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

  • Difficulty Level : Medium
  • Last Updated : 20 Jul, 2021

Given an array A[] consisting 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.
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}

 

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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, and both of these problems are variation of famous Dutch national flag problem.
    The problem was posed with three colours, here `0′, `1′ and `2′. The array is divided into four sections: 
    1. a[1..Lo-1] zeroes (red)
    2. a[Lo..Mid-1] ones (white)
    3. a[Mid..Hi] unknown
    4. a[Hi+1..N] twos (blue)
    5. If the ith element is 0 then swap the element to the low range, thus shrinking the unknown range.
    6. Similarly, if the element is 1 then keep it as it is but shrink the unknown range.
    7. If the element is 2 then swap it with an element in high range.
  • Algorithm: 
    1. 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).
    2. Traverse the array from start to end and mid is less than high. (Loop counter is i)
    3. If the element is 0 then swap the element with the element at index low and update low = low + 1 and mid = mid + 1
    4. If the element is 1 then update mid = mid + 1
    5. 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
    6. Print the output 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]:
     

DNF1



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++
 

DNF2

Case (1) a[Mid] is white, Mid++
 

DNF3

Case (2) a[Mid] is blue, swap a[Mid] and a[Hi]; Hi–
 

DNF4

Continue until Mid>Hi. 
 

  •  
  • Implementation: 
     

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);
 
    cout << "array after segregation ";
 
    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;
 
    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);
 
    printf("array after segregation ");
    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;
        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);
        System.out.println("Array after seggregation ");
        printArray(arr, arr_size);
    }
}
/*This code is contributed by Devesh Agrawal*/

Python




# 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
    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 array
def printArray( a):
    for k in a:
        print k,
   
# Driver Program
arr = [0, 1, 1, 0, 1, 2, 1, 2, 0, 0, 0, 1]
arr_size = len(arr)
arr = sort012( arr, arr_size)
print "Array after segregation :\n",
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;
 
        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);
 
        Console.Write("Array after seggregation ");
 
        printArray(arr, arr_size);
    }
}
 
// This code is contributed by Sam007

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;
 
    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);
 
echo "array after segregation ";
printArray($arr, $arr_size);
 
// This code is contributed
// by ChitraNayal
?>

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;
        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);
    document.write("Array after seggregation <br>")
    printArray(arr, arr_size);
     
    // This code is contributed by rag2127
</script>
  • Output: 
     
array after segregation
 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: 
      1. Keep three counter c0 to count 0s, c1 to count 1s and c2 to count 2s
      2. 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
      3. 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++




// 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

Python




# 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
     
    # Prthe 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

 

  •  
  • Complexity Analysis: 
    • Time Complexity: O(n). 
      Only two traversals of the array is needed.
    • Space Complexity: O(1). 
      As no extra space is required.



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