Check whether bitwise OR of N numbers is Even or Odd

• Difficulty Level : Basic
• Last Updated : 16 Apr, 2021

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++

 // C++ implementation to check whether// bitwise OR of n numbers is even or odd#include using namespace std; // Function to check if bitwise OR// of n numbers is even or oddbool 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 Codeint main(){    int arr[] = { 3, 9, 12, 13, 15 };    int n = sizeof(arr) / sizeof(arr);     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 oddclass GFG{ // Function to check if bitwise OR// of n numbers is even or oddstatic 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 Codepublic 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 odddef 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 codeif __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 oddusing System; class GFG{ // Function to check if bitwise OR// of n numbers is even or oddstatic 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 Codepublic 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



Javascript


Output:
Odd Bit-wise OR

Time Complexity: O(N) in worst case.

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