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}