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};

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++

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// 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 :\n";
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
    return 0;
}

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Java














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/* 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 */



















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Python3














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# 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" 



















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C#














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/* 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 */



















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PHP














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<?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 


?> 



















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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)




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.



          
          
          
          

            

    

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