Solution Explanation: Product Sales Analysis V

This problem requires calculating the total spending of each user based on the Sales and Product tables. The solution involves joining the two tables, calculating the total spending for each user, and then ordering the results.

Approach

1. JOIN the tables: We need to combine information from both the Sales and Product tables. We use an INNER JOIN based on the product_id column, which is common to both tables. This ensures that we only consider sales for products that exist in the Product table. The USING clause simplifies the join syntax when joining on columns with the same name.

2. Calculate spending: For each sale, we multiply the quantity from the Sales table with the price from the Product table. This gives us the total spending for each individual sale.

3. Aggregate by user: We use the GROUP BY clause to group the results by user_id. The SUM() aggregate function calculates the sum of the spending for each user. GROUP BY 1 is shorthand for GROUP BY user_id.

4. Order the results: Finally, we order the results first by spending in descending order (ORDER BY 2 DESC) and then by user_id in ascending order (ORDER BY 1) to handle ties. ORDER BY 2 DESC, 1 is shorthand for ordering by the second and first columns respectively in that order.

MySQL Code

SELECT user_id, SUM(quantity * price) AS spending
FROM
    Sales
    JOIN Product USING (product_id)
GROUP BY 1
ORDER BY 2 DESC, 1;

Time Complexity Analysis

• JOIN: The time complexity of the JOIN operation depends on the implementation of the database system and the indexing of the tables. In a well-optimized database with appropriate indexes, it should be close to O(N log N) or even O(N) in the best case.
• GROUP BY and SUM: The GROUP BY operation and the SUM() aggregation have a time complexity of O(N), where N is the number of rows in the joined table.
• ORDER BY: The sorting operation (ORDER BY) typically has a time complexity of O(N log N) in the worst case, but can be better with optimized algorithms and data structures.

Therefore, the overall time complexity of the query is dominated by the sorting step and is approximately O(N log N). The space complexity is O(N) to store the intermediate results. This assumes that the input datasets are of size N. In practice, performance can be influenced by factors like table size, indexing, and database system optimizations.