# 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 functionint 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 endvoid 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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