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 = 8Output:8

(9 AND 8) * (8 AND 8) = 8 * 8 = 64 (maximum possible)Input:A = 11 and B = 13Output: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 impliesa * b ≤ (a + b)^{2}/ 4

Hence, it concludes thata * bwill be maximum whena * b = (a + b)^{2}/ 4

which impliesa = 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 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` |

## Javascript

`<script>` `// Javascript implementation of the approach` `// Function to find X according` `// to the given conditions` `function` `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>` |

**Output:**

9

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