Move all zeroes to end of array | Set-2 (Using single traversal)

Given an array of n numbers. The problem is to move all the 0’s to the end of the array while maintaining the order of the other elements. Only single traversal of the array is required.

Examples:

Input : arr[]  = {1, 2, 0, 0, 0, 3, 6}
Output : 1 2 3 6 0 0 0

Input: arr[] = {0, 1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9}
Output: 1 9 8 4 2 7 6 9 0 0 0 0 0

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Algorithm:

moveZerosToEnd(arr, n)
    Initialize count = 0
    for i = 0 to n-1
        if (arr[i] != 0) then
            swap(arr[count++], arr[i])

CPP

// 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) 
{ 
    // Count of non-zero elements 
    int count = 0; 
  
    // Traverse the array. If arr[i] is non-zero, then 
    // swap the element at index 'count' with the 
    // element at index 'i' 
    for (int i = 0; i < n; i++) 
        if (arr[i] != 0) 
            swap(arr[count++], arr[i]); 
} 
  
// function to print the array elements 
void printArray(int arr[], int n) 
{ 
    for (int i = 0; i < n; i++) 
        cout << arr[i] << " "; 
} 
  
// Driver program to test above 
int main() 
{ 
    int arr[] = { 0, 1, 9, 8, 4, 0, 0, 2, 
                         7, 0, 6, 0, 9 }; 
    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 
import java.io.*; 
  
class GFG { 
  
// function to move all zeroes at 
// the end of array 
static void moveZerosToEnd(int arr[], int n) { 
      
    // Count of non-zero elements 
    int count = 0; 
    int temp; 
  
    // Traverse the array. If arr[i] is  
    // non-zero, then swap the element at  
    // index 'count' with the element at  
    // index 'i' 
    for (int i = 0; i < n; i++) { 
    if ((arr[i] != 0)) { 
        temp = arr[count]; 
        arr[count] = arr[i]; 
        arr[i] = temp; 
        count = count + 1; 
    } 
    } 
} 
  
// 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 program to test above 
public static void main(String args[]) { 
    int arr[] = {0, 1, 9, 8, 4, 0, 0, 2, 
                         7, 0, 6, 0, 9}; 
    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 Nikita Tiwari. 

Python3

# Python implementation to move all zeroes at 
# the end of array 
  
# function to move all zeroes at 
# the end of array 
def moveZerosToEnd (arr, n): 
  
    # Count of non-zero elements 
    count = 0; 
  
    # Traverse the array. If arr[i] is non-zero, then 
    # swap the element at index 'count' with the 
    # element at index 'i' 
    for i in range(0, n): 
        if (arr[i] != 0): 
            arr[count], arr[i] = arr[i], arr[count] 
            count+=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 = [ 0, 1, 9, 8, 4, 0, 0, 2, 
    7, 0, 6, 0, 9 ] 
n = len(arr) 
  
print("Original array:", end=" ") 
printArray(arr, n) 
  
moveZerosToEnd(arr, n) 
  
print("\nModified array: ", end=" ") 
printArray(arr, n) 
  
# This code is contributed by 
# Smitha Dinesh Semwal 

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) 
    { 
        // Count of non-zero elements 
        int count = 0; 
        int temp; 
  
        // Traverse the array. If arr[i] is 
        // non-zero, then swap the element at 
        // index 'count' with the element at 
        // index 'i' 
        for (int i = 0; i < n; i++) 
        { 
            if ((arr[i] != 0))  
            { 
                temp = arr[count]; 
                arr[count] = arr[i]; 
                arr[i] = temp; 
                count = count + 1; 
            } 
        } 
    } 
  
    // 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 program to test above 
    public static void Main() 
    { 
        int[] arr = { 0, 1, 9, 8, 4, 0, 0, 2, 
                    7, 0, 6, 0, 9 }; 
        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 Sam007 

Output:

Original array: 0 1 9 8 4 0 0 2 7 0 6 0 9 
Modified array: 1 9 8 4 2 7 6 9 0 0 0 0 0

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

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

CPP

Java

Python3

C#

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