Maximum XOR pair product with a given value

Given a positive integer value V, the task is to find the maximum XOR of two numbers such that their product is equal to the given value V.

Examples:

Input: V = 20
Output: 7
Explanation: For the given value of 20, there are several pairs of numbers whose product is 20: (1, 20), (2, 10), and (4, 5). The maximum XOR among these pairs is 7, which corresponds to the pair (1, 20).

Input: V = 36
Output: 39
Explanation: For the given value of 36, there are several pairs of numbers whose product is 36: (1, 36), (2, 18), (3, 12), (4, 9), and (6, 6). The maximum XOR among these pairs is 39, which corresponds to the pair (3, 12).

Approach: This can be solved with the following idea:

The approach takes advantage of the fact that if a and b are two numbers such that a * b = value, then the XOR of a and b will be maximum when a and b are as close to each other as possible. By iterating through the numbers up to the square root of the given value, we cover all possible pairs of factors and find the maximum XOR among them.

Below is the code for the above approach:

C++

// C++ code for the above approach:
#include <cmath>
#include <iostream>

// Find maximum XOR pair value
int find_max_xor_product(int value)
{
    int max_xor = 0;

    // Calculate the square
    // root of the given value
    int sqrt_value = sqrt(value);

    // Iterate from 1 to the square root
    // of the value
    for (int num = 1; num <= sqrt_value; ++num) {

        // Check if num is a divisor
        // of the value
        if (value % num == 0) {

            // Calculate the
            // corresponding divisor
            int div = value / num;
            max_xor = std::max(max_xor, num ^ div);
        }
    }

    // Return the maximum XOR value
    return max_xor;
}

// Driver code
int main()
{
    int value = 36;

    // Function call
    int max_xor = find_max_xor_product(value);
    std::cout << max_xor << std::endl;

    return 0;
}

Java

// Java code for the above approach
import java.util.*;

public class GFG {
    // Find maximum XOR pair value
    public static int findMaxXorProduct(int value)
    {
        int max_xor = 0;

        // Calculate the square root of the given value
        int sqrt_value = (int)Math.sqrt(value);

        // Iterate from 1 to the square root of the value
        for (int num = 1; num <= sqrt_value; ++num) {

            // Check if num is a divisor of the value
            if (value % num == 0) {

                // Calculate the corresponding divisor
                int div = value / num;
                max_xor = Math.max(max_xor, num ^ div);
            }
        }

        // Return the maximum XOR value
        return max_xor;
    }

    public static void main(String[] args)
    {
        int value = 36;

        // Function call
        int max_xor = findMaxXorProduct(value);
        System.out.println(max_xor);
    }
}

// This code is contributed by Susobhan Akhuli

Python3

#Python code for the above approach:
import math

# Find maximum XOR pair value
def find_max_xor_product(value):
    max_xor = 0

    # Calculate the square root of the given value
    sqrt_value = int(math.sqrt(value))

    # Iterate from 1 to the square root of the value
    for num in range(1, sqrt_value + 1):

        # Check if num is a divisor of the value
        if value % num == 0:

            # Calculate the corresponding divisor
            div = value // num
            max_xor = max(max_xor, num ^ div)

    # Return the maximum XOR value
    return max_xor

# Driver code
if __name__ == '__main__':
    value = 36

    # Function call
    max_xor = find_max_xor_product(value)
    print(max_xor)

C#

// C# code for the above approach
using System;

public class GFG {
    // Find maximum XOR pair value
    static int FindMaxXorProduct(int value)
    {
        int max_xor = 0;

        // Calculate the square root
        // of the given value
        int sqrt_value = (int)Math.Sqrt(value);

        // Iterate from 1 to the square root
        // of the value
        for (int num = 1; num <= sqrt_value; ++num) {
            // Check if num is a divisor
            // of the value
            if (value % num == 0) {
                // Calculate the corresponding divisor
                int div = value / num;
                max_xor = Math.Max(max_xor, num ^ div);
            }
        }

        // Return the maximum XOR value
        return max_xor;
    }

    // Driver code
    static void Main(string[] args)
    {
        int value = 36;

        // Function call
        int max_xor = FindMaxXorProduct(value);
        Console.WriteLine(max_xor);
    }
}

// This code is contributed by Susobhan Akhuli

Javascript

// JavaScript code for the above approach

// Find maximum XOR pair value
function find_max_xor_product(value) {
    let max_xor = 0;

    // Calculate the square root of the given value
    let sqrt_value = Math.sqrt(value);

    // Iterate from 1 to the square root of the value
    for (let num = 1; num <= sqrt_value; ++num) {
        // Check if num is a divisor of the value
        if (value % num === 0) {
            // Calculate the corresponding divisor
            let div = value / num;
            max_xor = Math.max(max_xor, num ^ div);
        }
    }

    // Return the maximum XOR value
    return max_xor;
}

// Driver code

let value = 36;

// Function call
let max_xor = find_max_xor_product(value);
console.log(max_xor);

// This code is contributed by Susobhan Akhuli

Output

37

Time complexity: O(sqrt(N))
Auxiliary Space: O(1)