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

Last Updated : 17 Aug, 2023

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)

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 <iostream> 
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[0]); 
  
    fun(A, n); 
  
    for (int i = 0; i < n; i++) { 
        printf("%d ", A[i]); 
    } 
  
    return 0; 
}