从1到N的三元组的计数，以使中间元素始终最大

给定整数N ，任务是计算在[1，N]内以三元组( a ， b ， c )排列方式的数量，以使中间元素始终大于左右元素。

Example:
Input: N = 4
Output: 8
Explaination
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

天真的方法：使用三个嵌套循环检查是否满足给定条件的所有三元组，并在每次三元组满足条件时不断增加其计数。
时间复杂度： O(N ³ )
辅助空间： O(1)

高效方法：

1. 检查中间元素的所有可能性，并尝试找到可能的布置的数量，使每个元素一个接一个地固定。
2. 我们可以观察到[3，N]之间的所有数字都可以占据中间位置。

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

下面是上述方法的实现。

C++

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

// 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

# 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#

// 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

输出：

240

时间复杂度： O(1)
辅助空间： O(1)