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📜  从1到N的三元组的计数,以使中间元素始终最大

📅  最后修改于: 2021-04-27 05:20:26             🧑  作者: Mango

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

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。

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)