Given two integers X and Y, the task is to find the two integers having sum X and Bitwise XOR equal to Y.
Examples:
Input: X = 17, Y = 13
Output: 2 15
Explanation: 2 + 15 = 17 and 2 ^ 15 = 13Input: X = 1870807699, Y = 259801747
Output: 805502976 1065304723
Naive Approach: Refer to the previous post of this article for the simplest approach to solve the problem.
Time Complexity: O(log N)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized based on the following observations:
A + B = (A ^ B) + 2 * (A & B)
=> X = Y + 2 * (A & B)While calculating sum, if both bits are 1(i.e., AND is 1), the resultant bit is 0, and 1 is added as carry, which means every bit in AND is left-shifted by 1, i.e. value of AND is multiplied by 2 and added.
Rearranging the terms, the expression (A&B) = (X – Y) / 2 is obtained.
This verifies the above observation.
There exist the following cases:
- If X < Y: In this case, the solution does not exist because (A & B) becomes negative which is not possible.
- If X – Y is odd: In this case, the solution does not exist because (X – Y) is not divisible by 2.
- If X = Y: In this case, A & B = 0. Therefore, the minimum value of A should be 0 and the value of B should be Y to satisfy the given equations.
- Otherwise: A&B = (X – Y)/2 is satisfied, only when ((X – Y)/2) & Y equals 0. If true, the A = (X – Y)/2 and B = A + Y. Otherwise, A = -1 and B = -1.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the value of A and // B whose sum is X and xor is Y void findNums( int X, int Y)
{ // Initialize the two numbers
int A, B;
// Case 1: X < Y
if (X < Y) {
A = -1;
B = -1;
}
// Case 2: X-Y is odd
else if ( abs (X - Y) & 1) {
A = -1;
B = -1;
}
// Case 3: If both Sum and XOR
// are equal
else if (X == Y) {
A = 0;
B = Y;
}
// Case 4: If above cases fails
else {
// Update the value of A
A = (X - Y) / 2;
// Check if A & Y value is 0
if ((A & Y) == 0) {
// If true, update B
B = (A + Y);
}
// Otherwise assign -1 to A,
// -1 to B
else {
A = -1;
B = -1;
}
}
// Print the numbers A and B
cout << A << " " << B;
} // Driver Code int main()
{ // Given Sum and XOR of 2 numbers
int X = 17, Y = 13;
// Function Call
findNums(X, Y);
return 0;
} |
// Java program for the above approach import java.util.*;
class GFG{
// Function to find the value of A and // B whose sum is X and xor is Y static void findNums( int X, int Y)
{ // Initialize the two numbers
int A, B;
// Case 1: X < Y
if (X < Y)
{
A = - 1 ;
B = - 1 ;
}
// Case 2: X-Y is odd
else if (((Math.abs(X - Y)) & 1 ) != 0 )
{
A = - 1 ;
B = - 1 ;
}
// Case 3: If both Sum and XOR
// are equal
else if (X == Y)
{
A = 0 ;
B = Y;
}
// Case 4: If above cases fails
else
{
// Update the value of A
A = (X - Y) / 2 ;
// Check if A & Y value is 0
if ((A & Y) == 0 )
{
// If true, update B
B = (A + Y);
}
// Otherwise assign -1 to A,
// -1 to B
else
{
A = - 1 ;
B = - 1 ;
}
}
// Print the numbers A and B
System.out.print(A + " " + B);
} // Driver Code public static void main(String[] args)
{ // Given Sum and XOR of 2 numbers
int X = 17 , Y = 13 ;
// Function Call
findNums(X, Y);
} } // This code is contributed by susmitakundugoaldanga |
# Python program for the above approach # Function to find the value of A and # B whose sum is X and xor is Y def findNums(X, Y):
# Initialize the two numbers
A = 0 ;
B = 0 ;
# Case 1: X < Y
if (X < Y):
A = - 1 ;
B = - 1 ;
# Case 2: X-Y is odd
elif ((( abs (X - Y)) & 1 ) ! = 0 ):
A = - 1 ;
B = - 1 ;
# Case 3: If both Sum and XOR
# are equal
elif (X = = Y):
A = 0 ;
B = Y;
# Case 4: If above cases fails
else :
# Update the value of A
A = (X - Y) / / 2 ;
# Check if A & Y value is 0
if ((A & Y) = = 0 ):
# If True, update B
B = (A + Y);
# Otherwise assign -1 to A,
# -1 to B
else :
A = - 1 ;
B = - 1 ;
# Print the numbers A and B
print A;
print B;
# Driver Code if __name__ = = '__main__' :
# Given Sum and XOR of 2 numbers
X = 17 ;
Y = 13 ;
# Function Call
findNums(X, Y);
# This code is contributed by shikhasingrajput |
// C# program for the above approach using System;
class GFG{
// Function to find the value of A and // B whose sum is X and xor is Y static void findNums( int X, int Y)
{ // Initialize the two numbers
int A, B;
// Case 1: X < Y
if (X < Y)
{
A = -1;
B = -1;
}
// Case 2: X-Y is odd
else if (((Math.Abs(X - Y)) & 1) != 0)
{
A = -1;
B = -1;
}
// Case 3: If both Sum and XOR
// are equal
else if (X == Y)
{
A = 0;
B = Y;
}
// Case 4: If above cases fails
else
{
// Update the value of A
A = (X - Y) / 2;
// Check if A & Y value is 0
if ((A & Y) == 0)
{
// If true, update B
B = (A + Y);
}
// Otherwise assign -1 to A,
// -1 to B
else
{
A = -1;
B = -1;
}
}
// Print the numbers A and B
Console.Write(A + " " + B);
} // Driver Code public static void Main(String[] args)
{ // Given Sum and XOR of 2 numbers
int X = 17, Y = 13;
// Function Call
findNums(X, Y);
} } // This code is contributed by Rajput-Ji |
<script> // JavaScript program to implement the above approach // Function to find the value of A and // B whose sum is X and xor is Y function findNums(X, Y)
{ // Initialize the two numbers
let A, B;
// Case 1: X < Y
if (X < Y) {
A = -1;
B = -1;
}
// Case 2: X-Y is odd
else if (Math.abs(X - Y) & 1) {
A = -1;
B = -1;
}
// Case 3: If both Sum and XOR
// are equal
else if (X == Y) {
A = 0;
B = Y;
}
// Case 4: If above cases fails
else {
// Update the value of A
A = (X - Y) / 2;
// Check if A & Y value is 0
if ((A & Y) == 0) {
// If true, update B
B = (A + Y);
}
// Otherwise assign -1 to A,
// -1 to B
else {
A = -1;
B = -1;
}
}
// Print the numbers A and B
document.write(A + " " + B);
} // Driver Code // Given Sum and XOR of 2 numbers
let X = 17, Y = 13;
// Function Call
findNums(X, Y);
</script> |
2 15
Time Complexity: O(1)
Auxiliary Space: O(1)