Solution Explanation for LeetCode 1892: Page Recommendations II

This problem involves recommending pages to users based on their friends' likes. The solution uses SQL queries to efficiently process the Friendship and Likes tables.

Approach:

The core idea is to join the Friendship and Likes tables, counting how many friends like each page for each user. We then filter out pages the user already likes.

MySQL Solution:

The MySQL solution uses a WITH clause (Common Table Expression or CTE) to create a simplified Friendship table (S) that includes both directions of friendships (A is friends with B and B is friends with A). This simplifies the joining process later. The query then performs the following steps:

1. WITH S AS (...): This CTE combines both directions of friendships using UNION so that we don't need to handle separate user1_id and user2_id columns in the Friendship table.

2. SELECT user1_id AS user_id, page_id, COUNT(1) AS friends_likes: This selects the user ID, page ID, and counts the number of friends who liked that page, which is aliased as friends_likes.

3. FROM S AS s LEFT JOIN Likes AS l ON s.user2_id = l.user_id: This joins the simplified Friendship table (s) with the Likes table (l). A LEFT JOIN is used to include all users from the Friendship table, even if they don't have friends who liked any pages.

4. WHERE NOT EXISTS (...): This crucial part filters out recommendations for pages that the user already likes. It checks if there's no entry in the Likes table (l2) where the user_id matches the current user (user1_id) and the page_id matches the page being considered (l.page_id).

5. GROUP BY user1_id, page_id: This groups the results by user and page ID so that COUNT(1) correctly counts friends' likes for each page per user.

Time Complexity Analysis:

The time complexity of this SQL query depends on the size of the Friendship and Likes tables. The UNION operation in the CTE has a complexity proportional to the size of the Friendship table. The LEFT JOIN and NOT EXISTS operations also have complexities that depend on the sizes of the tables. In the worst case, the time complexity could be O(n*m), where 'n' is the number of rows in Friendship and 'm' is the number of rows in Likes. However, database optimizers can significantly reduce this in practice through indexing and efficient join algorithms. The exact performance will vary based on the database system and its indexing strategy.

Space Complexity Analysis:

The space complexity is primarily determined by the intermediate results generated during the query execution. The CTE (S) uses space proportional to the size of the Friendship table. The final result set will store the recommendations, which could range in size depending on the number of recommendations made. In the worst case, it would be proportional to the number of users multiplied by the number of pages. Again, the exact space usage will depend on the database system and its query optimization.