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

11 Jul, 2022
Given an array A[] consisting 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}
A simple solution is discussed in this(Sort an array of 0s, 1s and 2s (Simple Counting)) post.Method 1 
Approach: The problem is similar to our old post “Segregate 0s and 1s in an array”.The problem was posed with three colours, here `0′, `1′ and `2′. The array is divided into four sections: a[1..Lo-1] zeroes (red)a[Lo..Mid-1] ones (white)a[Mid..Hi] unknowna[Hi+1..N] twos (blue)If the ith element is 0 then swap the element to the low range, thus shrinking the unknown range.Similarly, if the element is 1 then keep it as it is but shrink the unknown range.If the element is 2 then swap it with an element in high range.
a[1..Lo-1] zeroes (red)a[Lo..Mid-1] ones (white)a[Mid..Hi] unknowna[Hi+1..N] twos (blue)If the ith element is 0 then swap the element to the low range, thus shrinking the unknown range.Similarly, if the element is 1 then keep it as it is but shrink the unknown range.If the element is 2 then swap it with an element in high range.
a[1..Lo-1] zeroes (red)
a[Lo..Mid-1] ones (white)
a[Mid..Hi] unknown
a[Hi+1..N] twos (blue)
If the ith element is 0 then swap the element to the low range, thus shrinking the unknown range.
Similarly, if the element is 1 then keep it as it is but shrink the unknown range.
If the element is 2 then swap it with an element in high range.
Algorithm: 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 + 1If the element is 1 then update mid = mid + 1If 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 processedPrint the array.
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 + 1If the element is 1 then update mid = mid + 1If 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 processedPrint the array.
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.
Dry Run: Part way through the process, some red, white and blue elements are known and are in the “right” place. The section of unknown elements, a[Mid..Hi], is shrunk by examining a[Mid]: 
Examine a[Mid]. There are three possibilities: a[Mid] is (0) red, (1) white or (2) blue. Case (0) a[Mid] is red, swap a[Lo] and a[Mid]; Lo++; Mid++ 
Case (1) a[Mid] is white, Mid++ 
Case (2) a[Mid] is blue, swap a[Mid] and a[Hi]; Hi– 
Continue until Mid>Hi.   
Implementation:
C++
C
Java
Python3
C#
PHP
Javascript
// 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 Codeint 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 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;      while (mid <= hi) {        switch (a[mid]) {        case 0:            swap(&a[lo++], &a[mid++]);            break;        case 1:            mid++;            break;        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]);    printf("n");}  /* 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 program to sort an array of 0, 1 and 2import 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;        while (mid <= hi) {            switch (a[mid]) {            case 0: {                temp = a[lo];                a[lo] = a[mid];                a[mid] = temp;                lo++;                mid++;                break;            }            case 1:                mid++;                break;            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*/
# Python program to sort an array with # 0, 1 and 2 in a single pass  # Function to sort arraydef sort012( a, arr_size):    lo = 0    hi = arr_size - 1    mid = 0    while mid <= hi:        if a[mid] == 0:            a[lo], a[mid] = a[mid], a[lo]            lo = lo + 1            mid = mid + 1        elif a[mid] == 1:            mid = mid + 1        else:            a[mid], a[hi] = a[hi], a[mid]             hi = hi - 1    return a      # Function to print arraydef printArray( a):    for k in a:        print(k,end=' ')    # Driver Programarr = [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# program to sort an// array of 0, 1 and 2using 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;          while (mid <= hi) {            switch (a[mid]) {            case 0: {                temp = a[lo];                a[lo] = a[mid];                a[mid] = temp;                lo++;                mid++;                break;            }            case 1:                mid++;                break;            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
<?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;    $mid = 0;      while ($mid <= $hi)    {        switch ($a[$mid])        {        case 0:            swap($a[$lo++], $a[$mid++]);            break;        case 1:            $mid++;            break;        case 2:            swap($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?>
<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;         while (mid <= hi)        {            if(a[mid] == 0)            {                temp = a[lo];                 a[lo] = a[mid];                 a[mid] = temp;                 lo++;                 mid++;             }            else if(a[mid] == 1)            {                mid++;             }            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>
0 0 0 0 0 1 1 1 1 1 2 2 
Complexity Analysis: 
Time Complexity: O(n). Only one traversal of the array is needed.
Space Complexity: O(1). No extra space is required.
Approach: Count the number of 0s, 1s and 2s in the given array. Then store all the 0s in the beginning followed by all the 1s then all the 2s.
Algorithm: Keep three counter c0 to count 0s, c1 to count 1s and c2 to count 2sTraverse 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 2Now again traverse the array and replace first c0 elements with 0, next c1 elements with 1 and next c2 elements with 2.
Keep three counter c0 to count 0s, c1 to count 1s and c2 to count 2sTraverse 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 2Now again traverse the array and replace first c0 elements with 0, next c1 elements with 1 and next c2 elements with 2.
Keep three counter 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 first c0 elements with 0, next c1 elements with 1 and next c2 elements with 2.
Implementation: 
C++
Java
Python3
C#
Javascript
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;  // Utility function to print the contents of an arrayvoid printArr(int arr[], int n){    for (int i = 0; i < n; i++)        cout << arr[i] << " ";}  // Function to sort the array of 0s, 1s and 2svoid 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 codeint 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 implementation of the approachimport 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
# Python implementation of the approach  # Utility function to print contents of an arraydef printArr(arr, n):    for i in range(n):        print(arr[i],end=" ")    # Function to sort the array of 0s, 1s and 2sdef 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# implementation of the approachusing 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
<script>// javascript implementation of the approach  // Utility function to print the contents of an arrayfunction printArr( arr, n){    for (let i = 0; i < n; i++)        document.write(arr[i] + " ");}  // Function to sort the array of 0s, 1s and 2sfunction 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>
0 0 0 0 0 1 1 1 1 1 2 2 
Complexity Analysis: 
Time Complexity: O(n). Only non nested traversals of the array is needed.
Space Complexity: O(1). As no extra space is required.
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