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Problem solution · SQL

Find Interview Candidates

Find Interview Candidates: 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
37 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 Find Interview Candidates, 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

  • 37 lines of SQL from the credited upstream file 1811.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 codeFind Interview Candidates · SQLSQL
Use this to learn the idea, then write your own version.
WITH  UserToContest AS (    SELECT gold_medal AS user_id, contest_id FROM Contests    UNION ALL    SELECT silver_medal AS user_id, contest_id FROM Contests    UNION ALL    SELECT bronze_medal AS user_id, contest_id FROM Contests  ),  UserToContestWithGroupId AS (    SELECT      user_id,      contest_id - ROW_NUMBER() OVER(        PARTITION BY user_id        ORDER BY contest_id      ) AS group_id    FROM UserToContest  ),  CandidateUserIds AS (    -- consecutive medal winners    SELECT user_id    FROM UserToContestWithGroupId    GROUP BY user_id, group_id    HAVING COUNT(*) >= 3    UNION DISTINCT    -- gold medal winners    SELECT gold_medal AS user_id    FROM Contests    GROUP BY user_id    HAVING COUNT(*) >= 3  )SELECT  Users.name,  Users.mailFROM CandidateUserIdsINNER JOIN Users  USING (user_id);

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