Find two numbers with given sum and maximum possible LCM

Given an integer X, the task is to find two integers A and B such that sum of these two numbers is X and the LCM of A and B is maximum.

Examples:

Input: X = 15
Output: 7 8
Explanation: 
7 + 8 = 15 and LCM(7, 8) = 56 is the maximum possible. 

Input: X = 30 
Output: 13 17
Explanation:
13 + 17 = 30 and LCM(13, 17) = 221 is the maximum possible.

Naive Approach: The simplest approach is to use Two Pointers to find the pair of integers A and B with a given sum X and maximum possible LCM. Below are the steps:

• Initialize A and B as 1 and X–1 respectively.
• Run a loop, while, A is less than and equal to B.
• At each iteration calculate the LCM of A and B, then increment A by 1 and decrement B by 1.
• Print the A and B corresponding to the maximum LCM.

Time Complexity: O(N)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above naive approach the idea is to use some mathematical observations. The LCM of two co-prime integers is equal to the product of the two integers. Thus, the problem can be simplified to finding two co-prime integers A and B such that A+B = X and A×B is maximum. Below are the steps:

• If X is odd, then A = floor(X/2) and B = floor(X/2) + 1.
• Otherwise, if X is even, then
  • If floor(X/2) is even, then A = floor(X/2) – 1 and B = floor(X/2) + 1.
  • Otherwise, if floor(X/2) is odd, then A = floor(X/2) – 2 and B = floor(X/2) + 2.
    

Below is the implementation of the above approach:

C++

// C++ program of the above approach
#include <bits/stdc++.h>
using namespace std;

// Function that print two numbers with
// the sum X and maximum possible LCM
void maxLCMWithGivenSum(int X)
{
    // variables to store the result
    int A, B;

    // If X is odd
    if (X & 1) {
        A = X / 2;
        B = X / 2 + 1;
    }

    // If X is even
    else {

        // If floor(X/2) is even
        if ((X / 2) % 2 == 0) {
            A = X / 2 - 1;
            B = X / 2 + 1;
        }

        // If floor(X/2) is odd
        else {
            A = X / 2 - 2;
            B = X / 2 + 2;
        }
    }

    // Print the result
    cout << A << " " << B << endl;
}

// Driver Code
int main()
{
    // Given Number
    int X = 30;

    // Function call
    maxLCMWithGivenSum(X);
    return 0;
}

Java

// Java program of the above approach
import java.util.*;

class GFG{

// Function that print two numbers with
// the sum X and maximum possible LCM
static void maxLCMWithGivenSum(int X)
{

    // Variables to store the result
    int A, B;

    // If X is odd
    if ((X & 1) == 1)
    {
        A = X / 2;
        B = X / 2 + 1;
    }

    // If X is even
    else
    {

        // If floor(X/2) is even
        if ((X / 2) % 2 == 0)
        {
            A = X / 2 - 1;
            B = X / 2 + 1;
        }

        // If floor(X/2) is odd
        else
        {
            A = X / 2 - 2;
            B = X / 2 + 2;
        }
    }

    // Print the result
    System.out.println(A + " " + B);
}

// Driver code
public static void main(String[] args)
{

    // Given number
    int X = 30;

    // Function call
    maxLCMWithGivenSum(X);
}
}

// This code is contributed by offbeat

Python3

# Python3 program for the above approach

# Function that print two numbers with
# the sum X and maximum possible LCM
def maxLCMWithGivenSum(X):

    # If X is odd
    if X % 2 != 0:
        A = X / 2
        B = X / 2 + 1

    # If X is even
    else:

        # If floor(X/2) is even
        if (X / 2) % 2 == 0:
            A = X / 2 - 1
            B = X / 2 + 1

        # If floor(X/2) is odd
        else:
            A = X / 2 - 2
            B = X / 2 + 2

    # Print the result
    print(int(A), int(B), end = " ")

# Driver Code
if __name__ == '__main__':

    # Given Number
    X = 30

    # Function call
    maxLCMWithGivenSum(X)

# This code is contributed by virusbuddah_

C#

// C# program of the above approach
using System;
class GFG{

// Function that print two numbers with
// the sum X and maximum possible LCM
static void maxLCMWithGivenSum(int X)
{

    // Variables to store the result
    int A, B;

    // If X is odd
    if ((X & 1) == 1)
    {
        A = X / 2;
        B = X / 2 + 1;
    }

    // If X is even
    else
    {

        // If floor(X/2) is even
        if ((X / 2) % 2 == 0)
        {
            A = X / 2 - 1;
            B = X / 2 + 1;
        }

        // If floor(X/2) is odd
        else
        {
            A = X / 2 - 2;
            B = X / 2 + 2;
        }
    }

    // Print the result
    Console.WriteLine(A + " " + B);
}

// Driver code
public static void Main(String[] args)
{

    // Given number
    int X = 30;

    // Function call
    maxLCMWithGivenSum(X);
}
}

// This code is contributed by sapnasingh4991

Javascript

<script>

// Javascript program of the above approach

// Function that print two numbers with
// the sum X and maximum possible LCM
function maxLCMWithGivenSum(X)
{
    // variables to store the result
    let A, B;

    // If X is odd
    if (X & 1) {
        A = X / 2;
        B = X / 2 + 1;
    }

    // If X is even
    else {

        // If floor(X/2) is even
        if ((X / 2) % 2 == 0) {
            A = X / 2 - 1;
            B = X / 2 + 1;
        }

        // If floor(X/2) is odd
        else {
            A = X / 2 - 2;
            B = X / 2 + 2;
        }
    }

    // Print the result
    document.write(A + " " + B + "<br>");
}

// Driver Code

    // Given Number
    let X = 30;

    // Function call
    maxLCMWithGivenSum(X);

// This code is contributed by Manoj

</script>

Output:

13 17

Time Complexity: O(1)
Auxiliary Space: O(1)