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; ` `} ` |

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## 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 */` |

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## 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" ` |

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## 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 */` |

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## 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 ` `?> ` |

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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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