Generate a pair of integers from a range [L, R] whose LCM also lies within the range

Given two integers L and R, the task is to find a pair of integers from the range [L, R] having LCM within the range [L, R] as well. If no such pair can be obtained, then print -1. If multiple pairs exist, print any one of them.

Examples:

Input: L = 13, R = 69
Output: X =13, Y = 26
Explanation: LCM(x, y) = 26 which satisfies the conditions L ? x < y ? R and L <= LCM(x, y) <= R

Input: L = 1, R = 665
Output: X = 1, Y = 2

Naive Approach: The simplest approach is to generate every pair between L and R and compute their LCM. Print a pair having LCM between the range L and R. If no pair is found to have LCM in the given range, print "-1".

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

Efficient Approach: The problem can be solved by using the Greedy technique based on the observation that LCM(x, y) is at least equal to 2*x which is LCM of (x, 2*x). Below are the steps to implement the approach:

1. Select the value of x as L and compute the value of y as 2*x
2. Check if y is less than R or not.
3. If y is less than R then print the pair (x, y)
4. Else print "-1"

Below is the implementation of the above approach:

// C++ implementation of the above approach
#include <iostream>
using namespace std;

void lcmpair(int l, int r)
{
    int x, y;
    x = l;
    y = 2 * l;

    // Checking if any pair is possible
    // or not in range(l, r)
    if (y > r) {

        // If not possible print(-1)
        cout << "-1\n";
    }
    else {

        // Print LCM pair
        cout << "X = " << x << " Y = "
             << y << "\n";
    }
}

// Driver code
int main()
{
    int l = 13, r = 69;

    // Function call
    lcmpair(l, r);
    return 0;
}

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

class GFG{

static void lcmpair(int l, int r)
{
    int x, y;
    x = l;
    y = 2 * l;

    // Checking if any pair is possible
    // or not in range(l, r)
    if (y > r) 
    {
        
        // If not possible print(-1)
        System.out.print("-1\n");
    }
    else
    {
        
        // Print LCM pair
        System.out.print("X = " + x + 
                        " Y = " + y + "\n");
    }
}

// Driver code
public static void main(String[] args)
{
    int l = 13, r = 69;

    // Function call
    lcmpair(l, r);
}
}

// This code is contributed by 29AjayKumar 

# Python3 implementation of the above approach 
def lcmpair(l, r):

    x = l
    y = 2 * l

    # Checking if any pair is possible
    # or not in range(l, r)
    if(y > r):

        # If not possible print(-1)
        print(-1)
    else:
        
        # Print LCM pair
        print("X = {} Y = {}".format(x, y))

# Driver Code
l = 13
r = 69

# Function call
lcmpair(l, r)

# This code is contributed by Shivam Singh

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

static void lcmpair(int l, int r)
{
    int x, y;
    x = l;
    y = 2 * l;

    // Checking if any pair is possible
    // or not in range(l, r)
    if (y > r) 
    {        
        // If not possible print(-1)
        Console.Write("-1\n");
    }
    else
    {        
        // Print LCM pair
        Console.Write("X = " + x + 
                      " Y = " + y + "\n");
    }
}

// Driver code
public static void Main(String[] args)
{
    int l = 13, r = 69;

    // Function call
    lcmpair(l, r);
}
}

// This code is contributed by shikhasingrajput 

<script>
// javascript implementation of the above approach

    function lcmpair(l , r) {
        var x, y;
        x = l;
        y = 2 * l;

        // Checking if any pair is possible
        // or not in range(l, r)
        if (y > r) {

            // If not possible print(-1)
            document.write("-1\n");
        } else {

            // Print LCM pair
            document.write("X = " + x + " Y = " + y + "\n");
        }
    }

    // Driver code
        var l = 13, r = 69;

        // Function call
        lcmpair(l, r);

// This code is contributed by aashish1995
</script>

X = 13 Y = 26

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