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# C++ Program to Move all zeroes to end of array

• Difficulty Level : Easy
• Last Updated : 25 May, 2022

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);` `    ``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);`   `    ``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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