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Count of pairs having bit size at most X and Bitwise OR equal to X

  • Last Updated : 30 Apr, 2021

Given a number X, calculate number of possible pairs (a, b) such that bitwise or of a and b is equal to X and number of bits in both a and b is less than equal to number of bits in X.

Examples: 
 

Input: X = 6 
Output:
Explanation: 
The possible pairs of (a, b) are (4, 6), (6, 4), (6, 6), (6, 2), (4, 2), (6, 0), (2, 6), (2, 4), (0, 6).
Input: X = 21 
Output: 27 
Explanation: 
In total there are 27 pairs possible.

Approach: To solve the problem mentioned above follow the steps given below:

  • Iterate through every bit of given number X.
  • If the bit is 1 then from the truth table of Bitwise OR we know that there are 3 combinations possible for that given bit in number a and b that is (0, 1), (1, 0), (1, 1) that is 3 possible ways.
  • If the bit is 0 then from the truth table of Bitwise OR we know that there is only 1 combination possible for that given bit in number a and b that is (0, 0).
  • So our answer will be answer will be 3 ^ (number of on bits in X).

Below is the implementation of above approach:
 



C++




// C++ implementation to Count number of
// possible pairs of (a, b) such that
// their Bitwise OR gives the value X
 
#include <iostream>
using namespace std;
 
// Function to count the pairs
int count_pairs(int x)
{
    // Initializing answer with 1
    int ans = 1;
 
    // Iterating through bits of x
    while (x > 0) {
 
        // check if bit is 1
        if (x % 2 == 1)
 
            // multiplying ans by 3
            // if bit is 1
            ans = ans * 3;
 
        x = x / 2;
    }
    return ans;
}
 
// Driver code
int main()
{
    int X = 6;
 
    cout << count_pairs(X)
         << endl;
 
    return 0;
}

Java




// Java implementation to count number of
// possible pairs of (a, b) such that
// their Bitwise OR gives the value X
class GFG{
 
// Function to count the pairs
static int count_pairs(int x)
{
     
    // Initializing answer with 1
    int ans = 1;
 
    // Iterating through bits of x
    while (x > 0)
    {
         
        // Check if bit is 1
        if (x % 2 == 1)
 
            // Multiplying ans by 3
            // if bit is 1
            ans = ans * 3;
 
        x = x / 2;
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int X = 6;
 
    System.out.print(count_pairs(X) + "\n");
}
}
 
// This code is contributed by amal kumar choubey

Python3




# Python3 implementation to count number of
# possible pairs of (a, b) such that
# their Bitwise OR gives the value X
 
# Function to count the pairs
def count_pairs(x):
 
    # Initializing answer with 1
    ans = 1;
 
    # Iterating through bits of x
    while (x > 0):
 
        # Check if bit is 1
        if (x % 2 == 1):
 
            # Multiplying ans by 3
            # if bit is 1
            ans = ans * 3;
 
        x = x // 2;
     
    return ans;
 
# Driver code
if __name__ == '__main__':
     
    X = 6;
 
    print(count_pairs(X));
 
# This code is contributed by amal kumar choubey

C#




// C# implementation to count number of
// possible pairs of (a, b) such that
// their Bitwise OR gives the value X
using System;
class GFG{
 
  // Function to count the pairs
  static int count_pairs(int x)
  {
 
    // Initializing answer with 1
    int ans = 1;
 
    // Iterating through bits of x
    while (x > 0)
    {
 
      // Check if bit is 1
      if (x % 2 == 1)
 
        // Multiplying ans by 3
        // if bit is 1
        ans = ans * 3;
 
      x = x / 2;
    }
    return ans;
  }
 
  // Driver code
  public static void Main(String[] args)
  {
    int X = 6;
 
    Console.Write(count_pairs(X) + "\n");
  }
}
 
// This code is contributed by sapnasingh4991

Javascript




<script>
 
// javascript implementation to count number of
// possible pairs of (a, b) such that
// their Bitwise OR gives the value X   
// Function to count the pairs
    function count_pairs(x) {
 
        // Initializing answer with 1
        var ans = 1;
 
        // Iterating through bits of x
        while (x > 0) {
 
            // Check if bit is 1
            if (x % 2 == 1)
 
                // Multiplying ans by 3
                // if bit is 1
                ans = ans * 3;
 
            x = parseInt(x / 2);
        }
        return ans;
    }
 
    // Driver code
     
        var X = 6;
 
        document.write(count_pairs(X) + "\n");
 
// This code contributed by Rajput-Ji
 
</script>
Output: 
9

 

Time complexity: O(log(X))
 

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