Solution Explanation and Code for LeetCode Problem 1070: Product Sales Analysis III

This problem requires finding the first year of sale for each product and returning the corresponding product ID, year, quantity, and price. We can solve this using two main approaches: using subqueries or using window functions.

Approach 1: Using Subqueries

This approach uses a subquery to find the minimum year for each product and then joins this information back to the original Sales table to retrieve the required details.

MySQL Code:

SELECT
    product_id,
    year AS first_year,
    quantity,
    price
FROM Sales
WHERE
    (product_id, year) IN (
        SELECT
            product_id,
            MIN(year) AS year
        FROM Sales
        GROUP BY product_id
    );

Explanation:

1. Inner Query: SELECT product_id, MIN(year) AS year FROM Sales GROUP BY product_id This subquery groups the Sales table by product_id and finds the minimum year (MIN(year)) for each product. This identifies the first year each product was sold.

2. Outer Query: SELECT product_id, year AS first_year, quantity, price FROM Sales WHERE (product_id, year) IN (...) This query selects the product_id, year (renamed as first_year), quantity, and price from the Sales table. The WHERE clause filters the results to only include rows where the (product_id, year) combination matches the (product_id, minimum year) combinations found in the inner query.

Time Complexity Analysis:

The inner query has a time complexity of O(N log N) due to the GROUP BY and MIN operations (assuming sorting is involved). The outer query then performs a nested loop join with the inner query result set, giving an overall time complexity of approximately O(N log N), where N is the number of rows in the Sales table.

Approach 2: Using Window Functions

This approach leverages the RANK() window function to assign a rank to each sale within a product's sales history based on the year. We can then filter for only the rows with rank 1, representing the first year of sale.

MySQL Code:

WITH
    T AS (
        SELECT
            *,
            RANK() OVER (
                PARTITION BY product_id
                ORDER BY year
            ) AS rk
        FROM Sales
    )
SELECT product_id, year AS first_year, quantity, price
FROM T
WHERE rk = 1;

Explanation:

1. Common Table Expression (CTE): WITH T AS (...) This defines a CTE named T that calculates the rank for each sale.

2. RANK() Function: RANK() OVER (PARTITION BY product_id ORDER BY year) This function assigns a rank to each row within each product_id partition, ordered by year in ascending order. The first sale for each product will have rank 1.

3. Final Query: SELECT product_id, year AS first_year, quantity, price FROM T WHERE rk = 1 This query selects the relevant columns from the CTE T and filters the results to only include rows where rk (rank) is equal to 1.

Time Complexity Analysis:

Window functions typically have a time complexity of O(N log N) because they may involve sorting. The final selection of rows where rk = 1 is O(N), making the overall time complexity O(N log N), similar to Approach 1.

In summary: Both approaches achieve the desired result. The choice between them depends on personal preference and the specific database system being used; window functions are often considered more elegant and efficient in more complex scenarios. The time complexity of both approaches is approximately O(N log N).