在数组中找到一个三元组，使得arr [i] <arr [j]> arr [k]和i <j <k

给定一个数组arr [] ，该数组包含前N个自然数的排列，任务是从给定数组中找到一个三元组(i，j，k) ，以使arr [i] arr [k] ，其中(i 。如果存在多个三元组，则打印索引的任何有效三元组。否则，打印-1 。

例子：

Input: arr[] = {2, 1, 4, 3} 
Output: 1 2 3 
Explanation: Triplet that satisfy the given condition is (arr[1], arr[2], arr[3]) 
Therefore, the required output is 1 2 3.

Input: arr[] = {1, 2, 3, 4, 5} 
Output: -1

天真的方法：解决此问题的最简单方法是遍历数组并生成给定数组的所有可能的三元组，对于每个三元组，检查其是否满足给定条件。如果发现是真实的，则打印该三元组。否则，打印-1 。

时间复杂度： O(N ³ )
辅助空间： O(1)

高效方法：为了优化上述方法，该思想基于以下观察结果：

• 如果给定数组从索引范围[1，N – 2]升序或降序排序，则解决方案不存在。
• 否则，给定数组中至少存在一个索引，以使该索引之前和之后的元素小于当前元素。

请按照以下步骤解决问题：

• 遍历给定数组，并为每个数组索引检查当前索引之前和之后的元素是否小于当前元素。如果发现是真的，则打印该三元组(i – 1，i，i + 1) 。
• 否则，打印-1 。

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 a triplet
// that satisfy the conditions
void FindTrip(int arr[], int N)
{
      
    // Traverse the given array
    for(int i = 1; i < N - 1; i++)
    {
          
        // Stores current element
        int p = arr[i - 1];
          
        // Stores element just before
        // the current element
        int q = arr[i];
          
        // Stores element just after
        // the current element
        int r = arr[i + 1];
          
        // Check the given conditions
        if (p < q && q > r) 
        {
              
            // Print a triplet
            cout << i - 1 << " "
                 << i << " " << i + 1;
              
            return;
        }
    }
      
    // If no triplet found
    cout << -1;
}
  
// Driver Code
int main()
{
    int arr[] = { 2, 1, 4, 3 };
      
    int N = sizeof(arr) / sizeof(arr[0]);
      
    FindTrip(arr, N); 
      
    return 0;
}
  
// This code is contributed by jyoti369

Java

// Java program to implement 
// the above approach 
import java.util.*; 
class GFG { 
  
    // Function to find a triplet 
    // that satisfy the conditions 
    static void FindTrip(int arr[], 
                        int N) 
    { 
        // Traverse the given array 
        for (int i = 1; i < N - 1; 
            i++) { 
  
            // Stores current element 
            int p = arr[i - 1]; 
  
            // Stores element just before 
            // the current element 
            int q = arr[i]; 
  
            // Stores element just after 
            // the current element 
            int r = arr[i + 1]; 
  
            // Check the given conditions 
            if (p < q && q > r) { 
  
                // Print a triplet 
                System.out.println( 
                    (i - 1) + " "
                    + (i) + " "
                    + (i + 1)); 
                return; 
            } 
        } 
  
        // If no triplet found 
        System.out.println(-1); 
    } 
  
    // Driver Code 
    public static void main(String args[]) 
    { 
        int arr[] = { 2, 1, 4, 3 }; 
  
        int N = arr.length; 
        FindTrip(arr, N); 
    } 
}

Python3

# Python3 program to implement
# the above approach
  
# Function to find a triplet
# that satisfy the conditions
def FindTrip(arr, N):
      
    # Traverse the given array
    for i in range(1, N - 1):
          
        # Stores current element
        p = arr[i - 1]
  
        # Stores element just before
        # the current element
        q = arr[i]
  
        # Stores element just after
        # the current element
        r = arr[i + 1]
  
        # Check the given conditions
        if (p < q and q > r):
  
            # Print a triplet
            print(i - 1, i, i + 1)
  
            return
  
    # If no triplet found
    print(-1)
  
# Driver Code
if __name__ == '__main__':
      
    arr = [ 2, 1, 4, 3 ]
  
    N = len(arr)
  
    FindTrip(arr, N)
  
# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach  
using System;
  
class GFG{ 
   
// Function to find a triplet 
// that satisfy the conditions 
static void FindTrip(int[] arr, int N) 
{ 
      
    // Traverse the given array 
    for(int i = 1; i < N - 1; i++)
    { 
          
        // Stores current element 
        int p = arr[i - 1]; 
  
        // Stores element just before 
        // the current element 
        int q = arr[i]; 
  
        // Stores element just after 
        // the current element 
        int r = arr[i + 1]; 
  
        // Check the given conditions 
        if (p < q && q > r)
        { 
              
            // Print a triplet 
            Console.WriteLine((i - 1) + " " + 
                              (i) + " " + (i + 1)); 
            return; 
        } 
    } 
  
    // If no triplet found 
    Console.WriteLine(-1); 
} 
  
// Driver Code 
public static void Main() 
{ 
    int[] arr = { 2, 1, 4, 3 }; 
    int N = arr.Length; 
      
    FindTrip(arr, N); 
} 
}
  
// This code is contributed by code_hunt

输出：

1 2 3

时间复杂度： O(N)

辅助空间： O(1)