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.



Time Complexity: O(N)

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 number 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 whole array will be odd otherwise even.

Below is the implementation of the above approach:

C++

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// 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;

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Java

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// 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

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Python3

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# 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

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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 [tabby title="PHP"]

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<?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
?>

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Output:

Odd Bit-wise OR

Time Complexity: O(N) in worst case.



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Improved By : shrikanth13, jit_t, 29AjayKumar