Number of special pairs possible from the given two numbers

Given two numbers A, B. The task is to find the numbers of special pairs of A, B. A special pair of two numbers A, B is a pair of numbers X, Y which satisfies the both of the given conditions – A = X | Y, B = X & Y.

Examples:

Input: A = 3, B = 0
Output: 2
(0, 3), (1, 2) will satisfy the conditions

Input:  A = 5, B = 7
Output: 0

Approach: The key observation here is that if we want the OR of two numbers, X, Y to be equal to A. Then both X, Y has to be less than or equal to A. If anyone is greater A then there OR won’t be equal to A. This will give us the limits where our loop will terminate, rest we will try and check if two pairs meet the given condition, then we will increment the counter.

Below is the required implementation:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of above approach
#include <iostream>
using namespace std;
  
// Function to count the pairs
int countPairs(int A, int B)
{
  
    // Variable to store a number of special pairs
    int cnt = 0;
  
    for (int i = 0; i <= A; ++i) {
        for (int j = i; j <= A; ++j) {
            // Calculating AND of i, j
            int AND = i & j;
  
            // Calculating OR of i, j
            int OR = i | j;
  
            // If the conditions are met,
            // then increment the count of special pairs
            if (OR == A and AND == B) {
                cnt++;
            }
        }
    }
    return cnt;
}
  
// Driver code
int main()
{
    int A = 3, B = 0;
    cout << countPairs(A, B);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of above approach
class GFG
{
  
// Function to count the pairs
static int countPairs(int A, int B)
{
  
    // Variable to store a number 
    // of special pairs
    int cnt = 0;
  
    for (int i = 0; i <= A; ++i) 
    {
        for (int j = i; j <= A; ++j) 
        {
            // Calculating AND of i, j
            int AND = i & j;
  
            // Calculating OR of i, j
            int OR = i | j;
  
            // If the conditions are met,
            // then increment the count
            // of special pairs
            if (OR == A && AND == B) 
            {
                cnt++;
            }
        }
    }
    return cnt;
}
  
// Driver code
public static void main(String [] args)
{
    int A = 3, B = 0;
    System.out.println(countPairs(A, B));
}
}
  
// This code is contributed by ihritik

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of above 
# approach
  
# Function to count the pairs
def countPairs(A,B):
  
    # Variable to store a number 
    # of special pairs
    cnt=0
    for i in range(0,A+1):
        for j in range(i,A+1):
  
            # Calculating AND of i, j
            AND = i&j
            OR = i|j
  
            # If the conditions are met,
            # then increment the count of 
            # special pairs
            if(OR==A and AND==B):
                cnt +=1
    return cnt
  
if __name__=='__main__':
    A = 3
    B = 0
    print(countPairs(A,B))
  
# This code is contributed by 
# Shrikant13

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of above approach
using System;
  
class GFG
{
      
// Function to count the pairs
static int countPairs(int A, int B)
{
  
    // Variable to store a number
    // of special pairs
    int cnt = 0;
  
    for (int i = 0; i <= A; ++i) 
    {
        for (int j = i; j <= A; ++j) 
        {
            // Calculating AND of i, j
            int AND = i & j;
  
            // Calculating OR of i, j
            int OR = i | j;
  
            // If the conditions are met,
            // then increment the count 
            // of special pairs
            if (OR == A && AND == B)
            {
                cnt++;
            }
        }
    }
    return cnt;
}
  
// Driver code
public static void Main()
{
    int A = 3, B = 0;
    Console.WriteLine(countPairs(A, B));
}
}
  
// This code is contributed by ihritik 

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP implementation of above approach
  
// Function to count the pairs
function countPairs($A, $B)
{
  
    // Variable to store a number
    // of special pairs
    $cnt = 0;
  
    for ($i = 0; $i <= $A; ++$i)
    {
        for ($j = $i; $j <= $A; ++$j
        {
            // Calculating AND of i, j
            $AND = $i & $j;
  
            // Calculating OR of i, j
            $OR = $i | $j;
  
            // If the conditions are met,
            // then increment the count 
            // of special pairs
            if ($OR == $A && $AND == $B)
            {
                $cnt++;
            }
        }
    }
    return $cnt;
}
          
// Driver code
$A = 3;
$B = 0;
echo countPairs($A, $B);
  
// This code is contributed by ihritik
?> 

chevron_right


Output:

2


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : shrikanth13, ihritik