# Solving Leetcode Interviews in Seconds with AI: Increasing Triplet Subsequence


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "334" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# 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.    Constraints:  1 <= nums.length <= 5 * 105 -231 <= nums[i] <= 231 - 1    Follow up: Could you implement a solution that runs in O(n) time complexity and O(1) space complexity?

	# Explanation
	Here's a breakdown of the optimal solution:

*   **Maintain two smallest values:** Keep track of the smallest element encountered so far (`smallest`) and the second smallest element encountered so far (`second_smallest`).
*   **Iterate and check:** Iterate through the array. If the current element is greater than `second_smallest`, we've found a triplet. If it's smaller than `smallest`, update `smallest`. Otherwise, if it's smaller than `second_smallest` but greater than `smallest`, update `second_smallest`.
*   **Early exit:** The algorithm allows for an early exit as soon as a triplet has been identified, therefore optimizing the efficiency of the function.

*   **Runtime & Storage Complexity**: O(n) time complexity, O(1) space complexity.

	
	# Code
	```python
	def increasingTriplet(nums):
    """
    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.

    Args:
        nums (list[int]): An integer array.

    Returns:
        bool: True if there exists a triplet, False otherwise.
    """
    smallest = float('inf')
    second_smallest = float('inf')

    for num in nums:
        if num > second_smallest:
            return True
        elif num < smallest:
            smallest = num
        elif num > smallest and num < second_smallest:
            second_smallest = num

    return False
	```