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).

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 C++ implementation of the above approach.

// 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.