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# Java Program to Move all zeroes to end of array

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.

## 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

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)

Please refer complete article on Move all zeroes to end of array for more details!

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