Choose X such that (A xor X) + (B xor X) is minimized
Given two integers A and B. The task is to choose an integer X such that (A xor X) + (B xor X) is minimum possible.
Examples:
Input: A = 2, B = 3
Output: X = 2, Sum = 1
Input: A = 7, B = 8
Output: X = 0, Sum = 15
A simple solution is to generate all possible sum by taking xor of A and B with all possible value of X ≤ min(A, B). To generate all possible sums it would take O(N) time where N = min(A, B).
An efficient solution is based on the fact that the number X will contain the set bits only at that index where both A and B contains a set bit such that after xor operation with X that bit will be unset. This would take only O(Log N) time.
Other cases: If at a particular index one or both the numbers contain 0 (unset bit) and the number X contains 1 (set bit) then 0 will be set after xor with X in A and B then the sum couldn't be minimized .
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <iostream> using namespace std; // Function to return the integer X such that // (A xor X) + (B ^ X) is minimized int findX(int A, int B) { int j = 0, x = 0; // While either A or B is non-zero while (A || B) { // Position at which both A and B // have a set bit if ((A & 1) && (B & 1)) { // Inserting a set bit in x x += (1 << j); } // Left shifting both numbers to // traverse all the bits A >>= 1; B >>= 1; j += 1; } return x; } // Driver code int main() { int A = 2, B = 3; int X = findX(A, B); cout << "X = " << X << ", Sum = " << (A ^ X) + (B ^ X); return 0; } |
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Java
// Java implementation of the approach class GFG { // Function to return the integer X such that // (A xor X) + (B ^ X) is minimized static int findX(int A, int B) { int j = 0, x = 0; // While either A or B is non-zero while (A != 0 || B != 0) { // Position at which both A and B // have a set bit if ((A % 2 == 1) && (B % 2 == 1)) { // Inserting a set bit in x x += (1 << j); } // Left shifting both numbers to // traverse all the bits A >>= 1; B >>= 1; j += 1; } return x; } // Driver code public static void main(String[] args) { int A = 2, B = 3; int X = findX(A, B); System.out.println("X = " + X + ", Sum = " + ((A ^ X) + (B ^ X))); } } // This code has been contributed by 29AjayKumar |
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Python3
# Python 3 implementation of the approach # Function to return the integer X such that # (A xor X) + (B ^ X) is minimized def findX(A,B): j = 0 x = 0 # While either A or B is non-zero while (A or B): # Position at which both A and B # have a set bit if ((A & 1) and (B & 1)): # Inserting a set bit in x x += (1 << j) # Left shifting both numbers to # traverse all the bits A >>= 1 B >>= 1 j += 1 return x # Driver code if __name__ == '__main__': A = 2 B = 3 X = findX(A, B) print("X =",X,", Sum =",(A ^ X) + (B ^ X)) # This code is contributed by # Surendra_Gangwar |
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C#
// C# implementation of the approach using System; class GFG { // Function to return the integer X such that // (A xor X) + (B ^ X) is minimized static int findX(int A, int B) { int j = 0, x = 0; // While either A or B is non-zero while (A != 0 || B != 0) { // Position at which both A and B // have a set bit if ((A % 2 == 1) && (B % 2 == 1)) { // Inserting a set bit in x x += (1 << j); } // Left shifting both numbers to // traverse all the bits A >>= 1; B >>= 1; j += 1; } return x; } // Driver code public static void Main(String[] args) { int A = 2, B = 3; int X = findX(A, B); Console.WriteLine("X = " + X + ", Sum = " + ((A ^ X) + (B ^ X))); } } // This code has been contributed by 29AjayKumar |
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PHP
<?php // PHP implementation of the approach // Function to return the integer X such that // (A xor X) + (B ^ X) is minimized function findX($A, $B) { $j = 0; $x = 0; // While either A or B is non-zero while ($A || $B) { // Position at which both A and B // have a set bit if (($A & 1) && ($B & 1)) { // Inserting a set bit in x $x += (1 << $j); } // Left shifting both numbers to // traverse all the bits $A >>= 1; $B >>= 1; $j += 1; } return $x; } // Driver code $A = 2; $B = 3; $X = findX($A, $B); echo "X = " , $X , ", Sum = ", ($A ^ $X) + ($B ^ $X); // This code is contributed by ajit. ?> |
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Output:
X = 2, Sum = 1
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