Find three integers less than or equal to N such that their LCM is maximum

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Given a number N(>=3). The task is to find the three integers (<=N) such that LCM of these three integers is maximum.
Examples:

Input: N = 3
Output: 1 2 3

Input: N = 5
Output: 3 4 5

Approach: Since the task is to maximize the LCM, so if all three numbers don’t have any common factor then the LCM will be the product of those three numbers and that will be maximum.

If n is odd then the answer will be n, n-1, n-2.

If n is even,

If gcd of n and n-3 is 1 then answer will be n, n-1, n-3.

Otherwise, n-1, n-2, n-3 will be required answer.

Below is the implementation of the above approach:

C++

// CPP Program to find three integers
// less than N whose LCM is maximum
#include <bits/stdc++.h>
using namespace std;
 
// function to find three integers
// less than N whose LCM is maximum
void MaxLCM(int n)
{
    // if n is odd
    if (n % 2 != 0)
        cout << n << " " << (n - 1) << " " << (n - 2);
 
    // if n is even and n, n-3 gcd is 1
    else if (__gcd(n, (n - 3)) == 1)
        cout << n << " " << (n - 1) << " " << (n - 3);
 
    else
        cout << (n - 1) << " " << (n - 2) << " " << (n - 3);
}
 
// Driver code
int main()
{
    int n = 12;
 
    // function call
    MaxLCM(n);
 
    return 0;
}

Java

// Java Program to find three integers
// less than N whose LCM is maximum
 
import java.io.*;
 
class GFG {
   // Recursive function to return gcd of a and b 
static int __gcd(int a, int b) 
{ 
    // Everything divides 0  
    if (a == 0) 
       return b; 
    if (b == 0) 
       return a; 
    
    // base case 
    if (a == b) 
        return a; 
    
    // a is greater 
    if (a > b) 
        return __gcd(a-b, b); 
    return __gcd(a, b-a); 
} 
 
// function to find three integers
// less than N whose LCM is maximum
static void MaxLCM(int n)
{
    // if n is odd
    if (n % 2 != 0)
        System.out.print(n + " " + (n - 1) + " " + (n - 2));
 
    // if n is even and n, n-3 gcd is 1
    else if (__gcd(n, (n - 3)) == 1)
        System.out.print( n + " " +(n - 1)+ " " + (n - 3));
 
    else
        System.out.print((n - 1) + " " + (n - 2) + " " + (n - 3));
}
 
// Driver code
public static void main (String[] args) {
    int n = 12;
 
    // function call
    MaxLCM(n);
    }
}
// This code is contributed by anuj_67..

Python3

# Python 3 Program to find three integers
# less than N whose LCM is maximum
from math import gcd
 
# function to find three integers
# less than N whose LCM is maximum
def MaxLCM(n) :
 
    # if n is odd
    if (n % 2 != 0) :
        print(n, (n - 1), (n - 2))
 
    # if n is even and n, n-3 gcd is 1
    elif (gcd(n, (n - 3)) == 1) :
        print(n, (n - 1), (n - 3))
 
    else :
        print((n - 1), (n - 2), (n - 3))
 
# Driver Code
if __name__ == "__main__" :
     
    n = 12
 
    # function call
    MaxLCM(n)
 
# This code is contributed by Ryuga

C#

// C# Program to find three integers
// less than N whose LCM is maximum
 
using System;
 
class GFG {
// Recursive function to return gcd of a and b 
static int __gcd(int a, int b) 
{ 
    // Everything divides 0 
    if (a == 0) 
    return b; 
    if (b == 0) 
    return a; 
     
    // base case 
    if (a == b) 
        return a; 
     
    // a is greater 
    if (a > b) 
        return __gcd(a-b, b); 
    return __gcd(a, b-a); 
} 
 
// function to find three integers
// less than N whose LCM is maximum
static void MaxLCM(int n)
{
    // if n is odd
    if (n % 2 != 0)
        Console.Write(n + " " + (n - 1) + " " + (n - 2));
 
    // if n is even and n, n-3 gcd is 1
    else if (__gcd(n, (n - 3)) == 1)
        Console.Write( n + " " +(n - 1)+ " " + (n - 3));
 
    else
        Console.Write((n - 1) + " " + (n - 2) + " " + (n - 3));
}
 
// Driver code
public static void Main () {
    int n = 12;
 
    // function call
    MaxLCM(n);
    }
}
// This code is contributed by anuj_67..

PHP

<?php
// PHP Program to find three integers 
// less than N whose LCM is maximum 
 
// Recursive function to return 
// gcd of a and b 
function __gcd($a, $b) 
{ 
    // Everything divides 0 
    if ($a == 0) 
        return $b; 
    if ($b == 0) 
        return $a; 
     
    // base case 
    if ($a == $b) 
        return $a; 
     
    // a is greater 
    if ($a > $b) 
        return __gcd($a - $b, $b); 
    return __gcd($a, $b - $a); 
}  
 
// function to find three integers 
// less than N whose LCM is maximum 
function MaxLCM($n) 
{ 
    // if n is odd 
    if ($n % 2 != 0) 
        echo $n , " " , ($n - 1) , 
                  " " , ($n - 2); 
 
    // if n is even and n, n-3 gcd is 1 
    else if (__gcd($n, ($n - 3)) == 1) 
        echo $n , " " , ($n - 1), 
                  " " , ($n - 3); 
  
    else
        echo ($n - 1) , " " , ($n - 2), 
                        " " , ($n - 3); 
} 
 
// Driver code 
$n = 12; 
 
// function call 
MaxLCM($n); 
 
// This code is contributed by Sachin
?>

Javascript

<script>
 
// JavaScript Program to find three integers
// less than N whose LCM is maximum
 
    // Recursive function to return gcd of a and b
    function __gcd(a , b) 
    {
        // Everything divides 0
        if (a == 0)
            return b;
        if (b == 0)
            return a;
 
        // base case
        if (a == b)
            return a;
 
        // a is greater
        if (a > b)
            return __gcd(a - b, b);
        return __gcd(a, b - a);
    }
 
    // function to find three integers
    // less than N whose LCM is maximum
    function MaxLCM(n) {
        // if n is odd
        if (n % 2 != 0)
            document.write(n + " " + (n - 1) + " " + (n - 2));
 
        // if n is even and n, n-3 gcd is 1
        else if (__gcd(n, (n - 3)) == 1)
            document.write(n + " " + (n - 1) + " " + (n - 3));
 
        else
            document.write((n - 1) + " " + (n - 2) + " " + (n - 3));
    }
 
    // Driver code
     
        var n = 12;
 
        // function call
        MaxLCM(n);
         
// This code contributed by Rajput-Ji 
 
</script>

Output:

11 10 9

Time Complexity: O(log(min(a, b))), where a and b are two parameters of gcd.

Auxiliary Space: O(log(min(a, b)))

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