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# C++ 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.

## C++

 `// A C++ program to move all zeroes at the end of array``#include ``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 :``";``    ``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)

OTHER APPROACH :

Assuming the start element as the pivot and while iterating through the array if a non-zero element is encountered then swap the element with the pivot and increment the index. This is continued till all the non-zero elements are placed towards the left and all the zeros are towards the right.

## C++

 `#include ``using` `namespace` `std;` `void` `swap(``int` `A[], ``int` `i, ``int` `j)``{``    ``int` `temp = A[i];``    ``A[i] = A[j];``    ``A[j] = temp;``}` `// Function to move all zeros present in an array to the end``void` `fun(``int` `A[], ``int` `n)``{``    ``int` `j = 0;` `    ``// when we encounter a non-zero, `j` is incremented, and``    ``// the element is placed before the pivot``    ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `(A[i] != 0) ``// pivot is 0``        ``{``            ``swap(A, i, j);``            ``j++;``        ``}``    ``}``}` `int` `main()``{``    ``int` `A[] = { 1, 0, 2, 0, 3, 0, 4, 0, 5, 0 };``    ``int` `n = ``sizeof``(A) / ``sizeof``(A[0]);` `    ``fun(A, n);` `    ``for` `(``int` `i = 0; i < n; i++) {``        ``printf``(``"%d "``, A[i]);``    ``}` `    ``return` `0;``}`

Output

`1 2 3 4 5 0 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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