Solution Explanation for LeetCode Problem 1083: Sales Analysis II

This problem requires finding buyers who purchased "S8" but not "iPhone". We achieve this by joining the Sales and Product tables, grouping results by buyer, and applying conditional aggregation.

Approach

1. Join Tables: We perform an INNER JOIN between the Sales and Product tables using product_id. This combines sales information with product names.

2. Group by Buyer: We use GROUP BY buyer_id to aggregate sales data for each buyer.

3. Conditional Aggregation: The HAVING clause filters the grouped results. SUM(product_name = 'S8') > 0 checks if the buyer purchased at least one "S8". SUM(product_name = 'iPhone') = 0 ensures they didn't purchase any "iPhone". MySQL treats boolean expressions as 1 (true) or 0 (false) in the sum.

4. Select Buyer IDs: Finally, we select the buyer_id for those buyers meeting the criteria.

MySQL Code Explanation

SELECT buyer_id
FROM
    Sales
    JOIN Product USING (product_id)
GROUP BY 1
HAVING SUM(product_name = 'S8') > 0 AND SUM(product_name = 'iPhone') = 0;

• SELECT buyer_id: Selects the buyer_id column.
• FROM Sales JOIN Product USING (product_id): Joins the Sales and Product tables on their common column product_id. USING (product_id) is a shorthand for ON Sales.product_id = Product.product_id.
• GROUP BY 1: Groups the results based on the first column in the SELECT statement (which is buyer_id).
• HAVING SUM(product_name = 'S8') > 0 AND SUM(product_name = 'iPhone') = 0: Filters the grouped results to include only those buyers who have purchased 'S8' (sum > 0) and have not purchased 'iPhone' (sum = 0).

Time and Space Complexity Analysis

• Time Complexity: The JOIN operation has a time complexity that depends on the size of the tables and the specific database implementation. In the worst case, it could be O(N*M), where N and M are the number of rows in Sales and Product respectively. The GROUP BY and HAVING operations add further complexity, but they're usually optimized in database systems. Overall, the complexity is largely dependent on the database engine's optimization strategies and can be considered close to linear with the size of the data.

• Space Complexity: The space complexity is primarily determined by the size of the intermediate result sets created during the JOIN, GROUP BY, and HAVING operations. It's again dependent on the database's implementation, but generally proportional to the size of the input data. In the worst case it could be O(N), where N is the number of rows in the joined table.

No other languages are specified in the prompt, so only the MySQL solution is provided. If you need solutions in other languages (e.g., Python with Pandas), please specify.