Move all zeroes to end of array

Difficulty Level : Easy
Last Updated : 18 Mar, 2021

Given an array of random numbers, Push all the zero’s of a given array to the end of the array. For example, if the given arrays is {1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0}, it should be changed to {1, 9, 8, 4, 2, 7, 6, 0, 0, 0, 0}. The order of all other elements should be same. Expected time complexity is O(n) and extra space is O(1).
Example:

Input :  arr[] = {1, 2, 0, 4, 3, 0, 5, 0};
Output : arr[] = {1, 2, 4, 3, 5, 0, 0};

Input : arr[]  = {1, 2, 0, 0, 0, 3, 6};
Output : arr[] = {1, 2, 3, 6, 0, 0, 0};

There can be many ways to solve this problem. Following is a simple and interesting way to solve this problem.
Traverse the given array ‘arr’ from left to right. While traversing, maintain count of non-zero elements in array. Let the count be ‘count’. For every non-zero element arr[i], put the element at ‘arr[count]’ and increment ‘count’. After complete traversal, all non-zero elements have already been shifted to front end and ‘count’ is set as index of first 0. Now all we need to do is that run a loop which makes all elements zero from ‘count’ till end of the array.
Below is the implementation of the above approach.

C++

// A C++ program to move all zeroes at the end of array
#include <iostream>
using namespace std;
 
// Function which pushes all zeros to end of an array.
void pushZerosToEnd(int arr[], int n)
{
    int count = 0;  // Count of non-zero elements
 
    // Traverse the array. If element encountered is non-
    // zero, then replace the element at index 'count'
    // with this element
    for (int i = 0; i < n; i++)
        if (arr[i] != 0)
            arr[count++] = arr[i]; // here count is
                                   // incremented
 
    // Now all non-zero elements have been shifted to
    // front and  'count' is set as index of first 0.
    // Make all elements 0 from count to end.
    while (count < n)
        arr[count++] = 0;
}
 
// Driver program to test above function
int main()
{
    int arr[] = {1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9};
    int n = sizeof(arr) / sizeof(arr[0]);
    pushZerosToEnd(arr, n);
    cout << "Array after pushing all zeros to end of array :\n";
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
    return 0;
}

Java

/* Java program to push zeroes to back of array */
import java.io.*;
 
class PushZero
{
    // Function which pushes all zeros to end of an array.
    static void pushZerosToEnd(int arr[], int n)
    {
        int count = 0;  // Count of non-zero elements
 
        // Traverse the array. If element encountered is
        // non-zero, then replace the element at index 'count'
        // with this element
        for (int i = 0; i < n; i++)
            if (arr[i] != 0)
                arr[count++] = arr[i]; // here count is
                                       // incremented
 
        // Now all non-zero elements have been shifted to
        // front and 'count' is set as index of first 0.
        // Make all elements 0 from count to end.
        while (count < n)
            arr[count++] = 0;
    }
 
    /*Driver function to check for above functions*/
    public static void main (String[] args)
    {
        int arr[] = {1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9};
        int n = arr.length;
        pushZerosToEnd(arr, n);
        System.out.println("Array after pushing zeros to the back: ");
        for (int i=0; i<n; i++)
            System.out.print(arr[i]+" ");
    }
}
/* This code is contributed by Devesh Agrawal */

Python3

# Python3 code to move all zeroes
# at the end of array
 
# Function which pushes all
# zeros to end of an array.
def pushZerosToEnd(arr, n):
    count = 0 # Count of non-zero elements
     
    # Traverse the array. If element
    # encountered is non-zero, then
    # replace the element at index
    # 'count' with this element
    for i in range(n):
        if arr[i] != 0:
             
            # here count is incremented
            arr[count] = arr[i]
            count+=1
     
    # Now all non-zero elements have been
    # shifted to front and 'count' is set
    # as index of first 0. Make all
    # elements 0 from count to end.
    while count < n:
        arr[count] = 0
        count += 1
         
# Driver code
arr = [1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9]
n = len(arr)
pushZerosToEnd(arr, n)
print("Array after pushing all zeros to end of array:")
print(arr)
 
# This code is contributed by "Abhishek Sharma 44"

C#

/* C# program to push zeroes to back of array */
using System;
 
class PushZero
{
    // Function which pushes all zeros
    // to end of an array.
    static void pushZerosToEnd(int []arr, int n)
    {
        // Count of non-zero elements
        int count = 0;
         
        // Traverse the array. If element encountered is
        // non-zero, then replace the element
        // at index â..countâ.. with this element
        for (int i = 0; i < n; i++)
        if (arr[i] != 0)
         
        // here count is incremented
        arr[count++] = arr[i];
         
        // Now all non-zero elements have been shifted to
        // front and â..countâ.. is set as index of first 0.
        // Make all elements 0 from count to end.
        while (count < n)
        arr[count++] = 0;
    }
     
    // Driver function
    public static void Main ()
    {
        int []arr = {1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9};
        int n = arr.Length;
        pushZerosToEnd(arr, n);
        Console.WriteLine("Array after pushing all zeros to the back: ");
        for (int i = 0; i < n; i++)
        Console.Write(arr[i] +" ");
    }
}
/* This code is contributed by Anant Agrawal */

PHP

<?php
// A PHP program to move all
// zeroes at the end of array
 
// Function which pushes all
// zeros to end of an array.
function pushZerosToEnd(&$arr, $n)
{
    // Count of non-zero elements
    $count = 0;
 
    // Traverse the array. If
    // element encountered is
    // non-zero, then replace
    // the element at index
    // 'count' with this element
    for ($i = 0; $i < $n; $i++)
        if ($arr[$i] != 0)
            // here count is incremented
            $arr[$count++] = $arr[$i];
 
    // Now all non-zero elements
    // have been shifted to front
    // and 'count' is set as index 
    // of first 0. Make all elements
    // 0 from count to end.
    while ($count < $n)
        $arr[$count++] = 0;
}
 
// Driver Code
$arr = array(1, 9, 8, 4, 0, 0,
             2, 7, 0, 6, 0, 9);
$n = sizeof($arr);
pushZerosToEnd($arr, $n);
echo "Array after pushing all " .
     "zeros to end of array :\n";
 
for ($i = 0; $i < $n; $i++)
echo $arr[$i] . " ";
 
// This code is contributed
// by ChitraNayal
?>

Javascript

<script>
// A JavaScript program to move all zeroes at the end of array
 
// Function which pushes all zeros to end of an array.
function pushZerosToEnd(arr, n)
{
    let count = 0; // Count of non-zero elements
 
    // Traverse the array. If element encountered is non-
    // zero, then replace the element at index 'count'
    // with this element
    for (let i = 0; i < n; i++)
        if (arr[i] != 0)
            arr[count++] = arr[i]; // here count is
                                // incremented
 
    // Now all non-zero elements have been shifted to
    // front and 'count' is set as index of first 0.
    // Make all elements 0 from count to end.
    while (count < n)
        arr[count++] = 0;
}
 
// Driver code
    let arr = [1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9];
    let n = arr.length;
    pushZerosToEnd(arr, n);
    document.write("Array after pushing all zeros to end of array :<br>");
    for (let i = 0; i < n; i++)
        document.write(arr[i] + " "); 
 
// This code is contributed by Surbhi Tyagi.
</script>

Output:

Array after pushing all zeros to end of array :
1 9 8 4 2 7 6 9 0 0 0 0

Time Complexity: O(n) where n is number of elements in input array.
Auxiliary Space: O(1)

This article is contributed by Chandra Prakash. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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