Introduction

In this blog post, we will explore how to solve the LeetCode problem "3404" 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

You are given an array nums consisting of positive integers. A special subsequence is defined as a subsequence of length 4, represented by indices (p, q, r, s), where p < q < r < s. This subsequence must satisfy the following conditions: nums[p] nums[r] == nums[q] nums[s] There must be at least one element between each pair of indices. In other words, q - p > 1, r - q > 1 and s - r > 1. Return the number of different special subsequences in nums. Example 1: Input: nums = [1,2,3,4,3,6,1] Output: 1 Explanation: There is one special subsequence in nums. (p, q, r, s) = (0, 2, 4, 6): This corresponds to elements (1, 3, 3, 1). nums[p] nums[r] = nums[0] nums[4] = 1 3 = 3 nums[q] nums[s] = nums[2] nums[6] = 3 1 = 3 Example 2: Input: nums = [3,4,3,4,3,4,3,4] Output: 3 Explanation: There are three special subsequences in nums. (p, q, r, s) = (0, 2, 4, 6): This corresponds to elements (3, 3, 3, 3). nums[p] nums[r] = nums[0] nums[4] = 3 3 = 9 nums[q] nums[s] = nums[2] nums[6] = 3 3 = 9 (p, q, r, s) = (1, 3, 5, 7): This corresponds to elements (4, 4, 4, 4). nums[p] nums[r] = nums[1] nums[5] = 4 4 = 16 nums[q] nums[s] = nums[3] nums[7] = 4 4 = 16 (p, q, r, s) = (0, 2, 5, 7): This corresponds to elements (3, 3, 4, 4). nums[p] nums[r] = nums[0] nums[5] = 3 4 = 12 nums[q] nums[s] = nums[2] nums[7] = 3 4 = 12 Constraints: 7 <= nums.length <= 1000 1 <= nums[i] <= 1000

Explanation

Here's a breakdown of the solution approach, complexity, and the Python code:

• High-Level Approach:

  • Iterate through all possible combinations of indices (p, q, r, s) that satisfy the conditions p < q < r < s, q - p > 1, r - q > 1, and s - r > 1.
  • For each combination, check if nums[p] * nums[r] == nums[q] * nums[s].
  • If the condition is met, increment the count of special subsequences.

• Complexity:

  • Runtime Complexity: O(n⁴) - due to the four nested loops.
  • Storage Complexity: O(1) - constant extra space.

Code

    def count_special_subsequences(nums):
    """
    Counts the number of special subsequences in the given array.

    Args:
        nums: A list of positive integers.

    Returns:
        The number of special subsequences in nums.
    """

    n = len(nums)
    count = 0

    for p in range(n - 3):
        for q in range(p + 2, n - 2):
            for r in range(q + 2, n - 1):
                for s in range(r + 2, n):
                    if nums[p] * nums[r] == nums[q] * nums[s]:
                        count += 1

    return count