Move all zeroes to end of array

• Difficulty Level : Easy
• Last Updated : 19 Jul, 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, 0};Input : arr[]  = {1, 2, 0, 0, 0, 3, 6};Output : arr[] = {1, 2, 3, 6, 0, 0, 0};

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution. 
 

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++
• Java
• Python3
• C#
• PHP
• Javascript

// 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;

}

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)