Find the value of X and Y such that X Bitwise XOR Y equals to (X+Y)/2

Given an integer N, representing X ⊕ Y where ⊕ represents Bitwise XOR, the task is to find the value of X and Y such that X ⊕ Y = (X+Y)/2. If no possible value of X and Y exist, print -1.

Examples:

Input: N = 20
Output: X = 30, Y = 10
Explanation: X ⊕ Y = 20 and (X+Y)/2 = 20

Input: N = 6
Output: -1
Explanation: No valid pair of X and Y exists.

Approach: The problem can be solved using the following approach:

Since it is given X ⊕ Y= (X + Y)/2,

=> 2 * (X ⊕ Y) = X + Y (Equation 1)

We know, A + B = A ⊕ B + 2 * (A & B), where & is the Bitwise AND operator

Thus Equation 1 can be modified to:

X ⊕ Y= 2 * (X & Y)

From the above equation we can say that X ⊕ Y should be even otherwise answer won’t exist. Let N =X ⊕ Y and M= (X ⊕ Y)/2, i.e., N represents Bitwise XOR and M represents Bitwise AND. It can also be observed that N and M cannot have a particular bit set simultaneously because if a particular bit is set in both N and M, then it would mean that the Bitwise AND as well as Bitwise XOR both have the bit as set which is impossible. Now, we will find X and Y bit by bit. For an ith bit, we will have 3 cases:

• Case 1: N have ith bit set: In this case, either of X or Y should have the ith bit set since N is bitwise XOR of X and Y. So will add (1<<i) to X.
• Case 2: M have ith bit set: In this case, both of X and Y should have the ith bit set since M is bitwise AND of X and Y. So will add (1<<i) to X and Y.
• Case 3: Neither N and M have ith bit set: In this case, both of X and Y will not have the ith bit set.

Below is the implementation of above approach:

C++

#include <bits/stdc++.h>

using namespace std;

void solve(int N)
{
      // (X XOR Y) should be an even number
    if (N % 2) {
        cout << -1 << "\n";
    }

      // M = X & Y
    int M = N / 2;

      // No bit should be set in both M and N
    if ((N & M) != 0) {
        cout << -1 << "\n";
    }

      int X = 0, Y = 0;
    for (int i = 0; i < 31; i++) {
          // If ith bit of (X XOR Y) is set
        if ((N & (1 << i)) != 0) {
            X = X + (1 << i);
        }
          // If ith bit of (X AND Y) is set
        else if ((M & (1 << i)) != 0) {
            X = X + (1 << i);
            Y = Y + (1 << i);
        }
    }
    cout << X  << " " << Y << "\n";
}

// Driver Code
int main()
{
    int N = 20;
    solve(N);
}

Java

public class Main {
    public static void solve(int N) {
        // (X XOR Y) should be an even number
        if (N % 2 != 0) {
            System.out.println(-1);
            return;
        }

        // M = X & Y
        int M = N / 2;

        // No bit should be set in both M and N
        if ((N & M) != 0) {
            System.out.println(-1);
            return;
        }

        int X = 0, Y = 0;
        for (int i = 0; i < 31; i++) {
            // If ith bit of (X XOR Y) is set
            if ((N & (1 << i)) != 0) {
                X = X + (1 << i);
            }
            // If ith bit of (X AND Y) is set
            else if ((M & (1 << i)) != 0) {
                X = X + (1 << i);
                Y = Y + (1 << i);
            }
        }
        System.out.println(X + " " + Y);
    }

    // Driver Code
    public static void main(String[] args) {
        int N = 20;
        solve(N);
    }
}


// This code is contributed by rambabuguphka

Python3

def solve(N):
    # (X XOR Y) should be an even number
    if N % 2:
        print(-1)
        return

    # M = X & Y
    M = N // 2

    # No bit should be set in both M and N
    if N & M:
        print(-1)
        return

    X, Y = 0, 0
    for i in range(31):
        # If ith bit of (X XOR Y) is set
        if N & (1 << i):
            X = X + (1 << i)
        # If ith bit of (X AND Y) is set
        elif M & (1 << i):
            X = X + (1 << i)
            Y = Y + (1 << i)

    print(X, Y)

# Driver Code
N = 20
solve(N)

C#

using System;

public class Program
{
    public static void Solve(int N)
    {
        // (X XOR Y) should be an even number
        if (N % 2 != 0)
        {
            Console.WriteLine(-1);
            return;
        }

        // M = X & Y
        int M = N / 2;

        // No bit should be set in both M and N
        if ((N & M) != 0)
        {
            Console.WriteLine(-1);
            return;
        }

        int X = 0, Y = 0;
        for (int i = 0; i < 31; i++)
        {
            // If ith bit of (X XOR Y) is set
            if ((N & (1 << i)) != 0)
            {
                X = X + (1 << i);
            }
            // If ith bit of (X AND Y) is set
            else if ((M & (1 << i)) != 0)
            {
                X = X + (1 << i);
                Y = Y + (1 << i);
            }
        }
        Console.WriteLine(X + " " + Y);
    }

    // Entry point
    public static void Main(string[] args)
    {
        int N = 20;
        Solve(N);
    }
}

Javascript

// javaScript code for the above approach

function solve(N) {
    // (X XOR Y) should be an even number
    if (N % 2) {
        console.log(-1);
        return;
    }

    // M = X & Y
    let M = N / 2;

    // No bit should be set in both M and N
    if ((N & M) !== 0) {
        console.log(-1);
        return;
    }

    let X = 0, Y = 0;
    for (let i = 0; i < 31; i++) {
        // If ith bit of (X XOR Y) is set
        if ((N & (1 << i)) !== 0) {
            X = X + (1 << i);
        }
        // If ith bit of (X AND Y) is set
        else if ((M & (1 << i)) !== 0) {
            X = X + (1 << i);
            Y = Y + (1 << i);
        }
    }
    console.log(X + " " + Y);
}

// Driver Code
let N = 20;
solve(N);

Output

30 10

Time Complexity: O(log N), where N is the input number ( X ⊕ Y )
Auxiliary Space: O(1)