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Count of pairs having bit size at most X and Bitwise OR equal to X
  • Last Updated : 26 Aug, 2020

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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 

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C#

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// 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

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Output: 

9

Time complexity: O(log(X))
 

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