# Maximize the expression (A AND X) * (B AND X) | Bit Manipulation

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `using` `namespace` `std; ` ` `  `#define MAX 32 ` ` `  `// Function to find X according ` `// to the given conditions ` `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 code ` `int` `main() ` `{ ` `    ``int` `A = 11, B = 13; ` ` `  `    ``cout << findX(A, B); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG ` `{ ` `static` `int` `MAX = ``32``; ` ` `  `// Function to find X according ` `// to the given conditions ` `static` `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 code ` `public` `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 approach ` `using` `System; ` `     `  `class` `GFG ` `{ ` `static` `int` `MAX = 32; ` ` `  `// Function to find X according ` `// to the given conditions ` `static` `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 code ` `public` `static` `void` `Main(String []args)  ` `{ ` `    ``int` `A = 11, B = 13; ` ` `  `    ``Console.WriteLine(findX(A, B)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```9
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Improved By : AnkitRai01, 29AjayKumar