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Problem solution · SQL

Find Candidates for Data Scientist Position II

Find Candidates for Data Scientist Position 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
42 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 Candidates for Data Scientist Position 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

  • 42 lines of SQL from the credited upstream file 3278.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 Candidates for Data Scientist Position II · SQLSQL
Use this to learn the idea, then write your own version.
WITH  ProjectSkills AS (    SELECT project_id, COUNT(skill) AS required_skills    FROM Projects    GROUP BY 1  ),  CandidateScores AS (    SELECT      Projects.project_id,      Candidates.candidate_id,      100 + SUM(        CASE          WHEN Candidates.proficiency > Projects.importance THEN 10          WHEN Candidates.proficiency < Projects.importance THEN -5          ELSE 0        END      ) AS score,      COUNT(Projects.skill) AS matched_skills    FROM Projects    INNER JOIN Candidates      USING (skill)    GROUP BY 1, 2  ),  RankedCandidates AS (    SELECT      CandidateScores.project_id,      CandidateScores.candidate_id,      CandidateScores.score,      RANK() OVER(        PARTITION BY CandidateScores.project_id        ORDER BY CandidateScores.score DESC, CandidateScores.candidate_id      ) AS `rank`    FROM CandidateScores    INNER JOIN ProjectSkills      USING (project_id)    WHERE CandidateScores.matched_skills = ProjectSkills.required_skills  )SELECT project_id, candidate_id, scoreFROM RankedCandidatesWHERE `rank` = 1ORDER BY 1;

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