Find the values of X and Y in the Given Equations – GeeksforGeeks

Given two numbers and . Find the values of X and Y in the equations.

1. A = X + Y
2. B = X xor Y

The task is to make ten a minimum as potential. If it is not possible to find any valid values for X and Y then print -1.
Examples:

Input : A = 12, B = 8
Output : X = 2, Y = 10

Input : A = 12, B = 9
Output : -1

Let ’ s take a look at some bite in X, which is peer to 1. If the respective bit in Y is peer to 0, then one can swap these two bits, frankincense reducing X and increasing Y without changing their sum and xor. We can conclude that if some moment in X is equal to 1 then the respective bit in Y is besides peer to 1. frankincense, Y = X + B. Taking into account that X + Y = X + X + B = A, one can obtain the follow rule for finding X and Y :

• X = (A – B) / 2
• Y = X + B = (A + B) / 2

One should besides notice that if A < B or A and B have different parity bit, then the suffice doesn ’ thyroxine exist and end product is -1. If x and ( A – X ) not equal to X then the answer is besides -1.
Below is the execution of the above approach :

C++

#include using namespace std; void findValues( int a, int b) {           if ((a - b) % 2 == 1) {          cout << "-1" ;          return ;      }           cout << (a - b) / 2 << " " << (a + b) / 2; } int main() {      int a = 12, b = 8;      findValues(a, b);      return 0; }

Java

import java.io.*; class GFG {      static void findValues( int a, int b) {           if ((a - b) % 2 == 1 )      {              System.out.println ( "-1" );          return ;      }           System.out.println (((a - b) / 2 )+ " " +                              ((a + b) / 2 )); }           public static void main (String[] args)      {          int a = 12 , b = 8 ;          findValues(a, b);      } }

Python3

def findValues(a, b):           if ((a - b) % 2 = = 1 ):          print ( "-1" );          return ;           print ((a - b) / / 2 , (a + b) / / 2 ); Read more: How to Go Sugar-Free for a Week Straight Without Hating Your Life a = 12 ; b = 8 ; findValues(a, b);

C#

using System; class GFG {          static void findValues( int a, int b) {           if ((a - b) % 2 == 1)      {              Console.Write ( "-1" );          return ;      }           Console.WriteLine(((a - b) / 2)+ " " +                          ((a + b) / 2)); } static public void Main () {      int a = 12, b = 8;      findValues(a, b); } }

PHP

function findValues( $a , $b ) {           if (( $a - $b ) % 2 == 1)      {          echo "-1" ;          return ;      }           echo ( $a - $b ) / 2, " " ,           ( $a + $b ) / 2; } $a = 12; $b = 8; findValues( $a , $b ); ?>

Javascript

Output: 

2 10

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

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