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}Â
Output: 1 2 3Â
Explanation: Triplet that satisfy the given condition is (arr[1], arr[2], arr[3])Â
Therefore, the required output is 1 2 3.Input: arr[] = {1, 2, 3, 4, 5}Â
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.Â
Time Complexity: O(N3)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach the idea is based on the following observations:Â
- 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 concerning 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++
// C++ program to implement // the above approach #include <iostream> using namespace std; Â
// 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; } Â
// Driver Code int main() { Â Â Â Â int arr[] = { 2, 1, 4, 3 }; Â Â Â Â Â Â Â Â Â int N = sizeof (arr) / sizeof (arr[0]); Â Â Â Â Â Â Â Â Â FindTrip(arr, N); Â Â Â Â Â Â Â Â Â return 0; } Â
// This code is contributed by jyoti369 |
Java
// Java program to implement // the above approach import java.util.*; class GFG { Â
    // 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++) { Â
            // Stores current element             int p = arr[i - 1 ]; Â
            // Stores element just before             // the current element             int q = arr[i]; Â
            // Stores element just after             // the current element             int r = arr[i + 1 ]; Â
            // Check the given conditions             if (p < q && q > r) { Â
                // Print a triplet                 System.out.println(                     (i - 1 ) + " "                     + (i) + " "                     + (i + 1 ));                 return ;             }         } Â
        // If no triplet found         System.out.println(- 1 );     } Â
    // Driver Code     public static void main(String args[])     {         int arr[] = { 2 , 1 , 4 , 3 }; Â
        int N = arr.length;         FindTrip(arr, N);     } } |
Python3
# Python3 program to implement # the above approach Â
# 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 ] Â
        # Stores element just before         # the current element         q = arr[i] Â
        # Stores element just after         # the current element         r = arr[i + 1 ] Â
        # Check the given conditions         if (p < q and q > r): Â
            # Print a triplet             print (i - 1 , i, i + 1 ) Â
            return Â
    # If no triplet found     print ( - 1 ) Â
# Driver Code if __name__ = = '__main__' : Â Â Â Â Â Â Â Â Â arr = [ 2 , 1 , 4 , 3 ] Â
    N = len (arr) Â
    FindTrip(arr, N) Â
# This code is contributed by mohit kumar 29 |
C#
// C# program to implement // the above approach using System; Â
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]; Â
        // Stores element just before         // the current element         int q = arr[i]; Â
        // Stores element just after         // the current element         int r = arr[i + 1]; Â
        // Check the given conditions         if (p < q && q > r)         {                          // Print a triplet             Console.WriteLine((i - 1) + " " +                               (i) + " " + (i + 1));             return ;         }     } Â
    // If no triplet found     Console.WriteLine(-1); } Â
// Driver Code public static void Main() { Â Â Â Â int [] arr = { 2, 1, 4, 3 }; Â Â Â Â int N = arr.Length; Â Â Â Â Â Â Â Â Â FindTrip(arr, N); } } Â
// This code is contributed by code_hunt |
Javascript
<script> // Javascript program to implement // the above approach Â
    // Function to find a triplet     // that satisfy the conditions     function FindTrip(arr, N)     {         // Traverse the given array         for (let i = 1; i < N - 1;             i++) {                // Stores current element             let p = arr[i - 1];                // Stores element just before             // the current element             let q = arr[i];                // Stores element just after             // the current element             let r = arr[i + 1];                // Check the given conditions             if (p < q && q > r) {                    // Print a triplet                 document.write(                     (i - 1) + " "                     + (i) + " "                     + (i + 1));                 return ;             }         }            // If no triplet found         document.write(-1);     }          // Driver Code                  let arr = [2, 1, 4, 3 ];            let N = arr.length;         FindTrip(arr, N);    </script> |
Output:
1 2 3
Time Complexity: O(N)
Auxiliary Space: O(1)
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