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

  • Difficulty Level : Easy
  • Last Updated : 23 Dec, 2021

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 <iostream>
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)
 

Please refer complete article on Move all zeroes to end of array for more details!


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