Last updated on October 5th, 2024 at 04:55 pm

Here, We see **Increasing Triplet Subsequence LeetCode Solution**. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

*List of all LeetCode Solution*

*List of all LeetCode Solution*

## Topics

Array

## Companies

## Level of Question

Medium

**Increasing Triplet Subsequence LeetCode Solution**

## Table of Contents

**Problem Statement**

Given an integer array `nums`

, return `true`

* if there exists a triple of indices *`(i, j, k)`

* such that *`i < j < k`

* and *`nums[i] < nums[j] < nums[k]`

. If no such indices exists, return `false`

.

**Example 1:****Input:** nums = [1,2,3,4,5] **Output:** true **Explanation:** Any triplet where i < j < k is valid.

**Example 2:****Input:** nums = [5,4,3,2,1] **Output:** false **Explanation:** No triplet exists.

**Example 3:****Input:** nums = [2,1,5,0,4,6] **Output:** true **Explanation:** The triplet (3, 4, 5) is valid because nums[3] == 0 < nums[4] == 4 < nums[5] == 6.

**1. Increasing Triplet Subsequence LeetCode Solution C++**

class Solution { public: bool increasingTriplet(vector<int>& nums) { int n=nums.size(); if(n<3)return false; int low=INT_MAX, mid=INT_MAX; for(int i=0;i<n;i++) { if(nums[i]>mid) return true; else if(nums[i]<low) low=nums[i]; else if(nums[i]> low and nums[i]<mid) mid=nums[i]; } return false; } };

**2. Increasing Triplet Subsequence LeetCode Solution Java**

class Solution { public boolean increasingTriplet(int[] nums) { int min = Integer.MAX_VALUE, secondMin = Integer.MAX_VALUE; for(int num : nums){ if(num <= min) min = num; else if(num < secondMin) secondMin = num; else if(num > secondMin) return true; } return false; } }

**3. Increasing Triplet Subsequence Solution JavaScript**

var increasingTriplet = function(nums) { let firstNumber = Infinity; let secondNumber = Infinity; for (let currentNumber of nums) { if (currentNumber > secondNumber && currentNumber > firstNumber) { return true; } if (currentNumber > firstNumber) { secondNumber = currentNumber; } else { firstNumber = currentNumber; } } return false; };

**4. Increasing Triplet Subsequence Solution Python**

class Solution(object): def increasingTriplet(self, nums): first = second = float('inf') for n in nums: if n <= first: first = n elif n <= second: second = n else: return True return False