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.

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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