Find all possible pairs with given Bitwise OR and Bitwise XOR values

Given two positive integers A and B representing Bitwise XOR and Bitwise OR of two positive integers, the task is to find all possible pairs (x, y) such that x ^ y is equal to A and x | y is equal to B.

Examples:

Input: A = 5, B = 7
Output:
2 7
3 6
6 3
7 2
Explanation:
7( XOR )2 = 5 and 7( OR )2 = 7
3( XOR )6 = 5 and 3( OR )6 = 7

Input: A = 8, B = 10
Output:
2 10
10 2

Brute Force Approach:

The brute force approach to solve this problem is to generate all possible pairs of positive integers and check if their bitwise XOR is equal to A and bitwise OR is equal to B. This can be done using two nested loops where the outer loop iterates through all positive integers up to B and the inner loop iterates through all positive integers up to the current index of the outer loop.

Here are the steps of approach:

1. Iterate through all positive integers up to B using a for loop.
2. For each integer i, iterate through all positive integers up to i using another for loop.
3. Check if the bitwise XOR of i and j is equal to A and the bitwise OR of i and j is equal to B.
4. If the condition is true, print the pair (j, i).
5. Continue with the outer loop until all positive integers up to B have been processed.
6. The output will be all the pairs of positive integers (x, y) such that x^y is equal to A and x|y is equal to B.

Below is the implementation of the above approach:

// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find pairs with
// XOR equal to A and OR equal to B
void findPairs(int A, int B)
{
    for(int i=1; i<=B; i++){
        for(int j=1; j<=i; j++){
            if((i^j)==A && (i|j)==B){
                cout<<j<<" "<<i<<endl;
                if(i!=j)
                    cout<<i<<" "<<j<<endl;
            }
        }
    }
}



// Driver Code
int main()
{
    int A = 8, B = 10;

    findPairs(A, B);
    return 0;
}

// Java program for the above approach
import java.util.*;

class GFG {

    // Function to find pairs with
    // XOR equal to A and OR equal to B
    static void findPairs(int A, int B)
    {
        for (int i = 1; i <= B; i++) {
            for (int j = 1; j <= i; j++) {
                if ((i ^ j) == A && (i | j) == B) {
                    System.out.println(j + " " + i + "\n");
                    if (i != j)
                        System.out.println(i + " " + j
                                           + "\n");
                }
            }
        }
    }

    // Driver Code
    public static void main(String[] args)
    {
        int A = 8, B = 10;

        findPairs(A, B);
    }
}

# Python3 code for the above approach

# Function to find pairs with
# XOR equal to A and OR equal to B
def findPairs(A, B):
    for i in range(1, B+1):
        for j in range(1, i+1):
            if (i ^ j) == A and (i | j) == B:
                print(j, i)
                if i != j:
                    print(i, j)

A = 8
B = 10
findPairs(A, B)

// C# code for the above approach
using System;

public class GFG
{   
    // Function to find pairs with
    // XOR equal to A and OR equal to B
    public static void FindPairs(int A, int B)
    {
        for (int i = 1; i <= B; i++)
        {
            for (int j = 1; j <= i; j++)
            {
                if ((i ^ j) == A && (i | j) == B)
                {
                    Console.WriteLine(j + " " + i);
                    if (i != j)
                        Console.WriteLine(i + " " + j);
                }
            }
        }
    }

    // Driver Code
    public static void Main()
    {
        int A = 8, B = 10;
        FindPairs(A, B);
    }
}

// Javascript code for the above approach

// Function to find pairs with
// XOR equal to A and OR equal to B
function findPairs(A, B) {
  for (let i = 1; i <= B; i++) {
    for (let j = 1; j <= i; j++) {
      if ((i ^ j) === A && (i | j) === B) {
        console.log(j + " " + i);
        if (i !== j) console.log(i + " " + j);
      }
    }
  }
}

// Driver Code
const A = 8, B = 10;
findPairs(A, B);

2 10
10 2

Time Complexity: O(B^2), as we are using nested loops to iterate through all possible pairs of positive integers up to B.
Space Complexity: O(1), as we are not using any extra space.

Efficient Approach: The idea is to traverse through all possible values of x and use the property of XOR that if x ^ y = A, then x ^ A = y to find all possible values of y. Follow the steps below to solve the problem:

• Iterate from 1 to B using a variable, say i, and perform the following operations: 
  
  • Initialize a variable y as i ^ A.
  • Check if the value of y is greater than 0 and (i | y) is equal to B or not.
  • If found to be true, then print the values of i and y.

Below is the implementation of the above approach:

// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find pairs with
// XOR equal to A and OR equal to B
void findPairs(int A, int B)
{
    // Iterate from 1 to B
    for (int i = 1; i <= B; i++) {
        int y = A ^ i;

        // Check if (i OR y) is B
        if (y > 0 and (i | y) == B) {
            cout << i << " " << y << endl;
        }
    }
}

// Driver Code
int main()
{
    int A = 8, B = 10;

    findPairs(A, B);
    return 0;
}

// Java code for the above approach
import java.util.*;

class GFG{

// Function to find pairs with
// XOR equal to A and OR equal to B
static void findPairs(int A, int B)
{
    
    // Iterate from 1 to B
    for(int i = 1; i <= B; i++) 
    {
        int y = A ^ i;

        // Check if (i OR y) is B
        if (y > 0 && (i | y) == B) 
        {
            System.out.println(i + " " + y);
        }
    }
}

// Driver Code
public static void main(String[] args)
{
    int A = 8, B = 10;

    findPairs(A, B);
}
}

// This code is contributed by Hritik

# Python3 code for the above approach

# Function to find pairs with
# XOR equal to A and OR equal to B
def findPairs(A, B):
    
    # Iterate from 1 to B
    for i in range(1, B + 1):
        
        y = A ^ i
        
        # Check if (i OR y) is B
        if (y > 0 and (i | y) == B):
            print(i, " ", y)

# Driver Code
A = 8
B = 10

findPairs(A, B)

# This code is contributed by amreshkumar3

// C# code for the above approach
using System;
class GFG
{
    
    // Function to find pairs with
    // XOR equal to A and OR equal to B
    static void findPairs(int A, int B)
    {
         
        // Iterate from 1 to B
        for(int i = 1; i <= B; i++)
        {
            int y = A ^ i;
     
            // Check if (i OR y) is B
            if (y > 0 && (i | y) == B)
            {
                Console.WriteLine(i + " " + y);
            }
        }
    }

  // Driver code
  static void Main ()
  {
    int A = 8, B = 10;
 
    findPairs(A, B);
  }
}

// This code is contributed by suresh07.

<script>

// JavaScript code for the above approach


// Function to find pairs with
// XOR equal to A and OR equal to B
function findPairs(A, B) {
    // Iterate from 1 to B
    for (let i = 1; i <= B; i++) {
        let y = A ^ i;

        // Check if (i OR y) is B
        if (y > 0 && (i | y) == B) {
            document.write(i + " " + y + "<br>");
        }
    }
}

// Driver Code

let A = 8, B = 10;

findPairs(A, B);


// This code is contributed by gfgking

</script>

2 10
10 2

Time Complexity: O(B)
Auxiliary Space: O(1)

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Article Tags :

• Bit Magic
• Searching
• Mathematical
• Competitive Programming
• DSA
• Bitwise-XOR
• Bitwise-OR

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