Find a triplet in an array such that arr[i] arr[k] and i < j < k

Given an array arr[] consisting of a permutation of first N natural numbers, the task is to find a triplet (i, j, k) from the given array such that arr[i] < arr[j] > arr[k], where (i < j < k). If multiple triplets exist, then print any valid triplet of indices. Otherwise, print -1.

Examples:

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

 

Naive Approach: The simplest approach to solve this problem is to traverse the array and generate all possible triplets of the given array and for each triplet, check if it satisfies the given conditions or not. If found to be true, then print that triplet. Otherwise, print -1



Time Complexity: O(N3)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach the idea is based on the following observations: 

  • If the given array is sorted in ascending order or descending order from the index range [1, N – 2], then the solution does not exist.
  • Otherwise, at least one index exists in the given array such that the element just before and after that concerned index is less than the current element.

Follow the steps below to solve the problem:

  • Traverse the given array and for each array index, check if the element just before and after the current index is less than the current element or not. If found to be true, then print that the triplet (i – 1, i, i + 1)
     
  • Otherwise, print -1.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to implement
// the above approach
#include <iostream>
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

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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); 
    
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Output:

1 2 3

Time Complexity: O(N)

Auxiliary Space: O(1)
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.