Solution Explanation for LeetCode 2324: Product Sales Analysis IV

This problem requires analyzing sales data to determine which product each user spent the most money on. The solution involves joining two tables (Sales and Product), calculating total spending per product per user, and then identifying the product(s) with the maximum spending for each user.

Approach:

The solution uses a common table expression (CTE) in SQL to achieve this efficiently. Here's a breakdown:

1. Join Sales and Product Tables: We start by joining the Sales and Product tables using the product_id to combine sales quantity with product price.

2. Calculate Total Spending: A GROUP BY clause aggregates the data to calculate the total spending (SUM(quantity * price)) for each user and product combination.

3. Rank Spending: The RANK() window function assigns a rank to each product based on the total spending for each user. The PARTITION BY user_id ensures ranking is done separately for each user, and ORDER BY SUM(quantity * price) DESC ranks products in descending order of spending. Products with the same total spending will receive the same rank.

4. Filter for Top Products: Finally, we filter the results to include only those products with rank 1 (WHERE rk = 1), representing the product(s) with the highest spending for each user.

Time Complexity Analysis:

The time complexity is dominated by the JOIN and GROUP BY operations. In the worst case, the join operation can be O(nm), where 'n' is the number of rows in Sales and 'm' is the number of rows in Product. The GROUP BY operation is at worst O(n), where n is the number of rows in the intermediate result after the join. The RANK() operation is linear in the number of rows being ranked within each partition. Therefore the overall time complexity is approximately **O(nm)**, but it will be significantly faster with proper indexing on the product_id column in both tables.

Space Complexity Analysis:

The space complexity depends on the size of the intermediate tables created during the query execution. It's primarily determined by the size of the joined table and the CTE (T). In the worst-case scenario, the space complexity is proportional to the size of the input tables, making it approximately O(n+m), where 'n' and 'm' are the sizes of the Sales and Product tables respectively.

MySQL Code:

WITH
    T AS (
        SELECT
            user_id,
            product_id,
            RANK() OVER (
                PARTITION BY user_id
                ORDER BY SUM(quantity * price) DESC
            ) AS rk
        FROM
            Sales
            JOIN Product USING (product_id)
        GROUP BY 1, 2
    )
SELECT user_id, product_id
FROM T
WHERE rk = 1;

This SQL query directly implements the described approach and efficiently solves the problem. Note that other SQL dialects might have slightly different syntax, but the core logic remains the same.