Hello Everyone,

Question: 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;`

`}`

**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)