Count of triplets of numbers 1 to N such that middle element is always largest

Given an integer N, the task is to count the number of ways to arrange triplets (a, b, c) within [1, N] in such a way such that the middle element is always greater than left and right elements.
 

Example: 
Input: N = 4 
Output: 8 
Explanation 
For the given input N = 4 number of possible triplets are:{1, 3, 2}, {2, 3, 1}, {2, 4, 3}, {3, 4, 2}, {1, 4, 3}, {3, 4, 1}, {2, 4, 1}, {1, 4, 2}.
Input: 10 
Output: 240 
 

Naive Approach: Check for all triplets whether it satisfies the given condition using three nested loops and keep incrementing their count every time a triplet satisfies the condition. 
Time Complexity: O( N³ ) 
Auxiliary Space: O( 1 ) 
Efficient Approach: 
 

1. Check all the possibilities for middle element and try to find the number of possible arrangements keeping each of them fixed one by one.
2. We can observe that all the numbers between [3, N] can occupy the middle slot.

Possible arrangements for every middle element: 
On placing 3 at the middle, only 2 ( = 2 * 1) possible arrangements exist {1, 3, 2} and {2, 3, 1}. 
On placing 4 at the middle, 6 ( = 3 * 2) possible arrangements exist {1, 4, 3}, {1, 4, 2}, {2, 4, 1}, {2, 4, 3}, {3, 4, 1} and {3, 4, 2}. 
On placing 5 at the middle, 12 ( = 4 * 3) possible arrangements exist. 
. 
. 
. 
On placing N - 1 at the middle, (N-2) * (N-3) possible arrangements exist. 
On placing N at the middle, (N-1) * (N-2) possible arrangements exist.
Thus, Total possible arrangements = ( N * (N-1) * (N-2)) / 3 
 

Below is the implementation of the above approach.
 

// C++ program to implement
// the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find Number of triplets
// for given Number N such that
// middle element is always greater
// than left and right side element.
int findArrangement(int N)
{
    // check if arrangement is
    // possible or Not
    if (N < 3)
        return 0;

    // else return total ways
    return ((N) * (N - 1) * (N - 2)) / 3;
}

// Driver code.
int main()
{
    int N = 10;
    cout << findArrangement(N);
    return 0;
}

// Java program to implement
// the above approach
import java.io.*; 
import java.util.*; 

class GFG{ 
    
// Function to find number of triplets
// for given number N such that middle
// element is always greater than left  
// and right side element.
static int findArrangement(int N)
{ 
    
    // Check if arrangement 
    // is possible or not
    if (N < 3)
        return 0;

    // Else return total ways
    return ((N) * (N - 1) * (N - 2)) / 3;
} 

// Driver code 
public static void main(String[] args) 
{ 
    int N = 10;
    
    System.out.println(findArrangement(N)); 
} 
} 

// This code is contributed by coder001

# Python3 program to implement
# the above approach

# Function to find Number of triplets
# for given Number N such that middle
# element is always greater than left  
# and right side element.
def findArrangement(N):

    # Check if arrangement is
    # possible or Not
    if (N < 3):
        return 0;

    # Else return total ways
    return ((N) * (N - 1) * (N - 2)) // 3;

# Driver code.
N = 10;

print(findArrangement(N));

# This code is contributed by Akanksha_Rai

// C# program to implement
// the above approach
using System;
class GFG{ 
    
// Function to find number of triplets
// for given number N such that middle
// element is always greater than left 
// and right side element.
static int findArrangement(int N)
{ 
    
    // Check if arrangement 
    // is possible or not
    if (N < 3)
        return 0;

    // Else return total ways
    return ((N) * (N - 1) * (N - 2)) / 3;
} 

// Driver code 
public static void Main() 
{ 
    int N = 10;
    
    Console.Write(findArrangement(N)); 
} 
} 

// This code is contributed by Code_Mech

<script>
// Javascript program to implement
// the above approach

// Function to find number of triplets
// for given number N such that middle
// element is always greater than left  
// and right side element.
function findArrangement(N)
{ 
      
    // Check if arrangement 
    // is possible or not
    if (N < 3)
        return 0;
  
    // Else return total ways
    return ((N) * (N - 1) * (N - 2)) / 3;
}
 
  // Driver Code
    let N = 10;
   document.write(findArrangement(N)); 

// This code is contributed by susmitakundugoaldanga.
</script>

240

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