Check whether bitwise OR of N numbers is Even or Odd

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

Examples: 

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

Input : arr[] = { 3, 9, 12, 13, 15 }
Output : Odd Bit-wise OR

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

C++

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

// Function to check if bitwise OR
// of n numbers is even or odd
bool check(int arr[], int n)
{
    int x=arr[0];// assigining the biswise or 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[] = { 3, 9, 12, 13, 15 };
    int n = sizeof(arr) / sizeof(arr[0]);

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

Java

import java.util.*;

public class Gfg {

    // Function to check if bitwise OR
    // of n numbers is even or odd
    static boolean check(int[] arr, int n) {
        int x = arr[0]; // assigning the bitwise OR of given array to x.
        for (int i = 1; i < n; i++) {
            x = x | arr[i];
        }
        // checking if x is even or not.
        if (x % 2 == 0)
            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 = { 3, 9, 12, 13, 15 };
        int n = arr.length;

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

Python3

# Python implementation to check whether
# bitwise OR of n numbers is even or odd

# Function to check if bitwise OR
# of n numbers is even or odd
def check(arr, n):
    x = arr[0] # assigning the bitwise OR of given array to x.
    for i in range(1, n):
        x = x | arr[i]
    if x % 2 == 0: # checking if x is even or not.
        return False # if x is even then returning false.
    return True # if x is odd, returning true.

# Driver Code
arr = [3, 9, 12, 13, 15]
n = len(arr)

if check(arr, n):
    print("Odd Bit-wise OR")
else:
    print("Even Bit-wise OR")

C#

// C# implementation to check whether
// bitwise OR of n numbers is even or odd
using System;

class Gfg{
    // Function to check if bitwise OR
    // of n numbers is even or odd
    static bool check(int[] arr, int n)
    {
        int x=arr[0];// assigining the biswise or 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 = { 3, 9, 12, 13, 15 };
        int n = arr.Length;

        if (check(arr, n)) {
            Console.WriteLine("Odd Bit-wise OR");
        }
        else {
            Console.WriteLine("Even Bit-wise OR");
        }

    }
}

Javascript

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

// Function to check if bitwise OR
// of n numbers is even or odd
function check(arr) {
    let x = arr[0]; // assigning the bitwise or of given array to x.
    for (let i = 1; i < arr.length; i++) {
        x = x | arr[i];
    }
    if (x % 2 == 0) // checking if x 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 = [3, 9, 12, 13, 15];

if (check(arr))
console.log("Odd Bit-wise OR");
else
console.log("Even Bit-wise OR");

Output

Odd Bit-wise OR

Time Complexity: O(N)

Auxiliary Space: O(1)

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

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

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

Below is the implementation of the above approach: 

C++

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

// Function to check if bitwise OR
// of n numbers is even or odd
bool check(int arr[], int n)
{
    for (int i = 0; i < n; i++) {
        // if at least one odd number is found,
        // then the bitwise OR of all numbers will be odd
        if (arr[i] & 1)
            return true;
    }

    // Bitwise OR is an odd number
    return false;
}

// Driver Code
int main()
{
    int arr[] = { 3, 9, 12, 13, 15 };
    int n = sizeof(arr) / sizeof(arr[0]);

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

Java

// Java implementation to check whether
// bitwise OR of n numbers is even or odd
class GFG {

    // Function to check if bitwise OR
    // of n numbers is even or odd
    static boolean check(int arr[], int n)
    {
        for (int i = 0; i < n; i++) {
            // if at least one odd number is
            // found, then the bitwise OR of
            // all numbers will be odd
            if (arr[i] % 2 == 1) {
                return true;
            }
        }

        // Bitwise OR is an odd number
        return false;
    }

    // Driver Code
    public static void main(String args[])
    {
        int arr[] = { 3, 9, 12, 13, 15 };
        int n = arr.length;

        if (check(arr, n)) {
            System.out.println("Odd Bit-wise OR");
        }
        else {
            System.out.println("Even Bit-wise OR");
        }
    }
}

// This code is contributed
// by 29AjayKumar

Python3

# Python3 implementation to check whether
# bitwise OR of n numbers is even or odd

# Function to check if bitwise OR
# of n numbers is even or odd


def check(arr, n):
    for i in range(n):

        # if at least one odd number is found,
        # then the bitwise OR of all numbers
        # will be odd
        if arr[i] & 1:
            return True

    # Bitwise OR is an odd number
    return False


# Driver code
if __name__ == '__main__':
    arr = [3, 9, 12, 13, 15]
    n = len(arr)
    if check(arr, n):
        print("Odd Bit-wise OR")
    else:
        print("Even Bit-wise OR")

# This code is contributed by
# Shrikant13

C#

// C# implementation to check whether
// bitwise OR of n numbers is even or odd
using System;

class GFG {

    // Function to check if bitwise OR
    // of n numbers is even or odd
    static bool check(int[] arr, int n)
    {
        for (int i = 0; i < n; i++) {
            // if at least one odd number is
            // found, then the bitwise OR of
            // all numbers will be odd
            if (arr[i] % 2 == 1) {
                return true;
            }
        }

        // Bitwise OR is an odd number
        return false;
    }

    // Driver Code
    public static void Main()
    {
        int[] arr = { 3, 9, 12, 13, 15 };
        int n = arr.Length;

        if (check(arr, n)) {
            Console.WriteLine("Odd Bit-wise OR");
        }
        else {
            Console.WriteLine("Even Bit-wise OR");
        }
    }
}

// This code is contributed
// by 29AjayKumar

PHP

<?php
//PHp implementation to check whether
// bitwise OR of n numbers is even or odd

// Function to check if bitwise OR
// of n numbers is even or odd

function  check($arr, $n)
{
    for ($i = 0; $i < $n; $i++) {
        // if at least one odd number is found,
        // then the bitwise OR of all numbers will be odd
        if ($arr[$i] & 1)
            return true;
    }

    // Bitwise OR is an odd number
    return false;
}

// Driver Code

    $arr = array (3, 9, 12, 13, 15 );
     $n = sizeof($arr) / sizeof($arr[0]);

    if (check($arr, $n))
         echo  "Odd Bit-wise OR";
    else
        echo "Even Bit-wise OR";

// This code is contributed by ajit
?>

Javascript

<script>

// Javascript implementation to check whether
// bitwise OR of n numbers is even or odd

// Function to check if bitwise OR
// of n numbers is even or odd
function check(arr, n)
{
    for (let i = 0; i < n; i++)
    {

        // if at least one odd number is found,
        // then the bitwise OR of all numbers will be odd
        if (arr[i] & 1)
            return true;
    }

    // Bitwise OR is an odd number
    return false;
}

// Driver Code
    let arr = [ 3, 9, 12, 13, 15 ];
    let n = arr.length;

    if (check(arr, n))
        document.write("Odd Bit-wise OR");
    else
        document.write("Even Bit-wise OR");

// This code is contributed by rishavmahato348.
</script>

Output

Odd Bit-wise OR

Time Complexity: O(N) in the worst case.
Auxiliary Space: O(1)