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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 ``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);``    ``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

## 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

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## Javascript

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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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