Bitwise OR of Multiples in an Array

Given an array arr[] comprising of n non-zero, positive integers and an integer x, find the bitwise OR of all multiples of x present in the array.

The result of OR operation should be in decimal form. If no multiple of x is found, the answer should be 0.

Examples:

Input: arr[] = [3, 4, 3, 9], x = 2
Output: 4
Explanation: Only multiple of 2 in array is 4.

Input: arr[] = [9, 3, 1, 6, 1], x = 3
Output: 15
Explanation: Multiples of 3 in array are 9, 3 and 6. Their OR value is 15.

Using Bitwise OR - O(n) Time and O(1) Space

The idea is to traverse the array and consider only those elements that are divisible by x.

• Initialize ans as 0.
• For each element arr[i], check whether it is a multiple of x using: arr[i] % x == 0
• If it is a multiple of x, include it in the result using bitwise OR: ans = ans | arr[i]
• After traversing the entire array, ans contains the bitwise OR of all multiples of x.
• If no element is a multiple of x, ans remains 0, which is the required result.

Consider: arr[] = [9, 3, 1, 6, 1] and x = 3.

• Initialize ans = 0 and traverse the array from left to right.
• For 9, 9 % 3 == 0, so include it in the result: ans = 0 | 9 = 9
• For 3, 3 % 3 == 0, so include it: ans = 9 | 3 = 11
• For 1, 1 % 3 != 0, so skip it.
• For 6, 6 % 3 == 0, so include it: ans = 11 | 6 = 15
• For 1, 1 % 3 != 0, so skip it.

After processing all elements, ans becomes 15.

Therefore, the bitwise OR of all elements that are multiples of 3 is 15.

#include <bits/stdc++.h>
using namespace std;

int findOR(vector<int>& arr, int x) {
    int ans = 0;

    // Traverse the array and OR all elements that are divisible by x.
    for (int num : arr) {
        if (num % x == 0)
            ans |= num;
    }

    return ans;
}

int main() {
    vector<int> arr = {9, 3, 1, 6, 1};
    int x = 3;

    cout << findOR(arr, x);

    return 0;
}

class GFG {

    public static int findOR(int[] arr, int x) {
        int ans = 0;

        // Traverse the array and OR all elements that are divisible by x.
        for (int num : arr) {
            if (num % x == 0)
                ans |= num;
        }

        return ans;
    }

    public static void main(String[] args) {
        int[] arr = {9, 3, 1, 6, 1};
        int x = 3;

        System.out.println(findOR(arr, x));
    }
}

def findOR(arr, x):
    ans = 0

    # Traverse the array and OR all elements that are divisible by x.
    for num in arr:
        if num % x == 0:
            ans |= num

    return ans


if __name__ == "__main__":
    arr = [9, 3, 1, 6, 1]
    x = 3

    print(findOR(arr, x))

using System;

class GFG {

    static int findOR(int[] arr, int x) {
        int ans = 0;

        // Traverse the array and OR all elements that are divisible by x.
        foreach (int num in arr) {
            if (num % x == 0)
                ans |= num;
        }

        return ans;
    }

    static void Main() {
        int[] arr = {9, 3, 1, 6, 1};
        int x = 3;

        Console.WriteLine(findOR(arr, x));
    }
}

function findOR(arr, x) {
    let ans = 0;

    // Traverse the array and OR all elements that are divisible by x.
    for (let num of arr) {
        if (num % x === 0)
            ans |= num;
    }

    return ans;
}

// Driver code
let arr = [9, 3, 1, 6, 1];
let x = 3;
console.log(findOR(arr, x));

15