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Problem solution · SQL

Leetcodify Friends Recommendations

Leetcodify Friends Recommendations: 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
28 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 Leetcodify Friends Recommendations, 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

  • 28 lines of SQL from the credited upstream file 1917.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 codeLeetcodify Friends Recommendations · SQLSQL
Use this to learn the idea, then write your own version.
WITH  RecommendedUserPairs AS (    SELECT      Listen1.user_id AS user1_id,      Listen2.user_id AS user2_id    FROM Listens AS Listen1    INNER JOIN Listens AS Listen2      USING (song_id, day)    WHERE      Listen1.user_id < Listen2.user_id      AND NOT EXISTS(        SELECT * FROM Friendship        WHERE          Listen1.user_id = Friendship.user1_id          AND Listen2.user_id = Friendship.user2_id)    GROUP BY Listen1.user_id, Listen2.user_id, Listen1.day    HAVING COUNT(DISTINCT Listen1.song_id) >= 3  )SELECT  user1_id AS user_id,  user2_id AS recommended_idFROM RecommendedUserPairsUNIONSELECT  user2_id AS user_id,  user1_id AS recommended_idFROM RecommendedUserPairs;

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