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 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;// Function to find a triplet// that satisfy the conditionsvoid 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 Codeint 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 conditionsdef 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 Codeif __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 |
Javascript
<script>// Javascript program to implement// the above approach // 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)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
