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Move all zeroes to end of array using Two-Pointers

Given an array of random numbers, Push all the zero’s of the given array to the end of the array. For example, if the given arrays is {1, 0, 2, 6, 0, 4}, it should be changed to {1, 2, 6, 4, 0, 0}. The order of all other elements should be the same.
Examples: 
 

Input: arr[]={8, 9, 0, 1, 2, 0, 3}
Output: arr[]={8, 9, 1, 2, 3, 0, 0}
Explanation: 
Swap {0 ,1} -> Resulting array {8, 9, 1, 0, 2, 0, 3}
Swap {0 ,2} -> Resulting array {8, 9, 1, 2, 0, 0, 3}
Swap {0 ,3} -> Final array {8, 9, 1, 2, 3, 0, 0}

Input: arr[]={4, 5, 0, 0, 0, 0, 6, 7}
Output: arr[]={4, 5, 6, 7, 0, 0, 0, 0}

 

Approach: 
 

  1. Iterate the array from 0 to N
     
  2. Keep two pointers, one for zero elements and other for non-zero elements. 
     
  3. Swap every zero elements with the non-zero element that comes just after it. 
     

 

C




// C implementation to move all zeroes at
// the end of array
#include<stdio.h>
 
// Function to move all zeroes at
// the end of array
void moveZerosToEnd(int arr[], int n)
{
    int j=0, temp, i;
     
    // Traverse the array. If arr[i] is
    // non-zero and arr[j] is zero,
    // then swap both the element
    for(i=0;i<n;i++)
    {
        if(arr[i]!=0 && arr[j]==0)
            {
             temp=arr[i];
             arr[i]=arr[j];
             arr[j]=temp;
            }
        if(arr[j]!=0)
            j+=1;
    }
}
 
// Function to print the array elements
void printArray(int arr[], int n)
{
    for (int i = 0; i < n; i++)
         printf("%d ", arr[i]);
}
 
// Driver Code
int main()
{
    int arr[] = {8, 9, 0, 1, 2, 0, 3};
    int n = sizeof(arr) / sizeof(arr[0]);
 
    printf("Original array: ");
    printArray(arr, n);
 
    moveZerosToEnd(arr, n);
 
    printf("\nModified array: ");
    printArray(arr, n);
 
    return 0;
}


C++




// 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)
{
    int j=0, temp, i;
     
    // Traverse the array. If arr[i] is
    // non-zero and arr[j] is zero,
    // then swap both the element
    for(i=0;i<n;i++)
    {
        if(arr[i]!=0 && arr[j]==0)
            {
             swap(arr[i],arr[j]);
            }
        if(arr[j]!=0)
            j+=1;
    }
}
 
// Function to print the array elements
void printArray(int arr[], int n)
{
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}
 
// Driver Code
int main()
{
    int arr[] = {8, 9, 0, 1, 2, 0, 3};
    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
 
class GFG {
 
    // Function to move all zeroes at
    // the end of array
    static void moveZerosToEnd(int arr[], int n) {
        int j = 0, i;
 
        // Traverse the array. If arr[i] is
        // non-zero and arr[j] is zero,
        // then swap both the element
        for (i = 0; i < n; i++) {
            if (arr[i] != 0 && arr[j] == 0) {
                arr = swap(arr, i, j);
            }
            if (arr[j] != 0)
                j += 1;
        }
    }
 
    static int[] swap(int[] arr, int i, int j) {
        int temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
        return arr;
    }
 
    // 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 Code
    public static void main(String[] args)
    {
        int arr[] = { 8, 9, 0, 1, 2, 0, 3 };
        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 29AjayKumar


Python3




# Python3 implementation to move all zeroes at
# the end of array
 
# Function to move all zeroes at
# the end of array
def moveZerosToEnd(nums):
    j = 0
     
    for i in range(len(nums)):
        if nums[i]!= 0 and nums[j]== 0:
            nums[i], nums[j]= nums[j], nums[i]
        if nums[j]!= 0:
            j+= 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 = [8, 9, 0, 1, 2, 0, 3]
n = len(arr)
 
print("Original array:", end=" ")
printArray(arr, n)
 
moveZerosToEnd(arr)
 
print("\nModified array: ", end=" ")
printArray(arr, n)


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)
{
    int j = 0, i;
 
    // Traverse the array. If arr[i]
    // is non-zero and arr[j] is zero,
    // then swap both the element
    for(i = 0; i < n; i++)
    {
       if (arr[i] != 0 && arr[j] == 0)
       {
           arr = swap(arr, i, j);
       }
       if (arr[j] != 0)
           j += 1;
    }
}
 
static int[] swap(int[] arr, int i, int j)
{
    int temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
    return arr;
}
 
// 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 Code
public static void Main(String[] args)
{
    int []arr = { 8, 9, 0, 1, 2, 0, 3 };
    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 29AjayKumar


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)
{
    let j=0, temp, i;
     
    // Traverse the array. If arr[i] is
    // non-zero and arr[j] is zero,
    // then swap both the element
    for(i=0;i<n;i++)
    {
        if(arr[i]!=0 && arr[j]==0)
            {
                let temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        if(arr[j]!=0)
            j+=1;
    }
}
 
// Function to print the array elements
function printArray(arr, n)
{
    for (let i = 0; i < n; i++)
        document.write(arr[i] + " ");
}
 
// Driver Code
    let arr = [8, 9, 0, 1, 2, 0, 3];
    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 Surbhi Tyagi.
 
</script>


Output: 

Original array: 8 9 0 1 2 0 3 
Modified array: 8 9 1 2 3 0 0

 

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

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Ted Musemwa
As a software developer I’m interested in the intersection of computational thinking and design thinking when solving human problems. As a professional I am guided by the principles of experiential learning; experience, reflect, conceptualise and experiment.
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