Check whether bitwise AND of N numbers is Even or Odd

Given an array arr[] containing N numbers. The task is to check whether the bitwise-AND of the given N numbers is even or odd.
Examples: 

Input: arr[] = { 2, 12, 20, 36, 38 } 
Output: Even

Input: arr[] = { 3, 9, 17, 13, 15 } 
Output: Odd 

A Simple Solution is to first find the AND of the given N numbers, then check if this AND is even or odd.

C++

// C++ implementation to check whether
// bitwise and of n numbers is even or odd
#include <bits/stdc++.h>
using namespace std;

// Function to check if bitwise and
// of n numbers is even or odd
bool check(int arr[], int n)
{
    int x=arr[0];// assigining the biswise and of given array to x.
  for(int i=1;i<n;i++){
      x=x & arr[i];
  }
  if(x%2==0)// checking is even or not.
        return false;// if x is even then returning false.
  return true;// if x is odd returning true.
}

// Driver Code
int main()
{
    int arr[] ={ 2, 12, 20, 36, 38};
    int n = sizeof(arr) / sizeof(arr[0]);

    if (check(arr, n))
        cout << "Odd";
    else
        cout << "Even";
    return 0;
}

Java

import java.util.*;

public class Main {
  // Function to check if bitwise and
  // of n numbers is even or odd
  static boolean check(int arr[], int n)
  {
    int x = arr[0]; // assigining the biswise and of
    // given array to x.
    for (int i = 1; i < n; i++) {
      x = x & arr[i];
    }
    if (x % 2 == 0) // checking is even or not.
      return false; // if x is even then returning
    // false.
    return true; // if x is odd returning true.
  }

  // Driver Code
  public static void main(String[] args)
  {
    int arr[] = { 2, 12, 20, 36, 38 };
    int n = arr.length;

    if (check(arr, n))
      System.out.println("Odd");
    else
      System.out.println("Even");
  }
}

Python3

# Function to check if bitwise and
# of n numbers is even or odd
def check(arr, n):
    x = arr[0]
    for i in range(1, n):
        x = x & arr[i]
    if x % 2 == 0:
        return False
    return True

# Driver Code
if __name__ == '__main__':
    arr = [2, 12, 20, 36, 38]
    n = len(arr)
    if check(arr, n):
        print("Odd")
    else:
        print("Even")

Javascript

// JavaScript implementation to check whether
// bitwise and of n numbers is even or odd

// Function to check if bitwise and
// of n numbers is even or odd
function check(arr, n) {
    let x = arr[0]; // assigining the biswise and of given array to x.
    for (let i = 1; i < n; i++) {
        x = x & arr[i];
    }
    if (x % 2 == 0) // checking is even or not.
        return false; // if x is even then returning false.
    return true; // if x is odd returning true.
}

// Driver Code
let arr = [2, 12, 20, 36, 38];
let n = arr.length;

if (check(arr, n))
    console.log("Odd");
else
    console.log("Even");

C#

using System;

public class Program
{
    // Function to check if bitwise and
    // of n numbers is even or odd
    public static bool Check(int[] arr, int n)
    {
        int x = arr[0]; // assigining the biswise and of given array to x.
        for (int i = 1; i < n; i++)
        {
            x = x & arr[i];
        }
        if (x % 2 == 0) // checking is even or not.
            return false; // if x is even then returning false.
        return true; // if x is odd returning true.
    }

    // Driver Code
    public static void Main()
    {
        int[] arr = { 2, 12, 20, 36, 38 };
        int n = arr.Length;

        if (Check(arr, n))
            Console.WriteLine("Odd");
        else
            Console.WriteLine("Even");
    }
}

Output

Even

Time Complexity: O(N)

A Better Solution is based on bit manipulation and Mathematical facts. 

• Bitwise AND of any two even numbers is an even number.
• Bitwise AND of any two odd numbers is an odd number.
• Bitwise AND of an even and an odd number is an even number.

Based on the above facts, it can be deduced that if at least one even number is present in the array then the bitwise AND of whole array will be even otherwise odd.
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to check if the bitwise AND
// of the array elements is even or odd
void checkEvenOdd(int arr[], int n)
{
    for (int i = 0; i < n; i++) {

        // If at least an even element is present
        // then the bitwise AND of the
        // array elements will be even
        if (arr[i] % 2 == 0) {

            cout << "Even";
            return;
        }
    }

    cout << "Odd";
}

// Driver code
int main()
{
    int arr[] = { 2, 12, 20, 36, 38 };
    int n = sizeof(arr) / sizeof(arr[0]);

    checkEvenOdd(arr, n);

    return 0;
}

Java

// Java implementation of the approach
import java.io.*;

class GFG {

    // Function to check if the bitwise AND
    // of the array elements is even or odd
    static void checkEvenOdd(int[] arr, int n)
    {
        for (int i = 0; i < n; i++) {

            // If at least an even element is present
            // then the bitwise AND of the
            // array elements will be even
            if (arr[i] % 2 == 0) {
                System.out.print("Even");
                return;
            }
        }
        System.out.println("Odd");
    }

    // Driver code
    public static void main(String[] args)
    {
        int[] arr = { 2, 12, 20, 36, 38 };
        int n = arr.length;

        checkEvenOdd(arr, n);
    }
}

// This code is contributed by @tushil

Python3

# Python3 implementation of the approach

# Function to check if the bitwise AND
# of the array elements is even or odd


def checkEvenOdd(arr, n):

    for i in range(n):

        # If at least an even element is present
        # then the bitwise AND of the
        # array elements will be even
        if (arr[i] % 2 == 0):

            print("Even", end="")
            return

    print("Odd", end="")


# Driver code
if __name__ == "__main__":

    arr = [2, 12, 20, 36, 38]
    n = len(arr)

    checkEvenOdd(arr, n)

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;

class GFG {

    // Function to check if the bitwise AND
    // of the array elements is even or odd
    static void checkEvenOdd(int[] arr, int n)
    {
        for (int i = 0; i < n; i++) {

            // If at least an even element is present
            // then the bitwise AND of the
            // array elements will be even
            if (arr[i] % 2 == 0) {
                Console.Write("Even");
                return;
            }
        }
        Console.Write("Odd");
    }

    // Driver code
    static public void Main()
    {
        int[] arr = { 2, 12, 20, 36, 38 };
        int n = arr.Length;

        checkEvenOdd(arr, n);
    }
}

// This code is contributed by ajit..

Javascript

<script>
    // Javascript implementation of the approach

    // Function to check if the bitwise AND
    // of the array elements is even or odd
    function checkEvenOdd(arr, n)
    {
        for (let i = 0; i < n; i++)
        {

            // If at least an even element is present
            // then the bitwise AND of the
            // array elements will be even
            if (arr[i] % 2 == 0)
            {
                document.write ("Even");
                return;
            }
        }
        document.write ("Odd");
    }

    let arr = [ 2, 12, 20, 36, 38 ];
    let n = arr.length;

    checkEvenOdd(arr, n);

</script>

Output:

Even

Time Complexity: O(N)
Auxiliary Space: O(1)