Last updated on July 18th, 2024 at 03:45 am

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.

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Increasing Triplet Subsequence LeetCode Solution

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