# Solving Leetcode Interviews in Seconds with AI: Count Special Subsequences


	# 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](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
	> 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(n<sup>4</sup>) - due to the four nested loops.
    *   Storage Complexity: O(1) - constant extra space.

	
	# Code
	```python
	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
	```