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)