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

给定一个由前N 个自然数排列组成的数组arr[] ，任务是从给定数组中找到任何三元组(i, j, k)使得0 ≤ i < j < k ≤ (N – 1)和arr[i] < arr[j]和arr[j] > arr[k] 。如果不存在这样的三元组，则打印“-1” 。

例子：

Input: arr[] = {4, 3, 5, 2, 1, 6}
Output: 1 2 3
Explanation: For the triplet (1, 2, 3), arr[2] > arr[1]( i.e. 5 > 3) and arr[2] > arr[3]( i.e. 5 > 2).

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

朴素的方法：最简单的方法是从给定的数组arr[]生成所有可能的三元组，如果存在满足给定条件的任何三元组，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)

有效的方法：可以通过观察数组仅包含范围[1, N] 中的不同元素这一事实来优化上述方法。如果存在具有给定标准的任何三元组，则该三元组必须彼此相邻。

因此，想法是在范围[1, N – 2]上遍历给定数组arr[]并且如果存在任何索引 i 使得arr[i – 1] < arr[i]和arr[i] > arr [i + 1] ，然后打印三元组(i – 1, i, i + 1)作为结果。否则，打印“-1” 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find a triplet such
// that i < j < k and arr[i] < arr[j]
// and arr[j] > arr[k]
void print_triplet(int arr[], int n)
{
    // Traverse the array
    for (int i = 1; i <= n - 2; i++) {
 
        // Condition to satisfy for
        // the resultant triplet
        if (arr[i - 1] < arr[i]
            && arr[i] > arr[i + 1]) {
 
            cout << i - 1 << " "
                 << i << " " << i + 1;
            return;
        }
    }
 
    // Otherwise, triplet doesn't exist
    cout << -1;
}
 
// Driver Code
int main()
{
    int arr[] = { 4, 3, 5, 2, 1, 6 };
    int N = sizeof(arr) / sizeof(int);
    print_triplet(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to find a triplet such
// that i < j < k and arr[i] < arr[j]
// and arr[j] > arr[k]
static void print_triplet(int arr[], int n)
{
     
    // Traverse the array
    for(int i = 1; i <= n - 2; i++)
    {
         
        // Condition to satisfy for
        // the resultant triplet
        if (arr[i - 1] < arr[i] &&
            arr[i] > arr[i + 1])
        {
            System.out.print(i - 1 + " " + i + " " +
                            (i + 1));
            return;
        }
    }
     
    // Otherwise, triplet doesn't exist
    System.out.print(-1);
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 4, 3, 5, 2, 1, 6 };
    int N = arr.length;
     
    print_triplet(arr, N);
}
}
 
// This code is contributed by sanjoy_62

Python3

# Python3 program for the above approach
 
# Function to find a triplet such
# that i < j < k and arr[i] < arr[j]
# and arr[j] > arr[k]
def print_triplet(arr, n):
     
    # Traverse the array
    for i in range(1, n - 1):
         
        # Condition to satisfy for
        # the resultant triplet
        if (arr[i - 1] < arr[i] and
            arr[i] > arr[i + 1]):
            print(i - 1, i, i + 1)
            return
 
    # Otherwise, triplet doesn't exist
    print(-1)
 
# Driver Code
if __name__ == "__main__":
     
    arr = [ 4, 3, 5, 2, 1, 6 ]
    N = len(arr)
     
    print_triplet(arr, N)
 
# This code is contributed by chitranayal

C#

// C# program to implement
// the above approach
using System;
public class GFG
{
   
// Function to find a triplet such
// that i < j < k and arr[i] < arr[j]
// and arr[j] > arr[k]
static void print_triplet(int[] arr, int n)
{
     
    // Traverse the array
    for(int i = 1; i <= n - 2; i++)
    {
         
        // Condition to satisfy for
        // the resultant triplet
        if (arr[i - 1] < arr[i] &&
            arr[i] > arr[i + 1])
        {
            Console.Write(i - 1 + " " + i + " " +
                            (i + 1));
            return;
        }
    }
     
    // Otherwise, triplet doesn't exist
    Console.Write(-1);
}
 
// Driver Code
public static void Main(String[] args)
{
    int[] arr = { 4, 3, 5, 2, 1, 6 };
    int N = arr.Length;
     
    print_triplet(arr, N);
}
}
 
// This code is contributed by splevel62.

Javascript

 
      // JavaScript program to implement
      // the above approach
       
      // Function to find a triplet such
      // that i < j < k and arr[i] < arr[j]
      // and arr[j] > arr[k]
      function print_triplet(arr, n)
      {
        // Traverse the array
        for (var i = 1; i <= n - 2; i++)
        {
          // Condition to satisfy for
          // the resultant triplet
          if (arr[i - 1] < arr[i] && arr[i] > arr[i + 1])
          {
            document.write(i - 1 + " " + i + " " + (i + 1));
            return;
          }
        }
 
        // Otherwise, triplet doesn't exist
        document.write(-1);
      }
 
      // Driver Code
      var arr = [4, 3, 5, 2, 1, 6];
      var N = arr.length;
 
      print_triplet(arr, N);
       

输出：

1 2 3

时间复杂度： O(N)
辅助空间： O(1)

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