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 maximumvoid 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 codeint main(){ int n = 12; // function call MaxLCM(n); return 0;} |
Java
// Java Program to find three integers// less than N whose LCM is maximumimport 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 maximumstatic 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 codepublic 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 maximumfrom math import gcd# function to find three integers# less than N whose LCM is maximumdef 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 Codeif __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 maximumusing 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 maximumstatic 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 codepublic 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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