Solution Explanation for LeetCode 1159: Market Analysis II

This problem requires analyzing three tables: Users, Orders, and Items to determine if the brand of a user's second sold item matches their favorite brand. The solution uses SQL queries to achieve this.

Approach:

The core idea is to rank the orders for each seller by order date, then join the ranked orders with the Users and Items tables to compare the brand of the second item with the user's favorite brand.

SQL Solution (MySQL):

SELECT
    u.user_id AS seller_id,
    CASE
        WHEN u.favorite_brand = i.item_brand THEN 'yes'
        ELSE 'no'
    END AS 2nd_item_fav_brand
FROM
    users AS u
    LEFT JOIN (
        SELECT
            order_date,
            item_id,
            seller_id,
            RANK() OVER (
                PARTITION BY seller_id
                ORDER BY order_date
            ) AS rk
        FROM orders
    ) AS o
        ON u.user_id = o.seller_id AND o.rk = 2
    LEFT JOIN items AS i ON o.item_id = i.item_id;

Step-by-step explanation:

1. Inner Query (Ranked Orders):

   • SELECT order_date, item_id, seller_id, RANK() OVER (PARTITION BY seller_id ORDER BY order_date) AS rk FROM orders: This subquery selects relevant columns from the Orders table and uses the RANK() window function.
   • PARTITION BY seller_id: This partitions the data by seller ID, so the ranking is done separately for each seller.
   • ORDER BY order_date: This orders the orders within each partition by order date.
   • RANK(): This assigns a rank to each order within its partition. The earliest order gets rank 1, the next gets rank 2, and so on. If multiple orders have the same order date for a given seller, they'll receive the same rank.

2. Outer Query (Joining and Comparison):

   • FROM users AS u: This starts with the Users table, aliased as u.
   • LEFT JOIN (...) AS o ON u.user_id = o.seller_id AND o.rk = 2: This performs a LEFT JOIN with the inner query (ranked orders), aliased as o. The join condition ensures that only the second order (rk=2) for each seller is considered. A LEFT JOIN is crucial to include sellers who may have sold fewer than two items; their second order will be NULL.
   • LEFT JOIN items AS i ON o.item_id = i.item_id: This joins with the Items table to get the item brand. Again, LEFT JOIN handles cases where the second order might be missing.
   • CASE WHEN u.favorite_brand = i.item_brand THEN 'yes' ELSE 'no' END AS 2nd_item_fav_brand: This CASE statement compares the user's favorite brand (u.favorite_brand) with the brand of their second item (i.item_brand) and outputs 'yes' or 'no' accordingly. If a user has fewer than two orders or if the second order information is missing, i.item_brand will be NULL resulting in 'no'.

Time Complexity:

The time complexity is dominated by the RANK() window function and the joins. The RANK() function has a time complexity of O(N log N) in the worst case (N being the number of orders), while joins typically have a time complexity that depends on the indexing and database implementation, but it can be considered O(N) in an optimal scenario with proper indexing. Therefore, the overall time complexity is approximately O(N log N), where N is the number of orders.

Space Complexity:

The space complexity depends on the size of the intermediate result sets created during the query execution. It's largely determined by the size of the input tables and the amount of temporary space needed by the RANK() function and joins. In general, the space complexity is O(N) in the worst case, where N is the number of orders. Efficient indexing can help reduce this.