Count of possible pairs whose sum and bitwise XOR is given

Given two integers S and X representing the sum and bitwise XOR respectively of two integers, the task is to find the count of all such possible pairs such that their sum is equal to S and bitwise XOR is equal to X.

Examples:

Input: S = 9, X = 5
Output: 4
Explanation: (2, 7), (3, 6), (6, 3), (7, 2) completely satisfies the given conditions.
It can also be seen that no other pair satisfy the given condition.
So the total possibility is 4.

Input: S = 3, X = 3
Output: 4
Explanation: Only (1, 2), (2, 1), (3, 0) and (0, 3) satisfy the given condition.

Approach: The solution to the problem is based on the following observation:

Consider two numbers a and b. Their sum can be written as
a+b = (a XOR b) + (a AND b)*2

The above can be proved as follows:
If two bits x and y are added, the carry (say c) will be (x AND y) and it is to one position left of the single bit. So, c = 2 * (x AND y)
The value of the rightmost bit (i.e., at the same position as of the single bits) will be (x XOR y)
Therefore x + y = (x XOR y) + 2 * (x AND y)

So when adding a and b, this procedure repeats for each bit. Therefore it can be said that the sum will be 
a+b = (a XOR b) + (a AND b)*2 
S = X + 2*A   where A is the bitwise AND of the two numbers.
A = (S - X)/2

Based on the above observation the count can be obtained from the bit values of A and X as per the following case

• If the i^th bit of X and A both are 0, then there is only one possible pair for that bit position.
• If the i^th bit of X is 1 and of A is 0, then there are two possible pairs for that bit position.
• If the i^th bit of X is 0 and of A is 1, then there is only one possible pair for that bit position.
• There cannot be a situation where ith bit at X is 1 and i^th bit of A is also 1.

According to the multiplication principle of permutation, to get the number of possible pairs satisfying the given condition just multiply all possibilities. Follow the steps mentioned below to implement the idea:

• Find bitwise AND of two numbers by using the above formula.
• If it is not an integer then no solution exist 
• Otherwise for given value of the XOR (X) and AND (say A) check for each bit 
  • Find the number of possibilities for that bit using the above cases.
  • Multiply these possibilities with the final answer. 
• At the end the total number possibilities calculated as per the above conditions is the required answer.

   Below is the implementation of the above approach.

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to calculate
// total number of possible pairs
int count_all_possible_pair(int sum_value,
                            int xor_value)
{
    double and_value
        = ((sum_value - xor_value) / 2.0);

    // If and value is not an integer
    // then no pair is possible
    int check = and_value;
    if (check != and_value)
        return 0;

    int count = 1;

    // Traversing the bits of the sum_value
    // from MSB position
    for (int i = 1; i <= 32; i++) {
        int and_bit = (check >> i) & 1;
        int xor_bit = (xor_value >> i) & 1;

        // If both the bit is 0, only 1 possibility
        if (and_bit == 0 && xor_bit == 0)
            count = count * 1;

        // If Xor bit is 0 and And bit 1,
        // then only 1 possibility
        else if (xor_bit == 0 && and_bit == 1)
            count = count * 1;

        // If Xor bit is 1, And bit is 0,
        // then there are 2 possibilities
        else if (xor_bit == 1 && and_bit == 0)
            count = count * 2;

        // If Xor bit and And bit both 1,
        // no such case is possible
        else
            return 0;
    }

    // Return the count of possible pairs
    return count;
}

// Driver code
int main()
{
    int S = 9, X = 5;

    // Function call
    cout << count_all_possible_pair(S, X);
    return 0;
}

// Java code to implement the approach

public class GFG {
    
    // Function to calculate
    // total number of possible pairs
    static int count_all_possible_pair(int sum_value,
                                int xor_value)
    {
        double and_value
            = ((sum_value - xor_value) / 2.0);
    
        // If and value is not an integer
        // then no pair is possible
        int check = (int)and_value;
        if (check != and_value)
            return 0;
    
        int count = 1;
    
        // Traversing the bits of the sum_value
        // from MSB position
        for (int i = 1; i <= 32; i++) {
            int and_bit = (check >> i) & 1;
            int xor_bit = (xor_value >> i) & 1;
    
            // If both the bit is 0, only 1 possibility
            if (and_bit == 0 && xor_bit == 0)
                count = count * 1;
    
            // If Xor bit is 0 and And bit 1,
            // then only 1 possibility
            else if (xor_bit == 0 && and_bit == 1)
                count = count * 1;
    
            // If Xor bit is 1, And bit is 0,
            // then there are 2 possibilities
            else if (xor_bit == 1 && and_bit == 0)
                count = count * 2;
    
            // If Xor bit and And bit both 1,
            // no such case is possible
            else
                return 0;
        }
    
        // Return the count of possible pairs
        return count;
    }
    
    // Driver code
    public static void main (String[] args)
    {
        int S = 9, X = 5;
    
        // Function call
        System.out.println(count_all_possible_pair(S, X));
    }
}

// This code is contributed by AnkThon

# Python3 code to implement the approach

# Function to calculate
# total number of possible pairs
def count_all_possible_pair(sum_value, xor_value) :
    and_value = ((sum_value - xor_value) / 2.0);

    # If and value is not an integer
    # then no pair is possible
    check = int(and_value);
    if (check != and_value) :
        return 0;

    count = 1;

    # Traversing the bits of the sum_value
    # from MSB position
    for i in range(1 , 33) :
        and_bit = ((check >> i) & 1);
        xor_bit = ((xor_value >> i) & 1);

        # If both the bit is 0, only 1 possibility
        if (and_bit == 0 and xor_bit == 0) :
            count = count * 1;

        # If Xor bit is 0 and And bit 1,
        # then only 1 possibility
        elif (xor_bit == 0 and and_bit == 1) :
            count = count * 2;

        # If Xor bit is 1, And bit is 0,
        # then there are 2 possibilities
        elif (xor_bit == 1 and and_bit == 0) :
            count = count * 2;

        # If Xor bit and And bit both 1,
        # no such case is possible
        else :
            return 0;

    # Return the count of possible pairs
    return count;

# Driver code
if __name__ == "__main__" :

    S = 9; X = 5;

    # Function call
    print(count_all_possible_pair(S, X));

    # This code is contributed by AnkThon

// C# code to implement the approach
using System;
public class GFG {

  // Function to calculate
  // total number of possible pairs
  static int count_all_possible_pair(int sum_value,
                                     int xor_value)
  {
    double and_value
      = ((sum_value - xor_value) / 2.0);

    // If and value is not an integer
    // then no pair is possible
    int check = (int)and_value;
    if (check != and_value)
      return 0;

    int count = 1;

    // Traversing the bits of the sum_value
    // from MSB position
    for (int i = 1; i <= 32; i++) {
      int and_bit = (check >> i) & 1;
      int xor_bit = (xor_value >> i) & 1;

      Console.WriteLine("value:" + i);
      Console.WriteLine(and_bit+"test"+xor_bit);
      // If both the bit is 0, only 1 possibility
      if (and_bit == 0 && xor_bit == 0)
        count = count * 1;

      // If Xor bit is 0 and And bit 1,
      // then only 1 possibility
      else if (xor_bit == 0 && and_bit == 1)
        count = count * 1;

      // If Xor bit is 1, And bit is 0,
      // then there are 2 possibilities
      else if (xor_bit == 1 && and_bit == 0)
        count = count * 2;

      // If Xor bit and And bit both 1,
      // no such case is possible
      else
        return 0;
    }

    // Return the count of possible pairs
    return count;
  }

  // Driver code
  public static void Main ()
  {
    int S = 9, X = 5;

    // Function call
    Console.Write(count_all_possible_pair(S, X));
  }
}

// This code is contributed by gfgking

// JavaScript code to implement the approach

// Function to calculate
// total number of possible pairs
function count_all_possible_pair(sum_value, xor_value)
{
    var and_value = Math.floor((sum_value - xor_value) / 2.0);

    // If and value is not an integer
    // then no pair is possible
    var check = and_value;
    if (check != and_value)
        return 0;

    var count = 1;

    // Traversing the bits of the sum_value
    // from MSB position
    for (var i = 1; i <= 32; i++) {
        var and_bit = (check >> i) & 1;
        var xor_bit = (xor_value >> i) & 1;

        // If both the bit is 0, only 1 possibility
        if (and_bit == 0 && xor_bit == 0)
            count = count * 1;

        // If Xor bit is 0 and And bit 1,
        // then only 1 possibility
        else if (xor_bit == 0 && and_bit == 1)
            count = count * 1;

        // If Xor bit is 1, And bit is 0,
        // then there are 2 possibilities
        else if (xor_bit == 1 && and_bit == 0)
            count = count * 2;

        // If Xor bit and And bit both 1,
        // no such case is possible
        else
            return 0;
    }

    // Return the count of possible pairs
    return count;
}

// Driver code
var S = 9;
var X = 5;

// Function call
document.write(count_all_possible_pair(S, X));

// This code is contributed by phasing17

4

Time Complexity: O(N) where N is the number of bits in the given sum S
Auxiliary Space: O(1)

refertoyash

Follow

Improve

Article Tags :

• Bit Magic
• DSA
• permutation
• Bitwise-XOR

Practice Tags :

• Bit Magic
• permutation