Move all zeroes to end of array

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, 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 <stdio.h>

// 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);
    printf("%s\n", "Array after pushing all zeros to end of array:");
    for (int i = 0; i < n; i++)
      printf("%d ", arr[i]);
    return 0;
}

C++

#include <algorithm>
#include <iostream>
#include <vector>

void push_zeros_to_end(std::vector<int>& arr)
{
  std::stable_partition(arr.begin(),
            arr.end(),
            [](int n) { return n != 0; });
}

int main()
{
  std::vector<int> arr{1,9,8,4,0,0,2,7,0,6,0,9};

  push_zeros_to_end(arr);

  for(const auto& i : arr)
    std::cout << i << ' ';

  std::cout << "\n";

  return 0;
}

Java

/* Java program to push zeroes to back of array */
import java.io.*;

class PushZero
{
    // Function which pushes all zeros to end of an array.
    static 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 function to check for above functions*/
    public static void main (String[] args)
    {
        int arr[] = {1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9};
        int n = arr.length;
        pushZerosToEnd(arr, n);
        System.out.println("Array after pushing zeros to the back: ");
        for (int i=0; i<n; i++)
            System.out.print(arr[i]+" ");
    }
}
/* This code is contributed by Devesh Agrawal */

Python3

# Python3 code to move all zeroes
# at the end of array

# Function which pushes all
# zeros to end of an array.
def pushZerosToEnd(arr, n):
    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 i in range(n):
        if arr[i] != 0:

            # here count is incremented
            arr[count] = arr[i]
            count+=1

    # 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
        count += 1

# Driver code
arr = [1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9]
n = len(arr)
pushZerosToEnd(arr, n)
print("Array after pushing all zeros to end of array:")
print(arr)

# This code is contributed by "Abhishek Sharma 44"

C#

/* C# program to push zeroes to back of array */
using System;

class PushZero
{
    // Function which pushes all zeros
    // to end of an array.
    static void pushZerosToEnd(int []arr, int n)
    {
        // Count of non-zero elements
        int count = 0;

        // Traverse the array. If element encountered is
        // non-zero, then replace the element
        // at index â..countâ.. with this element
        for (int i = 0; i < n; i++)
        if (arr[i] != 0)

        // here count is incremented
        arr[count++] = arr[i];

        // Now all non-zero elements have been shifted to
        // front and â..countâ.. is set as index of first 0.
        // Make all elements 0 from count to end.
        while (count < n)
        arr[count++] = 0;
    }

    // Driver function
    public static void Main ()
    {
        int []arr = {1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9};
        int n = arr.Length;
        pushZerosToEnd(arr, n);
        Console.WriteLine("Array after pushing all zeros to the back: ");
        for (int i = 0; i < n; i++)
        Console.Write(arr[i] +" ");
    }
}
/* This code is contributed by Anant Agrawal */

PHP

<?php
// A PHP program to move all
// zeroes at the end of array

// Function which pushes all
// zeros to end of an array.
function pushZerosToEnd(&$arr, $n)
{
    // Count of non-zero elements
    $count = 0;

    // Traverse the array. If
    // element encountered is
    // non-zero, then replace
    // the element at index
    // 'count' with this element
    for ($i = 0; $i < $n; $i++)
        if ($arr[$i] != 0)
            // here count is incremented
            $arr[$count++] = $arr[$i];

    // 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 Code
$arr = array(1, 9, 8, 4, 0, 0,
             2, 7, 0, 6, 0, 9);
$n = sizeof($arr);
pushZerosToEnd($arr, $n);
echo "Array after pushing all " .
     "zeros to end of array :\n";

for ($i = 0; $i < $n; $i++)
echo $arr[$i] . " ";

// This code is contributed
// by ChitraNayal
?>

Javascript

<script>
// A JavaScript program to move all zeroes at the end of array

// Function which pushes all zeros to end of an array.
function pushZerosToEnd(arr, n)
{
    let 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 (let 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 code
    let arr = [1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9];
    let n = arr.length;
    pushZerosToEnd(arr, n);
    document.write("Array after pushing all zeros to end of array :<br>");
    for (let i = 0; i < n; i++)
        document.write(arr[i] + " ");

// This code is contributed by Surbhi Tyagi.
</script>

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)

Method 2: Partitioning the array

Approach: The approach is pretty simple. We will use 0 as a pivot element and whenever we see a non zero element we will swap it with the pivot element. So all the non zero element will come at the beginning.

Code:

C++

// C++ Program to move all zeros to the end

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int A[] = { 5, 6, 0, 4, 6, 0, 9, 0, 8 };
    int n = sizeof(A) / sizeof(A[0]);
    int j = 0;
    for (int i = 0; i < n; i++) {
        if (A[i] != 0) {
            swap(A[j], A[i]); // Partitioning the array
            j++;
        }
    }
    for (int i = 0; i < n; i++) {
        cout << A[i] << " "; // Print the array
    }

    return 0;
}

C

// C Program to move all zeros to the end
#include <stdio.h>
int main()
{
    int A[] = { 5, 6, 0, 4, 6, 0, 9, 0, 8 };
    int n = sizeof(A) / sizeof(A[0]);
    int j = 0;
    for (int i = 0; i < n; i++) {
        if (A[i] != 0) {
            int temp = A[i]; // Partitioning the array
            A[i] = A[j];
            A[j] = temp;
            j++;
        }
    }
    for (int i = 0; i < n; i++) {
        printf("%d ", A[i]); // Print the array
    }

    return 0;
}

Java

// Java Program to move all zeros to the end
import java.util.*;

public class Main {
    public static void main(String[] args)
    {
        int[] A = { 5, 6, 0, 4, 6, 0, 9, 0, 8 };
        int n = A.length;
        int j = 0;
        for (int i = 0; i < n; i++) {
            if (A[i] != 0) {
                //   Swap - A[j] , A[i]
                swap(A, j, i); // Partitioning the array
                j++;
            }
        }
        for (int i = 0; i < n; i++) {
            System.out.print(A[i] + " "); // Print the array
        }
    }

    // Utility function to swap two elements of an array
    public static void swap(int[] A, int a, int b)
    {
        int temp = A[a];
        A[a] = A[b];
        A[b] = temp;
    }
}

// This code is contributed by tapeshdua420.

Python3

# Python Program to move all zeros to the end
A = [5, 6, 0, 4, 6, 0, 9, 0, 8]
n = len(A)
j = 0
for i in range(n):
    if A[i] != 0:
        A[j], A[i] = A[i], A[j]  # Partitioning the array
        j += 1
print(A)  # Print the array

# This code is contributed by Tapesh(tapeshdua420)

C#

using System;
public static class GFG
{

    // C# Program to move all zeros to the end
    public static void Main()
    {
        int[] A = { 5, 6, 0, 4, 6, 0, 9, 0, 8 };
        int n = A.Length;
        int j = 0;
        for (int i = 0; i < n; i++) {
            if (A[i] != 0) {
                int temp = A[j];
                A[j] = A[i];
                A[i] = temp;
                j++;
            }
        }
        for (int i = 0; i < n; i++) {
            Console.Write(A[i]);
            Console.Write(" ");
        }
    }
}

// This code is contributed by Aarti_Rathi

Javascript

<script>
// Javascript Program to move all zeros to the end


let A = [5, 6, 0, 4, 6, 0, 9, 0, 8];
let n = A.length;
let j = 0;
for (let i = 0; i < n; i++) {
    if (A[i] != 0) {
        //   Swap - A[j] , A[i]
        swap(A, j, i); // Partitioning the array
        j++;
    }
}
for (let i = 0; i < n; i++) {
    document.write(A[i] + " "); // Print the array
}


// Utility function to swap two elements of an array
function swap(A, a, b) {
    let temp = A[a];
    A[a] = A[b];
    A[b] = temp;
}

// This code is contributed by Saurabh Jaiswal
</script>

Output

5 6 4 6 9 8 0 0 0 

Complexity Analysis:

Time Complexity: O(N)

Auxiliary Space: O(1)

Method 3: using C++ STL

In this approach we will traverse the whole array and will count the number of zeros present in array. While counting we will delete the zero from array.

After completing the above process, we will push back the count number of zeros into vector.

Below is the implementation of the above process:

C++

// C++ programm to shift all zeros
// to right most side of array
// whthout affecting order of non-zero
// elements.

#include <bits/stdc++.h>
using namespace std;
// function to shift zeros
void move_zeros_to_right(vector<int>& m)
{
    int count = 0;
    for (int i = 0; i < m.size(); i++) {
        if (m[i] == 0) {
            count++;
            // deleting the element from vector
            m.erase(m.begin() + i);
            i--;
        }
    }

    for (int i = 0; i < count; i++) {
        // inserting the zero into vector
        m.push_back(0);
    }
    cout << "array after shifting zeros to right side: "
         << endl;
    for (int i = 0; i < m.size(); i++) {
        // printing desired vector
        cout << m[i] << " ";
    }
}
// driver code
int main()
{
    vector<int> m{ 5, 6, 0, 4, 6, 0, 9, 0, 8 };
    // function call
    move_zeros_to_right(m);
    return 0;
}

// This code is contributed by Machhaliya Muhammad

Output

array after shifting zeros to right side: 
5 6 4 6 9 8 0 0 0 

Time complexity: O(N) where N is size of array.

Auxiliary space: O(1).

This article is contributed by Chandra Prakash. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.