# Move all Zeros

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 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]+` `" "` `);`

` ` `}`

`}`

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)