1, 1, 0, 1, 2, 1, 2, 0, 0, 0, 1}
Output: {0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2}

Sort an array of 0s, 1s, and 2s using the Pointer Approach: 

This approach is based on the following idea:

  • The problem is similar to “Segregate 0s and 1s in an array”.
  • The problem was posed with three colors, here `0′, `1′ and `2′. The array is divided into four sections: 
    • arr[1] to arr[low – 1]
    • arr[low] to arr[mid – 1]
    • arr[mid] to arr[high – 1]
    • arr[high] to arr[n]
  • If the ith element is 0 then swap the element to the low range.
  • Similarly, if the element is 1 then keep it as it is.
  • If the element is 2 then swap it with an element in high range.

Illustration:

arr[] = {0, 1, 2, 0, 1, 2}

lo = 0, mid = 0, hi = 5

Step – 1: arr[mid] == 0

  • swap(arr[lo], arr[mid])
  • lo = lo + 1 = 1
  • mid = mid + 1 = 1
  • arr[] = {0, 1, 2, 0, 1, 2}

Step – 2: arr[mid] == 1

  • mid = mid + 1 = 2
  • arr[] = {0, 1, 2, 0, 1, 2}

Step – 3: arr[mid] == 2

  • swap(arr[mid], arr[hi])
  • hi = hi – 1 = 4
  • arr[] = {0, 1, 2, 0, 1, 2}

Step – 4: arr[mid] == 2

  • swap(arr[mid], arr[hi])
  • hi = hi – 1 = 3
  • arr[] = {0, 1, 1, 0, 2, 2}

Step – 5: arr[mid] == 1

  • mid = mid + 1 = 3
  • arr[] = {0, 1, 1, 0, 2, 2}

Step – 6: arr[mid] == 0

  • swap(arr[lo], arr[mid])
  • lo = lo + 1 = 2
  • mid = mid + 1 = 4
  • arr[] = {0, 0, 1, 1, 2, 2}

Hence, arr[] = {0, 0, 1, 1, 2, 2}

Follow the steps below to solve the given problem:

  • Keep three indices low = 1, mid = 1, and high = N and there are four ranges, 1 to low (the range containing 0), low to mid (the range containing 1), mid to high (the range containing unknown elements) and high to N (the range containing 2).
  • Traverse the array from start to end and mid is less than high. (Loop counter is i)
  • If the element is 0 then swap the element with the element at index low and update low = low + 1 and mid = mid + 1
  • If the element is 1 then update mid = mid + 1
  • If the element is 2 then swap the element with the element at index high and update high = high – 1 and update i = i – 1. As the swapped element is not processed
  • Print the array.

C++

// C++ program to sort an array
// with 0, 1 and 2 in a single pass
#include <bits/stdc++.h>
using namespace std;
 
// Function to sort the input array,
// the array is assumed
// to have values in {0, 1, 2}
void sort012(int a[], int arr_size)
{
    int lo = 0;
    int hi = arr_size - 1;
    int mid = 0;
 
    // Iterate till all the elements
    // are sorted
    while (mid <= hi) {
        switch (a[mid]) {
 
        // If the element is 0
        case 0:
            swap(a[lo++], a[mid++]);
            break;
 
        // If the element is 1 .
        case 1:
            mid++;
            break;
 
        // If the element is 2
        case 2:
            swap(a[mid], a[hi--]);
            break;
        }
    }
}
 
// Function to print array arr[]
void printArray(int arr[], int arr_size)
{
    // Iterate and print every element
    for (int i = 0; i < arr_size; i++)
        cout << arr[i] << " ";
}
 
// Driver Code
int main()
{
    int arr[] = { 0, 1, 1, 0, 1, 2, 1, 2, 0, 0, 0, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    sort012(arr, n);
 
    printArray(arr, n);
 
    return 0;
}
 
// This code is contributed by Shivi_Aggarwal

                    

C

// C program to sort an array with 0, 1 and 2
// in a single pass
#include <stdio.h>
 
/* Function to swap *a and *b */
void swap(int* a, int* b);
 
// Sort the input array, the array is assumed to
// have values in {0, 1, 2}
void sort012(int a[], int arr_size)
{
    int lo = 0;
    int hi = arr_size - 1;
    int mid = 0;
    // Iterate till all the elements
    // are sorted
    while (mid <= hi) {
        switch (a[mid]) {
            // If the element is 0
        case 0:
            swap(&a[lo++], &a[mid++]);
            break;
            // If the element is 1
        case 1:
            mid++;
            break;
            // If the element is 2
        case 2:
            swap(&a[mid], &a[hi--]);
            break;
        }
    }
}
 
/* UTILITY FUNCTIONS */
void swap(int* a, int* b)
{
    int temp = *a;
    *a = *b;
    *b = temp;
}
 
/* Utility function to print array arr[] */
void printArray(int arr[], int arr_size)
{
    int i;
    for (i = 0; i < arr_size; i++)
        printf("%d ", arr[i]);
}
 
/* driver program to test */
int main()
{
    int arr[] = { 0, 1, 1, 0, 1, 2, 1, 2, 0, 0, 0, 1 };
    int arr_size = sizeof(arr) / sizeof(arr[0]);
    int i;
 
    sort012(arr, arr_size);
 
    printArray(arr, arr_size);
 
    getchar();
    return 0;
}

                    

Java

// Java program to sort an array of 0, 1 and 2
import java.io.*;
 
class countzot {
 
    // Sort the input array, the array is assumed to
    // have values in {0, 1, 2}
    static void sort012(int a[], int arr_size)
    {
        int lo = 0;
        int hi = arr_size - 1;
        int mid = 0, temp = 0;
        // Iterate till all the elements
        // are sorted
        while (mid <= hi) {
            switch (a[mid]) {
                // If the element is 0
            case 0: {
                temp = a[lo];
                a[lo] = a[mid];
                a[mid] = temp;
                lo++;
                mid++;
                break;
            }
                // If the element is 1
            case 1:
                mid++;
                break;
                // If the element is 2
            case 2: {
                temp = a[mid];
                a[mid] = a[hi];
                a[hi] = temp;
                hi--;
                break;
            }
            }
        }
    }
 
    /* Utility function to print array arr[] */
    static void printArray(int arr[], int arr_size)
    {
        int i;
        for (i = 0; i < arr_size; i++)
            System.out.print(arr[i] + " ");
        System.out.println("");
    }
 
    /*Driver function to check for above functions*/
    public static void main(String[] args)
    {
        int arr[] = { 0, 1, 1, 0, 1, 2, 1, 2, 0, 0, 0, 1 };
        int arr_size = arr.length;
        sort012(arr, arr_size);
        printArray(arr, arr_size);
    }
}
/*This code is contributed by Devesh Agrawal*/

                    

Python3

# Python program to sort an array with
# 0, 1 and 2 in a single pass
 
# Function to sort array
 
 
def sort012(a, arr_size):
    lo = 0
    hi = arr_size - 1
    mid = 0
    # Iterate till all the elements
    # are sorted
    while mid <= hi:
        # If the element is 0
        if a[mid] == 0:
            a[lo], a[mid] = a[mid], a[lo]
            lo = lo + 1
            mid = mid + 1
        # If the element is 1
        elif a[mid] == 1:
            mid = mid + 1
        # If the element is 2
        else:
            a[mid], a[hi] = a[hi], a[mid]
            hi = hi - 1
    return a
 
# Function to print array
 
 
def printArray(a):
    for k in a:
        print(k, end=' ')
 
 
# Driver Program
arr = [0, 1, 1, 0, 1, 2, 1, 2, 0, 0, 0, 1]
arr_size = len(arr)
arr = sort012(arr, arr_size)
printArray(arr)
 
# Contributed by Harshit Agrawal

                    

C#

// C# program to sort an
// array of 0, 1 and 2
using System;
 
class GFG {
    // Sort the input array, the array is assumed to
    // have values in {0, 1, 2}
    static void sort012(int[] a, int arr_size)
    {
        int lo = 0;
        int hi = arr_size - 1;
        int mid = 0, temp = 0;
        // Iterate till all the elements
        // are sorted
        while (mid <= hi) {
            switch (a[mid]) {
            // If the element is 0
            case 0: {
                temp = a[lo];
                a[lo] = a[mid];
                a[mid] = temp;
                lo++;
                mid++;
                break;
            }
            // If the element is 1
            case 1:
                mid++;
                break;
            // If the element is 2
            case 2: {
                temp = a[mid];
                a[mid] = a[hi];
                a[hi] = temp;
                hi--;
                break;
            }
            }
        }
    }
 
    /* Utility function to print array arr[] */
    static void printArray(int[] arr, int arr_size)
    {
        int i;
 
        for (i = 0; i < arr_size; i++)
            Console.Write(arr[i] + " ");
        Console.WriteLine("");
    }
 
    /*Driver function to check for above functions*/
    public static void Main()
    {
        int[] arr = { 0, 1, 1, 0, 1, 2, 1, 2, 0, 0, 0, 1 };
        int arr_size = arr.Length;
        sort012(arr, arr_size);
 
        printArray(arr, arr_size);
    }
}
 
// This code is contributed by Sam007

                    

Javascript

<script>
// Javascript program to sort an array of 0, 1 and 2
 
    // Sort the input array, the array is assumed to
    // have values in {0, 1, 2}
    function sort012(a,arr_size)
    {
         
        let lo = 0;
        let hi = arr_size - 1;
        let mid = 0;
        let temp = 0;
        // Iterate till all the elements
        // are sorted
        while (mid <= hi)
        {
            // If the element is 0
            if(a[mid] == 0)
            {
                temp = a[lo];
                a[lo] = a[mid];
                a[mid] = temp;
                lo++;
                mid++;
            }
            // If the element is 1
            else if(a[mid] == 1)
            {
                mid++;
            }
            // If the element is 2
            else
            {
                temp = a[mid];
                a[mid] = a[hi];
                a[hi] = temp;
                hi--;
            }
             
        }
    }
     
    /* Utility function to print array arr[] */
    function printArray(arr,arr_size)
    {
        let i;
        for (i = 0; i < arr_size; i++)
        {
            document.write(arr[i] + " ");
        }
        document.write("<br>");
    }
     
    /*Driver function to check for above functions*/
    let arr= [0, 1, 1, 0, 1, 2, 1, 2, 0, 0, 0, 1 ];
     
    let arr_size = arr.length;
    sort012(arr, arr_size);
    printArray(arr, arr_size);
     
    // This code is contributed by rag2127
</script>

                    

PHP

<?php
// PHP program to sort an array
// with 0, 1 and 2 in a single pass
 
// Sort the input array, the array is
// assumed to have values in {0, 1, 2}
function sort012(&$a, $arr_size)
{
    $lo = 0;
    $hi = $arr_size - 1;
    $mid = 0;
    // Iterate till all the elements
    // are sorted
    while ($mid <= $hi)
    {
        switch ($a[$mid])
        {
        // If the element is 0   
        case 0:
            swap($a[$lo++], $a[$mid++]);
            break;
        // If the element is 1   
        case 1:
            $mid++;
            break;
        // If the element is 2
        case 2:
            swap($a[$mid], $a[$hi--]);
            break;
        }
    }
}
 
/* UTILITY FUNCTIONS */
function swap(&$a, &$b)
{
    $temp = $a;
    $a = $b;
    $b = $temp;
}
 
/* Utility function to print array arr[] */
function printArray(&$arr, $arr_size)
{
    for ($i = 0; $i < $arr_size; $i++)
        echo $arr[$i]." ";
    echo "\n";
}
 
// Driver Code
$arr = array(0, 1, 1, 0, 1, 2,
             1, 2, 0, 0, 0, 1);
$arr_size = sizeof($arr);
 
sort012($arr, $arr_size);
 
printArray($arr, $arr_size);
 
// This code is contributed
// by ChitraNayal
?>

                    

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








Time Complexity: O(n), Only one traversal of the array is needed.
Space Complexity: O(1), No extra space is required.

Sort an array of 0s, 1s, and 2s using pointers:

Approach :

  • In this approach we will be using two pointers(l and r) along with an iterator(i). The entire array will be divided into three ranges:
    • Index 0 to l-1 (range containing 0)
    • Index l to r (range containing unknown elements)
    • Index r+1 to n (range containing 2)
  • Initialise l=0 and r=n-1.
  • Inside a for loop, make sure i<=r and do the following steps:
    • If the i-th element is 0, swap it with arr[l] and increment l and i.
    • If the i-th element is 2, swap it with arr[r] and decrement r (not i). The loop will automatically check for the next updated value of arr[i].
    • If the i-th element is arr[i], simply increment i and continue.

Illustration :

arr[] = {0, 1, 2, 0, 1, 2}

Step 1: l=0, r=5, i=0: arr[0]=0 –> Swap arr[0] with arr[0]. l=l+1 and i= i+1.

arr[] = {0, 1, 2, 0, 1, 2 }


Step 2: l=1, r=5, i=1: arr[1]=0 –> i= i+1.

arr[] = {0, 1, 2, 0, 1, 2 }


Step 3: l=1, r=5, i=2: arr[2]=2 –> Swap arr[2] with arr[5]. r=r-1.

arr[] = {0, 1, 2, 0, 1, 2 }


Step 4: l=1, r=4, i=2: arr[2]=2 –> Swap arr[2] with arr[4]. r=r-1.

arr[] = {0, 1, 1, 0, 2, 2 }


Step 5: l=1, r=3, i=2: arr[2]=1 –> i= i+1.

arr[] = {0, 1, 1, 0, 2, 2 }


Step 6: l=1, r=3, i=3: arr[3]=0 –> Swap arr[3] with arr[1]. l=l+1 and i= i+1.

arr[] = {0, 0, 1, 1, 2, 2 }

Follow the given steps to solve the problem :

  • Initialise l=0 and r=n-1.
  • Inside a for loop, make sure i<=r and do the following steps:
    • If the i-th element is 0, swap it with arr[l] and increment l and i.
    • If the i-th element is 2, swap it with arr[r] and decrement r (not i). The loop will automatically check for the next updated value of arr[i].
    • If the i-th element is arr[i], simply increment i and continue.

C++

#include <iostream>
using namespace std;
 
// Function to sort array containing 0, 1 and 2
void sort012(int arr[], int n)
{
    // Initialisation
    int l=0;
    int r=n-1;
     
    for(int i=0;i<n && i<=r;){
        // current element is 0
        if(arr[i]==0){
            swap(arr[l],arr[i]);
            l++;
            i++;
        }
        // current element is 2
        else if(arr[i]==2){
            swap(arr[i],arr[r]);
            r--;
        }
        // current element is 1
        else{
            i++;
        }
    }
}
 
// Function to print array arr[]
void printArray(int arr[], int arr_size)
{
    // Iterate and print every element
    for (int i = 0; i < arr_size; i++)
        cout << arr[i] << " ";
}
   
// Driver Code
int main()
{
    int arr[] = { 0, 1, 1, 0, 1, 2, 1, 2, 0, 0, 0, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
   
    sort012(arr, n);
   
    printArray(arr, n);
   
    return 0;
}

                    

Java

import java.util.Arrays;
 
public class Sort012 {
     
    // Function to sort array containing 0, 1, and 2
    static void sort012(int[] arr, int n) {
        // Initialization
        int low = 0;
        int high = n - 1;
 
        for (int i = 0; i < n && i <= high;) {
            // Current element is 0
            if (arr[i] == 0) {
                swap(arr, low, i);
                low++;
                i++;
            }
            // Current element is 2
            else if (arr[i] == 2) {
                swap(arr, i, high);
                high--;
            }
            // Current element is 1
            else {
                i++;
            }
        }
    }
 
    // Function to swap elements at indices i and j in the array
    static void swap(int[] arr, int i, int j) {
        int temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
    }
 
    // Function to print array arr[]
    static void printArray(int arr[], int arr_size) {
        // Iterate and print every element
        for (int i = 0; i < arr_size; i++)
            System.out.print(arr[i] + " ");
    }
 
    // Driver Code
    public static void main(String[] args) {
        int arr[] = { 0, 1, 1, 0, 1, 2, 1, 2, 0, 0, 0, 1 };
        int n = arr.length;
 
        sort012(arr, n);
 
        printArray(arr, n);
    }
}

                    

Python3

def sort012(arr, n):
    # Initialisation
    l = 0
    r = n - 1
 
    i = 0
    while i < n and i <= r:
        # current element is 0
        if arr[i] == 0:
            arr[l], arr[i] = arr[i], arr[l]
            l += 1
            i += 1
        # current element is 2
        elif arr[i] == 2:
            arr[i], arr[r] = arr[r], arr[i]
            r -= 1
        # current element is 1
        else:
            i += 1
 
 
def printArray(arr, arr_size):
    # Iterate and print every element
    for i in range(arr_size):
        print(arr[i], end=" ")
 
 
# Driver Code
if __name__ == "__main__":
    arr = [0, 1, 1, 0, 1, 2, 1, 2, 0, 0, 0, 1]
    n = len(arr)
 
    sort012(arr, n)
 
    printArray(arr, n)

                    

C#

using System;
 
namespace Sort012
{
    class Program
    {
        // Function to sort array containing 0, 1, and 2
        static void Sort012(int[] arr, int n)
        {
            // Initialization
            int l = 0;
            int r = n - 1;
 
            for (int i = 0; i < n && i <= r;)
            {
                // current element is 0
                if (arr[i] == 0)
                {
                    Swap(ref arr[l], ref arr[i]);
                    l++;
                    i++;
                }
                // current element is 2
                else if (arr[i] == 2)
                {
                    Swap(ref arr[i], ref arr[r]);
                    r--;
                }
                // current element is 1
                else
                {
                    i++;
                }
            }
        }
 
        // Function to swap two elements
        static void Swap(ref int a, ref int b)
        {
            int temp = a;
            a = b;
            b = temp;
        }
 
        // Function to print array arr[]
        static void PrintArray(int[] arr, int arr_size)
        {
            // Iterate and print every element
            for (int i = 0; i < arr_size; i++)
            {
                Console.Write(arr[i] + " ");
            }
        }
 
        // Driver Code
        static void Main(string[] args)
        {
            int[] arr = { 0, 1, 1, 0, 1, 2, 1, 2, 0, 0, 0, 1 };
            int n = arr.Length;
 
            Sort012(arr, n);
 
            PrintArray(arr, n);
        }
    }
}

                    

Javascript

function GFG(arr) {
    // Initialization
    let l = 0;
    let r = arr.length - 1;
    for (let i = 0; i < arr.length && i <= r;) {
        // Current element is 0
        if (arr[i] === 0) {
            [arr[l], arr[i]] = [arr[i], arr[l]];
            l++;
            i++;
        }
        // Current element is 2
        else if (arr[i] === 2) {
            [arr[i], arr[r]] = [arr[r], arr[i]];
            r--;
        }
        // Current element is 1
        else {
            i++;
        }
    }
}
// Function to print array arr[]
function printArray(arr) {
    // Iterate and print every element
    for (let i = 0; i < arr.length; i++)
        process.stdout.write(arr[i] + " ");
    console.log();
    // Add newline after printing the array
}
// Driver Code
let arr = [0, 1, 1, 0, 1, 2, 1, 2, 0, 0, 0, 1];
GFG(arr);
printArray(arr);

                    

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








Time Complexity: O(n), we iterate every element only once

Space Complexity: O(1)

Sort an array of 0s, 1s, and 2s using Counting Approach: 

This approach is based on the following idea:

Count the number of 0s, 1s, and 2s in the given array. Then store all the 0s at the beginning followed by all the 1s and then all the 2s.

Illustration:

arr[] = {0, 1, 2, 0, 1, 2}

cnt0 = 0, cnt1 = 0, cnt2 = 0

At i = 0: arr[0] == 0

  • cnt0 = cnt0 + 1 = 1

At i = 1: arr[1] == 1

  • cnt1 = cnt1 + 1 = 1

At i = 2: arr[2] == 2

  • cnt2 = cnt2 + 1 = 1

At i = 3: arr[3] == 0

  • cnt0 = cnt0 + 1 = 2

At i = 4: arr[4] == 1

  • cnt1 = cnt1 + 1 = 2

At i = 5: arr[5] == 2

  • cnt2 = cnt2 + 1 = 2

Replace cnt0 number of elements with 0 in arr

  • arr[] = {0, 0, 2, 0, 1, 2}

Replace cnt1 number of elements with 1 in arr

  • arr[] = {0, 0, 1, 1, 1, 2}

Replace cnt2 number of elements with 2 in arr

  • arr[] = {0, 0, 1, 1, 2, 2}

Hence, arr[] = {0, 0, 1, 1, 2, 2}

Follow the steps below to solve the given problem:  

  • Keep three counters c0 to count 0s, c1 to count 1s, and c2 to count 2s
  • Traverse through the array and increase the count of c0 if the element is 0, increase the count of c1 if the element is 1 and increase the count of c2 if the element is 2
  • Now again traverse the array and replace the first c0 elements with 0, the next c1 elements with 1, and the next c2 elements with 2.

Below is the implementation of the above idea : 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Utility function to print the contents of an array
void printArr(int arr[], int n)
{
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}
 
// Function to sort the array of 0s, 1s and 2s
void sortArr(int arr[], int n)
{
    int i, cnt0 = 0, cnt1 = 0, cnt2 = 0;
 
    // Count the number of 0s, 1s and 2s in the array
    for (i = 0; i < n; i++) {
        switch (arr[i]) {
        case 0:
            cnt0++;
            break;
        case 1:
            cnt1++;
            break;
        case 2:
            cnt2++;
            break;
        }
    }
 
    // Update the array
    i = 0;
 
    // Store all the 0s in the beginning
    while (cnt0 > 0) {
        arr[i++] = 0;
        cnt0--;
    }
 
    // Then all the 1s
    while (cnt1 > 0) {
        arr[i++] = 1;
        cnt1--;
    }
 
    // Finally all the 2s
    while (cnt2 > 0) {
        arr[i++] = 2;
        cnt2--;
    }
 
    // Print the sorted array
    printArr(arr, n);
}
 
// Driver code
int main()
{
    int arr[] = { 0, 1, 1, 0, 1, 2, 1, 2, 0, 0, 0, 1 };
    int n = sizeof(arr) / sizeof(int);
 
    sortArr(arr, n);
 
    return 0;
}

                    

Java

// Java implementation of the approach
import java.io.*;
 
class GFG {
    // Utility function to print the contents of an array
    static void printArr(int arr[], int n)
    {
        for (int i = 0; i < n; i++)
            System.out.print(arr[i] + " ");
    }
 
    // Function to sort the array of 0s, 1s and 2s
    static void sortArr(int arr[], int n)
    {
        int i, cnt0 = 0, cnt1 = 0, cnt2 = 0;
 
        // Count the number of 0s, 1s and 2s in the array
        for (i = 0; i < n; i++) {
            switch (arr[i]) {
            case 0:
                cnt0++;
                break;
            case 1:
                cnt1++;
                break;
            case 2:
                cnt2++;
                break;
            }
        }
 
        // Update the array
        i = 0;
 
        // Store all the 0s in the beginning
        while (cnt0 > 0) {
            arr[i++] = 0;
            cnt0--;
        }
 
        // Then all the 1s
        while (cnt1 > 0) {
            arr[i++] = 1;
            cnt1--;
        }
 
        // Finally all the 2s
        while (cnt2 > 0) {
            arr[i++] = 2;
            cnt2--;
        }
 
        // Print the sorted array
        printArr(arr, n);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 0, 1, 1, 0, 1, 2, 1, 2, 0, 0, 0, 1 };
        int n = arr.length;
        sortArr(arr, n);
    }
}
 
// This code is contributed by shubhamsingh10

                    

Python3

# Python implementation of the approach
 
# Utility function to print contents of an array
 
 
def printArr(arr, n):
    for i in range(n):
        print(arr[i], end=" ")
 
 
# Function to sort the array of 0s, 1s and 2s
def sortArr(arr, n):
    cnt0 = 0
    cnt1 = 0
    cnt2 = 0
 
    # Count the number of 0s, 1s and 2s in the array
    for i in range(n):
        if arr[i] == 0:
            cnt0 += 1
 
        elif arr[i] == 1:
            cnt1 += 1
 
        elif arr[i] == 2:
            cnt2 += 1
 
    # Update the array
    i = 0
 
    # Store all the 0s in the beginning
    while (cnt0 > 0):
        arr[i] = 0
        i += 1
        cnt0 -= 1
 
    # Then all the 1s
    while (cnt1 > 0):
        arr[i] = 1
        i += 1
        cnt1 -= 1
 
    # Finally all the 2s
    while (cnt2 > 0):
        arr[i] = 2
        i += 1
        cnt2 -= 1
 
    # Print the sorted array
    printArr(arr, n)
 
 
# Driver code
 
arr = [0, 1, 1, 0, 1, 2, 1, 2, 0, 0, 0, 1]
n = len(arr)
 
sortArr(arr, n)
 
# This code is contributed by shubhamsingh10

                    

C#

// C# implementation of the approach
using System;
 
class GFG {
    // Utility function to print the contents of an array
    static void printArr(int[] arr, int n)
    {
        for (int i = 0; i < n; i++)
            Console.Write(arr[i] + " ");
    }
 
    // Function to sort the array of 0s, 1s and 2s
    static void sortArr(int[] arr, int n)
    {
        int i, cnt0 = 0, cnt1 = 0, cnt2 = 0;
 
        // Count the number of 0s, 1s and 2s in the array
        for (i = 0; i < n; i++) {
            switch (arr[i]) {
            case 0:
                cnt0++;
                break;
            case 1:
                cnt1++;
                break;
            case 2:
                cnt2++;
                break;
            }
        }
 
        // Update the array
        i = 0;
 
        // Store all the 0s in the beginning
        while (cnt0 > 0) {
            arr[i++] = 0;
            cnt0--;
        }
 
        // Then all the 1s
        while (cnt1 > 0) {
            arr[i++] = 1;
            cnt1--;
        }
 
        // Finally all the 2s
        while (cnt2 > 0) {
            arr[i++] = 2;
            cnt2--;
        }
 
        // Print the sorted array
        printArr(arr, n);
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = { 0, 1, 1, 0, 1, 2, 1, 2, 0, 0, 0, 1 };
        int n = arr.Length;
 
        sortArr(arr, n);
    }
}
 
// This code is contributed by shubhamsingh10

                    

Javascript

<script>
// javascript implementation of the approach
 
// Utility function to print the contents of an array
function printArr( arr, n)
{
    for (let i = 0; i < n; i++)
        document.write(arr[i] + " ");
}
 
// Function to sort the array of 0s, 1s and 2s
function sortArr( arr,  n)
{
    let i, cnt0 = 0, cnt1 = 0, cnt2 = 0;
 
    // Count the number of 0s, 1s and 2s in the array
    for (i = 0; i < n; i++) {
        switch (arr[i]) {
        case 0:
            cnt0++;
            break;
        case 1:
            cnt1++;
            break;
        case 2:
            cnt2++;
            break;
        }
    }
 
    // Update the array
    i = 0;
 
    // Store all the 0s in the beginning
    while (cnt0 > 0) {
        arr[i++] = 0;
        cnt0--;
    }
 
    // Then all the 1s
    while (cnt1 > 0) {
        arr[i++] = 1;
        cnt1--;
    }
 
    // Finally all the 2s
    while (cnt2 > 0) {
        arr[i++] = 2;
        cnt2--;
    }
 
    // Print the sorted array
    printArr(arr, n);
}
 
    // Driver program to test above function
 
    let arr = [0, 1, 1, 0, 1, 2, 1,
                2, 0, 0, 0, 1];
    let n = arr.length;
     
    // Function calling
    sortArr(arr, n);
     
    // This code is contributed by jana_sayantan.
</script>

                    

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








Time Complexity: O(n), Only nonnested traversals of the array are needed.

Space Complexity: O(1)



Last Updated : 08 Dec, 2023
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