Check if an Array exists with Bitwise OR as A and Bitwise AND as B

Three integers N, A and B are given to you. you need to create an array arr of length ‘N‘ such that it satisfies the following conditions :

1. Each elements of arr should be unique.
2. arr[1] | arr[2] | arr[3] |….arr[N] = A
3. arr[1] & arr[2] & arr[3] &….arr[N] = B

Where ‘|’ Represents Bitwise OR and ‘&’ Represents Bitwise AND. Your task is to determine if it is possible to Create Such an array or not.

Example :

Input: ‘N’ = 3 , ‘A’ = 10, ‘B’ = 2
Output: YES
Explanation: one of the possible array arr is: arr = [2,3,8] .

Input: ‘N’ = 5 ‘A’ = 15 ‘B’ = 16
Output: NO.

Approach:

• ‘A‘ is the bitwise OR of all array elements, and ‘B‘ is the bitwise AND of all array elements.
• If the ‘i-th‘ bit of ‘B‘ is ‘1‘, then every element in array arr has ‘i-th‘ bit ‘1‘ in their binary representation.
• If the ‘i-th’ bit of ‘A’ is ‘1’, then some elements may have ‘i-th‘ bit ‘1‘ in their binary representation.
• Now, if the ‘i-th‘ bit of ‘A‘ is ‘0‘, but the ‘i-th‘ bit of ‘B‘ is ‘1‘, then it’s a contradiction, and it is not possible to create such array ‘arr’, then the answer will be ‘NO‘.
• If the ‘i-th‘ bit of ‘A‘ is ‘1‘, but the ‘i-th‘ bit of ‘B‘ is ‘0‘, we have a choice to choose a number, such that it may have the ‘i-th‘ bit ‘0‘ or ‘1‘. So, we have ‘2‘ choices for a number if the ‘i-th‘ bit of ‘A‘ is ‘1‘, but the ‘i-th’ bit of ‘B‘ is ‘0‘.
• So, let’s consider the count of such positions ‘ i ‘, with the ‘i-th’ bit of ‘A’ as ‘1’, but the ‘i-th’ bit of ‘B’ as ‘0’, as ‘cnt‘. Then there are total ‘2^cnt‘ possible different numbers.
• If ‘2^cnt‘ is less than ‘N‘, the answer will be ‘NO‘, Otherwise, the answer will be ‘YES‘.

Below is the Implementation of Above approach :

C++

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

string isPossible(int n, int a, int b)
{

    // If 'N' is equal to '1':
    if (n == 1) {

        // If 'X' is equal to 'Y', then return '1'.
        // Otherwise return '0'.
        if (a == b) {
            return "YES";
        }
        return "NO";
    }
    // Initialize an integer variable 'cnt' with '1'.
    int cnt = 1;

    // For 'i' in the range '0' to log2(maximum of X and
    // Y)':
    for (int i = 0; i <= log2(max(a, b)) + 1; i++) {

        // If the 'i-th' bit of 'X' is '0' and the 'i-th'
        // bit of 'Y' is '1', then return '0'.
        if (((a >> i) & 1) == 0 && ((b >> i) & 1) == 1) {
            return "NO";
        }

        // If the 'i-th' bit of 'X' is '1' and the 'i-th'
        // bit of 'Y' is '0', then multiply 'cnt' by '2'.
        if (((a >> i) & 1) == 1 && ((b >> i) & 1) == 0) {
            cnt *= 2;
        }
    }

    // If 'cnt' is less than 'N', then return '0'.
    if (cnt < n) {
        return "NO";
    }

    // Return '1'.
    return "YES";
}

int main()
{

    cout << isPossible(9, 11, 51);
    return 0;
}

Java

import java.io.*;

class GFG {
    static int isPossible(int n, int a, int b)
    {

        // If 'N' is equal to '1':
        if (n == 1) {

            // If 'X' is equal to 'Y', then return '1'.
            // Otherwise, return '0'.
            return (a == b) ? 1 : 0;
        }

        // Initialize an integer variable 'cnt' with '1'.
        int cnt = 1;

        // For 'i' in the range '0' to log2(maximum of X and
        // Y)':
        for (int i = 0; i <= log2(Math.max(a, b)) + 1;
             i++) {

            // If the 'i-th' bit of 'X' is '0' and the
            // 'i-th' bit of 'Y' is '1', then return '0'.
            if (((a >> i) & 1) == 0
                && ((b >> i) & 1) == 1) {
                return 0;
            }

            // If the 'i-th' bit of 'X' is '1' and the
            // 'i-th' bit of 'Y' is '0', then multiply 'cnt'
            // by '2'.
            if (((a >> i) & 1) == 1
                && ((b >> i) & 1) == 0) {
                cnt *= 2;
            }
        }

        // If 'cnt' is less than 'N', then return '0'.
        if (cnt < n) {
            return 0;
        }

        // Return '1'.
        return 1;
    }

    // Helper method to calculate the logarithm base 2
    static int log2(int num)
    {
        return (int)(Math.log(num) / Math.log(2));
    }
    public static void main(String[] args)
    {
        System.out.println(isPossible(9, 11, 51));
    }
}

Python3

# code by flutterfly
import math

def is_possible(n, a, b):
    # If 'N' is equal to '1':
    if n == 1:
        # If 'X' is equal to 'Y', then return '1'.
        # Otherwise, return '0'.
        return 1 if a == b else 0

    # Initialize an integer variable 'cnt' with '1'.
    cnt = 1

    # For 'i' in the range '0' to log2(maximum of X and Y)':
    for i in range(0, log2(max(a, b)) + 1):
        # If the 'i-th' bit of 'X' is '0' and the
        # 'i-th' bit of 'Y' is '1', then return '0'.
        if ((a >> i) & 1) == 0 and ((b >> i) & 1) == 1:
            return 0

        # If the 'i-th' bit of 'X' is '1' and the
        # 'i-th' bit of 'Y' is '0', then multiply 'cnt'
        # by '2'.
        if ((a >> i) & 1) == 1 and ((b >> i) & 1) == 0:
            cnt *= 2

    # If 'cnt' is less than 'N', then return '0'.
    if cnt < n:
        return 0

    # Return '1'.
    return 1

# Helper method to calculate the logarithm base 2
def log2(num):
    return int(math.log(num) / math.log(2))

# Main function
print(is_possible(9, 11, 51))

C#

//code by flutterfly
using System;

class GFG
{
    static int IsPossible(int n, int a, int b)
    {
        // If 'N' is equal to '1':
        if (n == 1)
        {
            // If 'X' is equal to 'Y', then return '1'.
            // Otherwise, return '0'.
            return (a == b) ? 1 : 0;
        }

        // Initialize an integer variable 'cnt' with '1'.
        int cnt = 1;

        // For 'i' in the range '0' to log2(maximum of X and Y)':
        for (int i = 0; i <= Log2(Math.Max(a, b)) + 1; i++)
        {
            // If the 'i-th' bit of 'X' is '0' and the
            // 'i-th' bit of 'Y' is '1', then return '0'.
            if (((a >> i) & 1) == 0 && ((b >> i) & 1) == 1)
            {
                return 0;
            }

            // If the 'i-th' bit of 'X' is '1' and the
            // 'i-th' bit of 'Y' is '0', then multiply 'cnt'
            // by '2'.
            if (((a >> i) & 1) == 1 && ((b >> i) & 1) == 0)
            {
                cnt *= 2;
            }
        }

        // If 'cnt' is less than 'N', then return '0'.
        if (cnt < n)
        {
            return 0;
        }

        // Return '1'.
        return 1;
    }

    // Helper method to calculate the logarithm base 2
    static int Log2(int num)
    {
        return (int)(Math.Log(num) / Math.Log(2));
    }

    public static void Main(string[] args)
    {
        Console.WriteLine(IsPossible(9, 11, 51));
    }
}

Javascript

//code by flutterfly
function isPossible(n, a, b) {
    // If 'N' is equal to '1':
    if (n === 1) {
        // If 'X' is equal to 'Y', then return '1'.
        // Otherwise, return '0'.
        return a === b ? 1 : 0;
    }

    // Initialize an integer variable 'cnt' with '1'.
    let cnt = 1;

    // For 'i' in the range '0' to log2(maximum of X and Y)':
    for (let i = 0; i <= log2(Math.max(a, b)) + 1; i++) {
        // If the 'i-th' bit of 'X' is '0' and the
        // 'i-th' bit of 'Y' is '1', then return '0'.
        if (((a >> i) & 1) === 0 && ((b >> i) & 1) === 1) {
            return 0;
        }

        // If the 'i-th' bit of 'X' is '1' and the
        // 'i-th' bit of 'Y' is '0', then multiply 'cnt'
        // by '2'.
        if (((a >> i) & 1) === 1 && ((b >> i) & 1) === 0) {
            cnt *= 2;
        }
    }

    // If 'cnt' is less than 'N', then return '0'.
    if (cnt < n) {
        return 0;
    }

    // Return '1'.
    return 1;
}

// Helper method to calculate the logarithm base 2
function log2(num) {
    return Math.floor(Math.log(num) / Math.log(2));
}

// Main function
console.log(isPossible(9, 11, 51));

Output

0

Time Complexity : O(log₂(z)) – where ‘z’ is the maximum of ‘A’ and ‘B’. (because we are traversing till the ‘log₂(z)^th bit .)
Auxiliary Space Complexity : O(1).