Count values whose Bitwise OR with A is equal to B
Last Updated :
15 Feb, 2024
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 ith bit in A can be converted to the ith 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;
}
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
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)
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.
// 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();
- Time Complexity: O(max(log a, log b))
Auxiliary Space: O(1)
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