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Calculate Bitwise OR of two integers from their given Bitwise AND and Bitwise XOR values

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Given two integers X and Y, representing Bitwise XOR and Bitwise AND of two positive integers, the task is to calculate the Bitwise OR value of those two positive integers.

Examples:

Input: X = 5, Y = 2 
Output:
Explanation: 
If A and B are two positive integers such that A ^ B = 5, A & B = 2, then the possible value of A and B is 3 and 6 respectively. 
Therefore, (A | B) = (3 | 6) = 7.

Input: X = 14, Y = 1 
Output: 15 
Explanation: 
If A and B are two positive integers such that A ^ B = 14, A & B = 1, then the possible value of A and B is 7 and 9 respectively. 
Therefore, (A | B) = (7 | 9) = 15.

Naive Approach: The simplest approach to solve this problem is to iterate up to the maximum of X and Y, say N, and generate all possible pairs of the first N natural numbers. For each pair, check if Bitwise XOR and the Bitwise AND of the pair is X and Y, respectively, or not. If found to be true, then print the Bitwise OR of that pair.

Below is the implementation of the above approach:

C++

// C++ program to implement the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate Bitwise OR from given
// bitwise XOR and bitwise AND values
int findBitwiseORGivenXORAND(int X, int Y)
{
    int range = X + Y;
    // Find the max range
    int ans = 0;
    // Traversing all the number from 0 to rangr
    for (int i = 1; i <= range; i++) {
        for (int j = 1; j <= range; j++) {
            // If X and Y satisfie
            if ((i ^ j) == X && (i & j) == Y) {
                ans = (i | j);
                // Calculate the OR
                break;
            }
        }
    }
    return ans;
}
 
// Driver Code
int main()
{
    int X = 5, Y = 2;
    cout << findBitwiseORGivenXORAND(X, Y);
}

                    

Java

// Java program to implement the above approach
import java.util.*;
 
public class GFG {
 
    // Function to calculate Bitwise OR from given
    // bitwise XOR and bitwise AND values
    static int findBitwiseORGivenXORAND(int X, int Y)
    {
        int range = X + Y;
        // Find the max range
        int ans = 0;
        // Traversing all the numbers from 0 to range
        for (int i = 1; i <= range; i++) {
            for (int j = 1; j <= range; j++) {
                // If X and Y satisfy the conditions
                if ((i ^ j) == X && (i & j) == Y) {
                    ans = (i | j);
                    // Calculate the OR
                    break;
                }
            }
        }
        return ans;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int X = 5, Y = 2;
        System.out.println(findBitwiseORGivenXORAND(X, Y));
    }
}
 
// This code is contributed by Susobhan Akhuli

                    

Python3

# Python program to implement the above approach
# Function to calculate Bitwise OR from given
# bitwise XOR and bitwise AND values
def findBitwiseORGivenXORAND(X, Y):
    range_val = X + Y
    # Find the max range
    ans = 0
    # Traversing all the numbers from 0 to range_val
    for i in range(1, range_val + 1):
        for j in range(1, range_val + 1):
            # If X and Y satisfy the conditions
            if (i ^ j) == X and (i & j) == Y:
                ans = (i | j)
                # Calculate the OR
                break
    return ans
 
# Driver Code
def main():
    X = 5
    Y = 2
    print(findBitwiseORGivenXORAND(X, Y))
 
if __name__ == "__main__":
    main()
 
# This code is contributed by Susobhan Akhuli

                    

C#

// C# program to implement the above approach
using System;
 
public class GFG {
    // Function to calculate Bitwise OR from given
    // bitwise XOR and bitwise AND values
    static int FindBitwiseORGivenXORAND(int X, int Y)
    {
        int range = X + Y;
        int ans = 0;
 
        // Traversing all the numbers from 0 to range
        for (int i = 1; i <= range; i++) {
            for (int j = 1; j <= range; j++) {
                // If X and Y satisfy the conditions
                if ((i ^ j) == X && (i & j) == Y) {
                    ans = (i | j); // Calculate the OR
                    break;
                }
            }
        }
        return ans;
    }
 
    static void Main()
    {
        int X = 5, Y = 2;
        Console.WriteLine(FindBitwiseORGivenXORAND(X, Y));
    }
}
 
// This code is contributed by Susobhan Akhuli

                    

Javascript

// Javascript program to implement the above approach
 
// Function to calculate Bitwise OR from given
// bitwise XOR and bitwise AND values
function findBitwiseORGivenXORAND(X, Y) {
    let range = X + Y;
    // Find the max range
    let ans = 0;
    // Traversing all the number from 0 to range
    for (let i = 1; i <= range; i++) {
        for (let j = 1; j <= range; j++) {
            // If X and Y satisfy
            if ((i ^ j) === X && (i & j) === Y) {
                ans = (i | j);
                // Calculate the OR
                break;
            }
        }
    }
    return ans;
}
 
// Driver Code
let X = 5, Y = 2;
console.log(findBitwiseORGivenXORAND(X, Y));

                    

Output
7







Time Complexity: O(N2), where N = (X+Y) 
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is based on the following observations:

(A ^ B) = (A | B) – (A & B) 
=> (A | B) = (A ^ B) + (A & B) = X + Y 
 

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate Bitwise OR from given
// bitwise XOR and bitwise AND values
int findBitwiseORGivenXORAND(int X, int Y) { return X + Y; }
 
// Driver Code
int main()
{
    int X = 5, Y = 2;
    cout << findBitwiseORGivenXORAND(X, Y);
}

                    

C

// C program to implement
// the above approach
#include <stdio.h>
 
// Function to calculate Bitwise OR from given
// bitwise XOR and bitwise AND values
int findBitwiseORGivenXORAND(int X, int Y)
{
  return X | Y;
}
 
// Driver Code
int main()
{
    int X = 5, Y = 2;
    printf("%d\n", findBitwiseORGivenXORAND(X, Y));
}
 
// This code is contributed by phalashi.

                    

Java

// Java program to implement
// the above approach
class GFG {
 
    // Function to calculate Bitwise OR from given
    // bitwise XOR and bitwise AND values
    static int findBitwiseORGivenXORAND(int X, int Y)
    {
        return X + Y;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int X = 5, Y = 2;
        System.out.print(findBitwiseORGivenXORAND(X, Y));
    }
}
 
// This code is contributed by AnkitRai01

                    

Python3

# Python3 program to implement
# the above approach
 
# Function to calculate Bitwise OR from
# given bitwise XOR and bitwise AND values
 
 
def findBitwiseORGivenXORAND(X, Y):
 
    return X + Y
 
 
# Driver Code
if __name__ == "__main__":
 
    X = 5
    Y = 2
 
    print(findBitwiseORGivenXORAND(X, Y))
 
# This code is contributed by AnkitRai01

                    

C#

// C# program to implement
// the above approach
using System;
 
class GFG {
 
    // Function to calculate Bitwise OR from given
    // bitwise XOR and bitwise AND values
    static int findBitwiseORGivenXORAND(int X, int Y)
    {
        return X + Y;
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        int X = 5, Y = 2;
 
        Console.Write(findBitwiseORGivenXORAND(X, Y));
    }
}
 
// This code is contributed by ipg2016107

                    

Javascript

<script>
// JavaScript program to implement
// the above approach
 
// Function to calculate Bitwise OR from given
// bitwise XOR and bitwise AND values
function findBitwiseORGivenXORAND(X, Y)
{
    return X + Y;
}
 
// Driver Code
 
    let X = 5, Y = 2;
    document.write(findBitwiseORGivenXORAND(X, Y));
 
// This code is contributed by Surbhi Tyagi.
 
</script>

                    

Output
7







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



Last Updated : 04 Dec, 2023
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