Maximum LCM among all pairs (i, j) of first N natural numbers

Given a positive integer N > 1, the task is to find the maximum LCM among all the pairs (i, j) such that i < j ? N.
Examples: 
 

Input: N = 3 
Output: 6 
LCM(1, 2) = 2 
LCM(1, 3) = 3 
LCM(2, 3) = 6
Input: N = 4 
Output: 12 
 

Approach: Since the LCM of two consecutive elements is equal to their multiples then it is obvious that the maximum LCM will be of the pair (N, N - 1) i.e. (N * (N - 1)).
Below is the implementation of the above approach: 
 

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

// Function to return the maximum LCM
// among all the pairs(i, j) of
// first n natural numbers
int maxLCM(int n)
{
    return (n * (n - 1));
}

// Driver code
int main()
{
    int n = 3;

    cout << maxLCM(n);

    return 0;
}

// Java implementation of the approach
class GFG
{
    
// Function to return the maximum LCM
// among all the pairs(i, j) of
// first n natural numbers
static int maxLCM(int n)
{
    return (n * (n - 1));
}

// Driver code
public static void main(String[] args)
{
    int n = 3;

    System.out.println(maxLCM(n));
}
}

// This code is contributed by Code_Mech

# Python3 implementation of the approach 

# Function to return the maximum LCM 
# among all the pairs(i, j) of 
# first n natural numbers 
def maxLCM(n) :

    return (n * (n - 1)); 

# Driver code 
if __name__ == "__main__" : 

    n = 3; 

    print(maxLCM(n)); 

# This code is contributed by AnkitRai01

// C# implementation of the approach 
using System;
    
class GFG
{
    
// Function to return the maximum LCM
// among all the pairs(i, j) of
// first n natural numbers
static int maxLCM(int n)
{
    return (n * (n - 1));
}

// Driver code
public static void Main(String[] args)
{
    int n = 3;

    Console.WriteLine(maxLCM(n));
}
}

// This code is contributed by Rajput-Ji

<script>

// Javascript implementation of the approach

// Function to return the maximum LCM
// among all the pairs(i, j) of
// first n natural numbers
function maxLCM(n)
{
    return (n * (n - 1));
}

// Driver code
var n = 3;
document.write(maxLCM(n));

</script>

6

Time Complexity: O(1)

Auxiliary Space: O(1)