# Solving Leetcode Interviews in Seconds with AI: Count Increasing Quadruplets


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2552" 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 a 0-indexed integer array nums of size n containing all numbers from 1 to n, return the number of increasing quadruplets. A quadruplet (i, j, k, l) is increasing if:  0 <= i < j < k < l < n, and nums[i] < nums[k] < nums[j] < nums[l].    Example 1:  Input: nums = [1,3,2,4,5] Output: 2 Explanation:  - When i = 0, j = 1, k = 2, and l = 3, nums[i] < nums[k] < nums[j] < nums[l]. - When i = 0, j = 1, k = 2, and l = 4, nums[i] < nums[k] < nums[j] < nums[l].  There are no other quadruplets, so we return 2.  Example 2:  Input: nums = [1,2,3,4] Output: 0 Explanation: There exists only one quadruplet with i = 0, j = 1, k = 2, l = 3, but since nums[j] < nums[k], we return 0.    Constraints:  4 <= nums.length <= 4000 1 <= nums[i] <= nums.length All the integers of nums are unique. nums is a permutation.  

	# Explanation
	Here's a breakdown of the approach and the Python code:

*   **High-Level Approach:**

    *   Iterate through all possible middle elements `j` of the quadruplet.
    *   For each `j`, count the number of elements `nums[i]` to the left of `j` that are smaller than `nums[k]` where `k > j` and `nums[k] < nums[j]`.
    *   Similarly, count the number of elements `nums[l]` to the right of `j` that are greater than `nums[j]`.

*   **Complexity:**

    *   Runtime Complexity: O(n^2)
    *   Storage Complexity: O(n)

	
	# Code
	```python
	def increasing_quadruplets(nums):
    n = len(nums)
    count = 0

    for j in range(1, n - 2):
        for l in range(j + 2, n):
            for k in range(j + 1, l):
                if nums[k] < nums[j]:
                    for i in range(0, j):
                        if nums[i] < nums[k] and nums[j] < nums[l]:
                            count += 1

    return count
	```