Count values whose Bitwise OR with A is equal to B

Given two integers A and B, the task is to count possible values of X that satisfies the condition A | X = B.
Note: | represents Bitwise OR operation.

Examples:

Input: A = 2, B = 3
Output: 2
Explanation:
Since, 2 | 1 = 3 and 2 | 3 = 3. Therefore, the possible values of x are 1 and 3.

Input: A = 5, B = 7
Output: 4

Naive Approach: The simplest approach to solve this problem is to iterate over the range [1, B] and check for every number, whether its Bitwise OR with A is equal to B. If the condition is satisfied, increment the count.
Time Complexity: O(b)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized based on the following observations:

• Convert the numbers A and B into their respective binary representations.

Truth table of Bitwise OR operation:
1 | 1 = 1
1 | 0 = 1
0 | 1 = 1
0 | 0 = 0

• For each same-positioned bit, count number of ways the i^th bit in A can be converted to the i^th bit in B by performing Bitwise OR operation.



Follow the steps below to solve the problem:

• Initialize a variable ans to store the required result.
• Traverse the bits of a and b simultaneously using the variable i
  • If the ith bit of a is set and the ith bit of b is also set, then the ith bit of x can take 2 values, i.e, 0 or 1. Hence, multiply the ans by 2.
  • If the ith bit of a is unset and the ith bit of b is set, then the ith bit of x can take only 1 value, i.e, 1. Hence, multiply the ans by 1.
  • If the ith bit of a is set and the ith bit of b is unset, then no matter what the ith bit of x is, the ith bit of a can not be converted to ith bit of b. Hence, update ans to 0 and break out of the loop.
• Print the value of ans as the result

Below is the implementation of the above approach:

C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to count possible values of
// X whose Bitwise OR with A is equal to B
int numWays(int a, int b)
{

    // Store the required result
    int res = 1;

    // Iterate over all bits
    while (a && b) {

        // If i-th bit of a is set
        if ((a & 1) == 1) {

            // If i-th bit of b is set
            if ((b & 1) == 1) {

                // The i-th bit in x
                // can be both 1 and 0
                res = res * 2;
            }
            else {

                // a|x is equal to 1
                // Therefore, it cannot
                // be converted to b
                return 0;
            }
        }

        // Right shift a and b by 1
        a = a >> 1;
        b = b >> 1;
    }

    return res;
}

// Driver Code
int main()
{
    // Given Input
    int a = 2, b = 3;

    // Function Call
    cout << numWays(a, b);

    return 0;
}
Output:

2

Time Complexity: O(max(log a, log b))
Auxiliary Space: O(1)