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.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.