Solution Explanation for Leetcodify Friends Recommendations

This problem requires finding pairs of users who are not friends but have listened to at least three common songs on the same day. The solution involves using SQL queries to process the Listens and Friendship tables.

Approach

The core idea is to perform a self-join on the Listens table to find pairs of users who listened to the same songs on the same day. Then, we filter out pairs who are already friends using a subquery or join with the Friendship table. Finally, we group the results and count the distinct songs to ensure at least three common songs are shared.

SQL Solution (MySQL)

The provided MySQL solution efficiently solves this problem in several steps:

1. Unioning Friendship Table: The Common Table Expression (CTE) T cleverly handles the unidirectional nature of friendships. It unions the Friendship table with itself, reversing user1_id and user2_id to represent both directions of a friendship. This avoids redundant checks later.

2. Self-Join on Listens: It performs a self-join on the Listens table (aliased as l1 and l2), comparing rows to find users who listened to the same song_id on the same day.

3. Filtering out Friends: The NOT EXISTS subquery checks if a pair of users (l1.user_id, l2.user_id) exists in the T (friendships) table. If it doesn't, it means they aren't friends.

4. Grouping and Counting: The GROUP BY clause groups the results by day, user1, and user2 to count distinct songs listened to in common.

5. Filtering by Common Songs: The HAVING clause filters out pairs with fewer than 3 common songs.

6. Distinct Result: SELECT DISTINCT ensures that only unique recommendations are returned.

Time Complexity Analysis

• Self-Join: The self-join on the Listens table has a time complexity of O(N^2) in the worst case, where N is the number of rows in Listens. However, the database optimizer might utilize indexes to improve performance.
• Subquery: The NOT EXISTS subquery adds an additional nested loop which contributes to the overall complexity. Again, database optimizations can significantly reduce this cost in practice.
• Grouping and Aggregation: The GROUP BY and HAVING clauses have a complexity dependent on the size of the intermediate results after the join and filtering.

Overall, the worst-case time complexity is approximately O(N^2), but the actual runtime is highly database-specific and influenced by the size of the tables and the presence of indexes. The use of indexes on user_id, song_id and day in Listens and on user1_id and user2_id in Friendship will greatly improve the performance.

Space Complexity Analysis

The space complexity is dominated by the intermediate results generated during the self-join. In the worst-case scenario, this could be O(N^2), representing all possible pairs of users. Again, database optimizations may significantly reduce this in practice. The final output has a complexity bounded by the number of recommendations, which is at most O(N^2).

In summary, while the theoretical complexity is high, a well-optimized database query can handle this problem efficiently for reasonably sized input tables. The specific performance heavily relies on the database system's query planner and available indexes.