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Problem solution · SQL

Page Recommendations II

Page Recommendations II: a SQL solution using relational aggregation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Relational aggregation
Source
walkccc LeetCode Solutions
Length
20 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Relational aggregation

For Page Recommendations II, the query transforms and combines relational rows, then filters or aggregates them into the requested result.

  1. Identify the source rows and join keys.
  2. Apply filters before aggregation when possible.
  3. Group, rank, or project the final columns required by the result.

Code notes

  • 20 lines of SQL from the credited upstream file 1892.sql.
  • The implementation keeps its working state in language-native values and containers.
  • No explicit loop blocks detected.

Complexity

Review join cardinality, grouping keys, and available indexes when estimating query cost.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codePage Recommendations II · SQLSQL
Use this to learn the idea, then write your own version.
WITH  UserToFriends AS (    SELECT user1_id AS user_id, user2_id AS friend_id FROM Friendship    UNION ALL    SELECT user2_id AS user_id, user1_id AS friend_id FROM friendship  )SELECT  UserToFriends.user_id,  FriendLikes.page_id,  COUNT(DISTINCT UserToFriends.friend_id) AS friends_likesFROM UserToFriendsLEFT JOIN Likes AS FriendLikes  ON (UserToFriends.friend_id = FriendLikes.user_id)LEFT JOIN Likes AS UserLikes  ON (    UserToFriends.user_id = UserLikes.user_id    AND FriendLikes.page_id = UserLikes.page_id)WHERE UserLikes.page_id IS NULLGROUP BY 1, 2;

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