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Problem solution · SQL

Friends With No Mutual Friends

Friends With No Mutual Friends: 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
23 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 Friends With No Mutual Friends, 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

  • 23 lines of SQL from the credited upstream file 3058.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 codeFriends With No Mutual Friends · SQLSQL
Use this to learn the idea, then write your own version.
WITH  TwoWayFriends AS (    SELECT user_id1 AS user_id, user_id2 AS friend_id FROM Friends    UNION ALL    SELECT user_id2, user_id1 FROM Friends  ),  UserToMutualFriend AS (    SELECT      User1.user_id,      User2.user_id AS friend_id    FROM TwoWayFriends AS User1    INNER JOIN TwoWayFriends AS User2      USING (friend_id)    WHERE User1.user_id != User2.user_id  )SELECT Friends.* FROM FriendsLEFT JOIN UserToMutualFriend  ON (    Friends.user_id1 = UserToMutualFriend.user_id    AND Friends.user_id2 = UserToMutualFriend.friend_id)WHERE UserToMutualFriend.friend_id IS NULLORDER BY 1, 2;

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