Solution Explanation for LeetCode Problem 1677: Product's Worth Over Invoices

This problem requires querying two tables, Product and Invoice, to aggregate invoice data for each product and present it in a summarized format. The solution involves joining the tables, grouping the results by product, and summing the relevant invoice columns.

Approach

1. Join the tables: We use a LEFT JOIN between the Product and Invoice tables using product_id. A LEFT JOIN ensures that all products are included in the result, even if they don't have any associated invoices.

2. Group by product: We use GROUP BY product_id to aggregate data for each product. This groups all invoices related to the same product together.

3. Aggregate invoice data: We use aggregate functions (SUM()) to calculate the total rest, paid, canceled, and refunded amounts for each product. The IFNULL() function handles cases where a product might not have any invoices (resulting in NULL values for the sums), replacing NULL with 0.

4. Order and Select: Finally, we select the name (from the Product table) along with the aggregated invoice amounts and order the results by name using ORDER BY name.

Time and Space Complexity Analysis

• Time Complexity: The time complexity is dominated by the JOIN and GROUP BY operations. In the worst case (all invoices associated with a single product), the time complexity would be O(N log N) or O(N), depending on the database's specific implementation of joins and grouping (e.g., using hash tables or sorting). 'N' represents the number of rows in the Invoice table. The SUM, IFNULL, and ORDER BY operations have a lower time complexity compared to the join and group by.

• Space Complexity: The space complexity is primarily determined by the size of the intermediate result set after the JOIN and GROUP BY operations. In the worst case, this could be proportional to the number of products (which is usually far less than the number of invoices). Therefore, the space complexity is O(P), where P is the number of unique products.

Code Implementation (MySQL)

SELECT
    name,
    IFNULL(SUM(rest), 0) AS rest,
    IFNULL(SUM(paid), 0) AS paid,
    IFNULL(SUM(canceled), 0) AS canceled,
    IFNULL(SUM(refunded), 0) AS refunded
FROM
    Product
    LEFT JOIN Invoice USING (product_id)
GROUP BY product_id
ORDER BY name;

This SQL query directly implements the approach described above. It's concise, efficient, and leverages SQL's built-in capabilities for joining, grouping, and aggregating data. The IFNULL function ensures that even products without invoices are included in the result set with zero values for the invoice amounts. The ORDER BY clause presents the final result in an organized manner.