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, 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

    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