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Problem solution · SQL

Find the Quiet Students in All Exams

Find the Quiet Students in All Exams: 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
16 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 the Quiet Students in All Exams, 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

  • 16 lines of SQL from the credited upstream file 1412.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 the Quiet Students in All Exams · SQLSQL
Use this to learn the idea, then write your own version.
WITH  RankedStudents AS (    SELECT      student_id,      RANK() OVER(PARTITION BY exam_id ORDER BY score ASC) rank_asc,      RANK() OVER(PARTITION BY exam_id ORDER BY score DESC) rank_desc    FROM Exam  )SELECT * FROM StudentWHERE student_id IN (  SELECT student_id FROM RankedStudents  GROUP BY 1  HAVING MIN(rank_asc) > 1 AND MIN(rank_desc) > 1)ORDER BY student_id;

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