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; }

  Java

  
  

  
  
  

  public class Main {     // Function to count possible values of X whose Bitwise OR with A is equal to B     static int numWays(int a, int b) {         // Store the required result         int res = 1;           // Iterate over all bits         while (a != 0 && b != 0) {             // 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 *= 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     public static void main(String[] args) {         // Given Input         int a = 2, b = 3;           // Function Call         System.out.println(numWays(a, b));     } } //This code is contributed by Utkarsh

  Python3

  
  

  
  
  

  def num_ways(a, b):     """     Counts the number of possible values of X whose bitwise OR with A is equal to B.       :param a: The first integer A.     :param b: The second integer B.     :return: The number of ways X can be formed.     """     # Store the required result     res = 1       # Iterate over all bits     while a != 0 and b != 0:         # 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 >>= 1         b >>= 1       return res   a = 2 b = 3   # Function Call result = num_ways(a, b) print(result)

  C#

  
  

  
  
  

  using System;   class Program {     // Function to count possible values of     // X whose Bitwise OR with A is equal to B     static int NumWays(int a, int b)     {         // Store the required result         int res = 1;           // Iterate over all bits         while (a != 0 && b != 0)         {             // 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 *= 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 >>= 1;             b >>= 1;         }           return res;     }       // Driver Code     static void Main()     {         // Given Input         int a = 2, b = 3;           // Function Call         Console.WriteLine(NumWays(a, b));     } } //THis code is contributed by Monu.

  Javascript

  
  

  
  
  

  // Function to count possible values of // X whose Bitwise OR with A is equal to B function numWays(a, b) {       // Store the required result     let 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 function main() {     // Given Input     let a = 2, b = 3;       // Function Call     console.log(numWays(a, b)); }   // Run the main function main();

• Output

  2
  
  
  
  
  
  

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