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C++ Program to Move all zeroes to end of array | Set-2 (Using single traversal)

  • Last Updated : 17 Jan, 2022

Given an array of n numbers. The problem is to move all the 0’s to the end of the array while maintaining the order of the other elements. Only single traversal of the array is required.
Examples: 
 

Input : arr[]  = {1, 2, 0, 0, 0, 3, 6}
Output : 1 2 3 6 0 0 0

Input: arr[] = {0, 1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9}
Output: 1 9 8 4 2 7 6 9 0 0 0 0 0

 

Algorithm: 
 

moveZerosToEnd(arr, n)
    Initialize count = 0
    for i = 0 to n-1
        if (arr[i] != 0) then
            swap(arr[count++], arr[i])

 

CPP




// C++ implementation to move all zeroes at
// the end of array
#include <iostream>
using namespace std;
  
// function to move all zeroes at
// the end of array
void moveZerosToEnd(int arr[], int n)
{
    // Count of non-zero elements
    int count = 0;
  
    // Traverse the array. If arr[i] is non-zero, then
    // swap the element at index 'count' with the
    // element at index 'i'
    for (int i = 0; i < n; i++)
        if (arr[i] != 0)
            swap(arr[count++], arr[i]);
}
  
// function to print the array elements
void printArray(int arr[], int n)
{
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}
  
// Driver program to test above
int main()
{
    int arr[] = { 0, 1, 9, 8, 4, 0, 0, 2,
                         7, 0, 6, 0, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << "Original array: ";
    printArray(arr, n);
  
    moveZerosToEnd(arr, n);
  
    cout << "
Modified array: ";
    printArray(arr, n);
  
    return 0;
}

Output: 
 

Original array: 0 1 9 8 4 0 0 2 7 0 6 0 9 
Modified array: 1 9 8 4 2 7 6 9 0 0 0 0 0

Time Complexity: O(n). 
Auxiliary Space: O(1).
 

Please refer complete article on Move all zeroes to end of array | Set-2 (Using single traversal) for more details!


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