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Problem solution · SQL

Find Students Who Improved

Find Students Who Improved: 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
24 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 Students Who Improved, 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

  • 24 lines of SQL from the credited upstream file 3421.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 Students Who Improved · SQLSQL
Use this to learn the idea, then write your own version.
WITH  RankedScores AS (    SELECT      student_id,      subject,      score,      exam_date,      RANK() OVER (PARTITION BY student_id, subject ORDER BY exam_date) AS rn_asc,      RANK() OVER (PARTITION BY student_id, subject ORDER BY exam_date DESC) AS rn_desc    FROM Scores  ),  FirstLastScores AS (    SELECT      student_id,      subject,      MIN(CASE WHEN rn_asc = 1 THEN score END) AS first_score,      MAX(CASE WHEN rn_desc = 1 THEN score END) AS latest_score    FROM RankedScores GROUP BY 1, 2    HAVING COUNT(*) > 1  )SELECT student_id, subject, first_score, latest_scoreFROM FirstLastScoresWHERE latest_score > first_score;

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