Check whether bitwise OR of N numbers is Even or Odd
Last Updated :
17 Apr, 2023
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;
}
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");
}
}
# 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# 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 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");
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 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 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# 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 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
?>
<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>
Time Complexity: O(N) in the worst case.
Auxiliary Space: O(1)
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