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Check whether there exists a triplet (i, j, k) such that arr[i] < arr[k] < arr[j] for i < j < k
• Last Updated : 06 Aug, 2020

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 ``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 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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