Check whether there exists a triplet (i, j, k) such that arr[i] < arr[k] < arr[j] for i < j < k

Given an array arr[], the task is to check that if there exist a triplet (i, j, k) such that arr[i]<arr[k]<arr[j] and i<j<k then print Yes else print No.

Examples:

Input: arr[] = {1, 2, 3, 4, 5}
Output: No
Explanation:
There is no such sub-sequence such that arr[i] < arr[k] < arr[j]

Input: arr[] = {3, 1, 5, 0, 4}
Output: Yes
Explanation:
There exist a triplet (3, 5, 4) which is arr[i] < arr[k] < arr[j]

Naive Approach: The idea is to generate all possible triplets and if any triplets satisfy the given conditions the print Yes else print No. 
Time Complexity: O(N³) 
Auxiliary Space: O(1) 

Efficient Approach: To optimize the above approach the idea is to use the stack to keep the track of the smaller elements in the right of every element in the array arr[]. Below are the steps:

• Traverse the array from the end and maintain a stack which stores the element in the decreasing order.
• To maintain the stack in decreasing order, pop the elements which are smaller than the current element. Then mark the popped element as the third element of the triplet.
• While traversing the array in reverse if any element is less than the last popped element(which is marked as the third element of the triplet). Then their exist a triplet which satisfy the given condition and print Yes.
• Otherwise print No.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check if there exist
// triplet in the array such that
// i < j < k and arr[i] < arr[k] < arr[j]
bool findTriplet(vector<int>& arr)
{
    int n = arr.size();
    stack<int> st;

    // Initialize the heights of h1 and h2
    // to INT_MAX and INT_MIN respectively
    int h3 = INT_MIN, h1 = INT_MAX;
    for (int i = n - 1; i >= 0; i--) {

        // Store the current element as h1
        h1 = arr[i];

        // If the element at top of stack
        // is less than the current element
        // then pop the stack top
        // and keep updating the value of h3
        while (!st.empty()
            && st.top() < arr[i]) {

            h3 = st.top();
            st.pop();
        }

        // Push the current element
        // on the stack
        st.push(arr[i]);

        // If current element is less
        // than h3, then we found such
        // triplet and return true
        if (h1 < h3) {
            return true;
        }
    }

    // No triplet found, hence return false
    return false;
}

// Driver Code
int main()
{
    // Given array
    vector<int> arr = { 4, 7, 5, 6 };

    // Function Call
    if (findTriplet(arr)) {
        cout << " Yes" << '\n';
    }
    else {
        cout << " No" << '\n';
    }
    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to check if there exist
// triplet in the array such that
// i < j < k and arr[i] < arr[k] < arr[j]
public static boolean findTriplet(int[] arr)
{
    int n = arr.length;
    Stack<Integer> st = new Stack<>();

    // Initialize the heights of h1 and h2
    // to INT_MAX and INT_MIN respectively
    int h3 = Integer.MIN_VALUE;
    int h1 = Integer.MAX_VALUE;

    for(int i = n - 1; i >= 0; i--)
    {

        // Store the current element as h1
        h1 = arr[i];

        // If the element at top of stack
        // is less than the current element
        // then pop the stack top
        // and keep updating the value of h3
        while (!st.empty() && st.peek() < arr[i])
        {
            h3 = st.peek();
            st.pop();
        }

        // Push the current element
        // on the stack
        st.push(arr[i]);

        // If current element is less
        // than h3, then we found such
        // triplet and return true
        if (h1 < h3)
        {
            return true;
        }
    }

    // No triplet found, hence return false
    return false;
}

// Driver code
public static void main(String[] args)
{

    // Given array
    int arr[] = { 4, 7, 5, 6 };

    // Function call
    if (findTriplet(arr))
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}
}

// This code is contributed by divyeshrabadiya07

Python3

# Python3 program for the above approach
import sys

# Function to check if there exist
# triplet in the array such that
# i < j < k and arr[i] < arr[k] < arr[j]
def findTriplet(arr):
    n = len(arr)
    st = []

    # Initialize the heights of h1 and h3
    # to INT_MAX and INT_MIN respectively
    h3 = -sys.maxsize - 1
    h1 = sys.maxsize

    for i in range(n - 1, -1, -1):

        # Store the current element as h1
        h1 = arr[i]

        # If the element at top of stack
        # is less than the current element
        # then pop the stack top
        # and keep updating the value of h3
        while (len(st) > 0 and st[-1] < arr[i]):
            h3 = st[-1]
            del st[-1]

        # Push the current element
        # on the stack
        st.append(arr[i])

        # If current element is less
        # than h3, then we found such
        # triplet and return true
        if (h1 < h3):
            return True

    # No triplet found, hence
    # return false
    return False

# Driver Code
if __name__ == '__main__':

    # Given array
    arr = [ 4, 7, 5, 6 ]

    # Function Call
    if (findTriplet(arr)):
        print("Yes")
    else:
        print("No")

# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function to check if there exist
// triplet in the array such that
// i < j < k and arr[i] < arr[k] < arr[j]
public static bool findTriplet(int[] arr)
{
    int n = arr.Length;
    Stack<int> st = new Stack<int>();

    // Initialize the heights of h1 and h2
    // to INT_MAX and INT_MIN respectively
    int h3 = int.MinValue;
    int h1 = int.MaxValue;

    for(int i = n - 1; i >= 0; i--)
    {

        // Store the current element as h1
        h1 = arr[i];

        // If the element at top of stack
        // is less than the current element
        // then pop the stack top
        // and keep updating the value of h3
        while (st.Count != 0 && st.Peek() < arr[i])
        {
            h3 = st.Peek();
            st.Pop();
        }

        // Push the current element
        // on the stack
        st.Push(arr[i]);

        // If current element is less
        // than h3, then we found such
        // triplet and return true
        if (h1 < h3)
        {
            return true;
        }
    }

    // No triplet found, hence return false
    return false;
}

// Driver code
public static void Main(String[] args)
{

    // Given array
    int []arr = { 4, 7, 5, 6 };

    // Function call
    if (findTriplet(arr))
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }
}
}

// This code is contributed by PrinciRaj1992

Output:

Yes

Time Complexity: O(N)
Auxiliary Space: O(N)

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