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

Difficulty Level : Easy
Last Updated : 03 May, 2021

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 << "\nModified array: ";
    printArray(arr, n);
 
    return 0;
}

Java

// Java implementation to move
// all zeroes at the end of array
import java.io.*;
 
class GFG {
 
// function to move all zeroes at
// the end of array
static void moveZerosToEnd(int arr[], int n) {
     
    // Count of non-zero elements
    int count = 0;
    int temp;
 
    // 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)) {
        temp = arr[count];
        arr[count] = arr[i];
        arr[i] = temp;
        count = count + 1;
    }
    }
}
 
// function to print the array elements
static void printArray(int arr[], int n) {
    for (int i = 0; i < n; i++)
    System.out.print(arr[i] + " ");
}
 
// Driver program to test above
public static void main(String args[]) {
    int arr[] = {0, 1, 9, 8, 4, 0, 0, 2,
                         7, 0, 6, 0, 9};
    int n = arr.length;
 
    System.out.print("Original array: ");
    printArray(arr, n);
 
    moveZerosToEnd(arr, n);
 
    System.out.print("\nModified array: ");
    printArray(arr, n);
}
}
 
// This code is contributed by Nikita Tiwari.

Python3

# Python implementation to move all zeroes at
# the end of array
 
# function to move all zeroes at
# the end of array
def moveZerosToEnd (arr, n):
 
    # Count of non-zero elements
    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 i in range(0, n):
        if (arr[i] != 0):
            arr[count], arr[i] = arr[i], arr[count]
            count+=1
 
 
# function to print the array elements
def printArray(arr, n):
 
    for i in range(0, n):
        print(arr[i],end=" ")
 
 
# Driver program to test above
arr = [ 0, 1, 9, 8, 4, 0, 0, 2,
    7, 0, 6, 0, 9 ]
n = len(arr)
 
print("Original array:", end=" ")
printArray(arr, n)
 
moveZerosToEnd(arr, n)
 
print("\nModified array: ", end=" ")
printArray(arr, n)
 
# This code is contributed by
# Smitha Dinesh Semwal

C#

// C# implementation to move
// all zeroes at the end of array
using System;
 
class GFG {
 
    // function to move all zeroes at
    // the end of array
    static void moveZerosToEnd(int[] arr, int n)
    {
        // Count of non-zero elements
        int count = 0;
        int temp;
 
        // 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))
            {
                temp = arr[count];
                arr[count] = arr[i];
                arr[i] = temp;
                count = count + 1;
            }
        }
    }
 
    // function to print the array elements
    static void printArray(int[] arr, int n)
    {
        for (int i = 0; i < n; i++)
            Console.Write(arr[i] + " ");
    }
 
    // Driver program to test above
    public static void Main()
    {
        int[] arr = { 0, 1, 9, 8, 4, 0, 0, 2,
                    7, 0, 6, 0, 9 };
        int n = arr.Length;
 
        Console.Write("Original array: ");
        printArray(arr, n);
 
        moveZerosToEnd(arr, n);
 
        Console.Write("\nModified array: ");
        printArray(arr, n);
    }
}
 
// This code is contributed by Sam007

Javascript

<script>
// JavaScript implementation to move all zeroes at
// the end of array
 
 
// function to move all zeroes at
// the end of array
function moveZerosToEnd(arr, n)
{
    // Count of non-zero elements
    let 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 (let i = 0; i < n; i++)
        if (arr[i] != 0)
        {
                temp = arr[count];
                arr[count] = arr[i];
                arr[i] = temp;
                count = count + 1;
         }
}
 
// function to print the array elements
function printArray(arr, n)
{
    for (let i = 0; i < n; i++)
        document.write(arr[i] + " ");
}
 
// Driver program to test above
 
    let arr = [ 0, 1, 9, 8, 4, 0, 0, 2,
                        7, 0, 6, 0, 9 ];
    let n = arr.length;
 
    document.write("Original array: ");
    printArray(arr, n);
 
    moveZerosToEnd(arr, n);
 
    document.write("<br>" + "Modified array: ");
    printArray(arr, n);
 
 
//This code is contributed by Mayank Tyagi
</script>

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).

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