# How to push all zeroes to the end of the array in C++

In this tutorial, we will learn to **push all zeroes to the end of an array in C++**. As we have given an array and we have to shift zeroes to the end of the array.

**For Example: **

- Given array= {1,0, 5, 7. 0, 12, 0,4, 0, 8}
- Push all zeroes to the end of the array.
- Output: {1, 5, 7, 12, 4, 8, 0, 0, 0 , 0}

Also, there are many ways to solve this given problem. But, we will use the most efficient of them.

**Approach:**

- Firstly, count the non-zero element.
- Now, traverse through the array element one by one.
- If the element is non-zero, then replace it with the element whose index is ‘count_non_zero’.
- In this process, here we will increase the count_non_zero.
- Finally, if count_on_zero is less than the total number of the element then, put 0 to the remaining indexes.
- The
*time complexity*of this method: O(n) where n is the total number of elements in the array. *Auxiliary Space:*O(1)

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## Push all zeroes to the end of the array in C++

You can see the implementation of the above approach here.

// A C++ program to shift zeroes at the end #include <iostream> using namespace std; // Function to shift zeroes void movezeroes(int arr[], int pu) { int count_non_zero = 0; // traverse through the array, and if the element is non zero then, //replace with the element. for (int i = 0; i < pu; i++) if (arr[i] != 0) arr[count_non_zero++] = arr[i]; //increment the count // check if count of non zero elements is less than total number //of elements, then put zero to the remaining indexes. while (count_non_zero < pu) arr[count_non_zero++] = 0; } int main() { int arr[] = {1, 8, 0, 2, 0, 0, 2, 1, 0, 0, 0, 9}; int pu = sizeof(arr) / sizeof(arr[0]); movezeroes(arr, pu); cout << "Array after shifting"<<endl ; for (int i = 0; i < pu; i++) cout << arr[i] << " "; return 0; }

**Output Explanation:**

INPUT: {1, 8, 0, 2, 0, 0, 2, 1, 0, 0, 0, 9} OUTPUT: {1, 8, 2, 2, 1, 9, 0, 0, 0, 0, 0, 0}

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