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};

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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 <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; 
} 

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 
?> 

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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My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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