Given two positive integers A and B such that A != B, the task is to find a positive integer X which maximizes the expression (A AND X) * (B AND X).
Example:
Input: A = 9 B = 8
Output: 8
(9 AND 8) * (8 AND 8) = 8 * 8 = 64 (maximum possible)
Input: A = 11 and B = 13
Output: 9
Naive approach: One can run a loop from 1 to max(A, B) and can easily find X which maximizes the given expression.
Efficient approach: It is known that,
(a – b)2 ? 0
which implies (a + b)2 – 4*a*b ? 0
which implies a * b ? (a + b)2 / 4
Hence, it concludes that a * b will be maximum when a * b = (a + b)2 / 4
which implies a = b
From the above result, (A AND X) * (B AND X) will be maximum when (A AND X) = (B AND X)
Now X can be found as:
A = 11 = 1011
B = 13 = 1101
X = ? = abcd
At 0th place: (1 AND d) = (1 AND d) implies d = 0, 1 but to maximize (A AND X) * (B AND X) d = 1
At 1st place: (1 AND d) = (0 AND d) implies c = 0
At 2nd place: (0 AND d) = (1 AND d) implies b = 0
At 3rd place: (1 AND d) = (1 AND d) implies a = 0, 1 but to maximize (A AND X) * (B AND X) a = 1
Hence, X = 1001 = 9
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define MAX 32// Function to find X according// to the given conditionsint findX(int A, int B){ int X = 0; // int can have 32 bits for (int bit = 0; bit < MAX; bit++) { // Temporary ith bit int tempBit = 1 << bit; // Compute ith bit of X according to // given conditions // Expression below is the direct // conclusion from the illustration // we had taken earlier int bitOfX = A & B & tempBit; // Add the ith bit of X to X X += bitOfX; } return X;}// Driver codeint main(){ int A = 11, B = 13; cout << findX(A, B); return 0;} |
C
// C implementation of the approach#include <stdio.h>#define MAX 32// Function to find X according// to the given conditionsint findX(int A, int B){ int X = 0; // int can have 32 bits for (int bit = 0; bit < MAX; bit++) { // Temporary ith bit int tempBit = 1 << bit; // Compute ith bit of X according to // given conditions // Expression below is the direct // conclusion from the illustration // we had taken earlier int bitOfX = A & B & tempBit; // Add the ith bit of X to X X += bitOfX; } return X;}// Driver codeint main(){ int A = 11, B = 13; printf("%d", findX(A, B)); return 0;}// This code is contributed by phalasi. |
Java
// Java implementation of the approachclass GFG{static int MAX = 32;// Function to find X according// to the given conditionsstatic int findX(int A, int B){ int X = 0; // int can have 32 bits for (int bit = 0; bit < MAX; bit++) { // Temporary ith bit int tempBit = 1 << bit; // Compute ith bit of X according to // given conditions // Expression below is the direct // conclusion from the illustration // we had taken earlier int bitOfX = A & B & tempBit; // Add the ith bit of X to X X += bitOfX; } return X;}// Driver codepublic static void main(String []args) { int A = 11, B = 13; System.out.println(findX(A, B));}}// This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach MAX = 32# Function to find X according # to the given conditions def findX(A, B) : X = 0; # int can have 32 bits for bit in range(MAX) : # Temporary ith bit tempBit = 1 << bit; # Compute ith bit of X according to # given conditions # Expression below is the direct # conclusion from the illustration # we had taken earlier bitOfX = A & B & tempBit; # Add the ith bit of X to X X += bitOfX; return X; # Driver code if __name__ == "__main__" : A = 11; B = 13; print(findX(A, B)); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System; class GFG{static int MAX = 32;// Function to find X according// to the given conditionsstatic int findX(int A, int B){ int X = 0; // int can have 32 bits for (int bit = 0; bit < MAX; bit++) { // Temporary ith bit int tempBit = 1 << bit; // Compute ith bit of X according to // given conditions // Expression below is the direct // conclusion from the illustration // we had taken earlier int bitOfX = A & B & tempBit; // Add the ith bit of X to X X += bitOfX; } return X;}// Driver codepublic static void Main(String []args) { int A = 11, B = 13; Console.WriteLine(findX(A, B));}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation of the approach// Function to find X according// to the given conditionsfunction findX( A, B){ var X = 0; var MAX = 32; // int can have 32 bits for (var bit = 0; bit < MAX; bit++) { // Temporary ith bit var tempBit = 1 << bit; // Compute ith bit of X according to // given conditions // Expression below is the direct // conclusion from the illustration // we had taken earlier var bitOfX = A & B & tempBit; // Add the ith bit of X to X X += bitOfX; } return X;} // Driver code var A = 11, B = 13; document.write(findX(A, B)); // This code is contributed by bunnyram19.</script> |
9
Time Complexity: O(MAX)
Auxiliary Space: O(1)
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
