# LCM of factorial and its neighbors

• Difficulty Level : Easy
• Last Updated : 23 Jun, 2022

Given a number, we need to find LCM of the factorial of the numbers and its neighbors. If the number is N, we need to find LCM of (N-1)!, N! and (N+1)!.Here N is always greater than or equal too 1
Examples :

```Input : N = 5
Output : 720
Explanation
Here the given number is 5, its neighbors
are 4 and 6. The factorial of these three
numbers are 24, 120, and 720.so the LCM
of 24, 120, 720 is 720.

Input : N = 3
Output : 24
Explanation
Here the given number is 3, its Neighbors
are 2 and 4.the factorial of these three
numbers are 2, 6, and 24. So the LCM of
2, 6 and 24 is 24.```

Method 1(Simple). We first calculate the factorial of number  and the factorial of its neighbor then
find the LCM of these factorials numbers.
Method 2(Efficient)
We can see that the LCM of (N-1)!, N! and (N+1)! is always (N-1)! * N! * (N+1)!
this can be written as (N-1)! * N*(N-1)! * (N+1)*N*(N-1)!
so the LCM become (N-1)! * N * (N+1)
which is (N+1)!
Example
N = 5
We need to find the LCM of 4!, 5!and 6!
LCM of 4!, 5!and 6!
= 4! * 5! * 6!
= 4! * 5*4! * 6*5*4!
= 6*5*4!
= 720
So we can say that LCM of the factorial of three consecutive numbers is always the factorial of the largest number.in this case (N+1)!.

## C++

 `// CPP program to calculate the LCM of N!``// and its neighbor (N-1)! and (N+1)!``#include ``using` `namespace` `std;` `// function to calculate the factorial``unsigned ``int` `factorial(unsigned ``int` `n)``{``    ``if` `(n == 0)``        ``return` `1;``    ``return` `n * factorial(n - 1);``}` `int` `LCMOfNeighbourFact(``int` `n)``{``    ``// returning the factorial of the``    ``// largest number in the given three``    ``// consecutive numbers``    ``return` `factorial(n + 1);``}` `// Driver code``int` `main()``{``    ``int` `N = 5;``    ``cout << LCMOfNeighbourFact(N) << ``"\n"``;``    ``return` `0;``}`

## Java

 `// Java program to calculate the LCM of N!``// and its neighbor (N-1)! and (N+1)!``import` `java.io.*;` `class` `GFG {` `    ``// function to calculate the factorial``    ``static` `int` `factorial(``int` `n)``    ``{``        ``if` `(n == ``0``)``            ``return` `1``;``            ` `        ``return` `n * factorial(n - ``1``);``    ``}` `    ``static` `int` `LCMOfNeighbourFact(``int` `n)``    ``{``        ` `        ``// returning the factorial of the``        ``// largest number in the given three``        ``// consecutive numbers``        ``return` `factorial(n + ``1``);``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `N = ``5``;``        ` `        ``System.out.println(LCMOfNeighbourFact(N));``    ``}``}` `/*This code is contributed by Nikita Tiwari.*/`

## Python3

 `# Python3 program to calculate the LCM of N!``# and its neighbor (N-1)! and (N+1)!` `# Function to calculate the factorial``def` `factorial(n):``    ``if` `(n ``=``=` `0``):``        ``return` `1``    ``return` `n ``*` `factorial(n ``-` `1``)` `def` `LCMOfNeighbourFact(n):` `    ``# returning the factorial of the``    ``# largest number in the given three``    ``# consecutive numbers``    ``return` `factorial(n ``+` `1``)` `# Driver code``N ``=` `5``print``(LCMOfNeighbourFact(N))` `# This code is contributed by Anant Agarwal.`

## C#

 `// Program to calculate the LCM``// of N! and its neighbor (N-1)!``// and (N+1)!``using` `System;` `class` `GFG``{``// function to calculate the factorial``static` `int` `factorial(``int` `n) {``    ` `    ``if` `(n == 0)``        ``return` `1;``    ``return` `n * factorial(n - 1);``}`` ` `static` `int` `LCMOfNeighbourFact(``int` `n) {` `    ``// returning the factorial of the``    ``// largest number in the given three``    ``// consecutive numbers``    ``return` `factorial(n + 1);``}` `// Driver code``public` `static` `void` `Main()``{`` ``int` `N = 5;`` ` ` ``Console.WriteLine(LCMOfNeighbourFact(N));``}``}`` ` `// This code is contributed by Anant Agarwal.`

## PHP

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## Javascript

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Output:

`720`

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

Please suggest if someone has a better solution which is more efficient in terms of space and time.