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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

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  • Difficulty Level : Hard
  • Last Updated : 10 Aug, 2022

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

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> nums)
{
    for (int i = 0; i < nums.size(); i++) {
        for (int j = i + 1; j < nums.size(); j++) {
            for (int k = j + 1; k < nums.size(); k++) {
                // Triplet found, hence return false
                if (nums[i] < nums[k] && nums[k] < nums[j])
                    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;
}


Python3




# Python3 program for the above approach
 
# Function to check if there exist
# triplet in the array such that
# i < j < k and arr[i] < arr[k] < arr[j]
def findTriplet(nums):
 
    for i in range(len(nums)):
        for j in range(i + 1, len(nums)):
            for k in range(j + 1, len(nums)):
               
                # Triplet found, hence return false
                if(nums[i] < nums[k] and nums[k] < nums[j]):
                    return True
                     
    # No triplet found, hence return false
    return False
 
# Driver Code
 
# Given array
arr = [ 4, 7, 5, 6 ]
 
# Function Call
if (findTriplet(arr)):
    print(" Yes")
 
else:
    print(" No")
 
 # This code is contributed by shinjanpatra


Javascript




<script>
 
// JavaScript program for the above approach
 
// Function to check if there exist
// triplet in the array such that
// i < j < k and arr[i] < arr[k] < arr[j]
function findTriplet(nums)
{
    for (let i = 0; i < nums.length; i++)
    {
        for (let j = i + 1; j < nums.length; j++)
        {
            for (let k = j + 1; k < nums.length; k++)
            {
             
                // Triplet found, hence return false
                if (nums[i] < nums[k] && nums[k] < nums[j])
                    return true;
            }
        }
    }
     
    // No triplet found, hence return false
    return false;
}
 
// Driver Code
 
// Given array
let arr = [ 4, 7, 5, 6 ];
 
// Function Call
if (findTriplet(arr)) {
    document.write(" Yes","</br>")
}
else {
    document.write(" No","</br>")
}
 
// This code is contributed by shinjanpatra
 
</script>


C#




// C# program for the above approach
     
using System;
 
public class HelloWorld
{
    // 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[] nums)
    {
        for (int i = 0; i < nums.Length; i++) {
            for (int j = i + 1; j < nums.Length; j++) {
                for (int k = j + 1; k < nums.Length; k++) {
                    // Triplet found, hence return false
                    if (nums[i] < nums[k] && nums[k] < nums[j])
                        return true;
                }
            }
        }
        // No triplet found, hence return false
        return false;
    }
     
     
    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 CodeWithMini


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[] nums)
{
    for (int i = 0; i < nums.length; i++) {
        for (int j = i + 1; j < nums.length; j++) {
            for (int k = j + 1; k < nums.length; k++) {
                // Triplet found, hence return false
                if (nums[i] < nums[k] && nums[k] < nums[j])
                    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 CodeWithMini


Output: 

Yes

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 h3
    // 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 h3
    // 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 h3
    // 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


Javascript




<script>
 
// JavaScript program for the above approach
 
// Function to check if there exist
// triplet in the array such that
// i < j < k and arr[i] < arr[k] < arr[j]
function findTriplet(arr)
{
    var n = arr.length;
    var st = [];
 
    // Initialize the heights of h1 and h3
    // to INT_MAX and INT_MIN respectively
    var h3 = -1000000000, h1 = 1000000000;
    for (var 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.length!=0
            && st[st.length-1] < arr[i]) {
 
            h3 = st[st.length-1];
            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
// Given array
var arr = [4, 7, 5, 6 ];
// Function Call
if (findTriplet(arr)) {
    document.write( " Yes");
}
else {
    document.write( " No" );
}
 
 
</script>


Output: 

Yes

 

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


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