CPMagical CP Library

Problem solution · SQL

Find Top Scoring Students II

Find Top Scoring Students 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
40 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 Top Scoring Students 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

  • 40 lines of SQL from the credited upstream file 3188.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 Top Scoring Students II · SQLSQL
Use this to learn the idea, then write your own version.
WITH  MandatoryMajors AS (    SELECT major, COUNT(course_id) AS course_count    FROM Courses    WHERE mandatory = 'Yes'    GROUP BY 1  ),  StudentsMetadata AS (    SELECT      Students.student_id,      Students.major,      SUM(        Students.major = Courses.major        AND Courses.mandatory = 'YES'        AND Enrollments.grade = 'A'      ) AS mandatory_grade_a_count,      SUM(        Students.major = Courses.major        AND Courses.mandatory = 'No'      ) AS elective_count,      ROUND(        SUM(Enrollments.GPA * Courses.credits) / SUM(Courses.credits),        1      ) AS avg_gpa    FROM Students    INNER JOIN Enrollments      USING (student_id)    INNER JOIN Courses      USING (course_id)    GROUP BY 1  )SELECT StudentsMetadata.student_idFROM StudentsMetadataINNER JOIN MandatoryMajors  ON (    StudentsMetadata.major = MandatoryMajors.major    AND StudentsMetadata.mandatory_grade_a_count = MandatoryMajors.course_count)WHERE StudentsMetadata.avg_gpa >= 2.5 AND StudentsMetadata.elective_count >= 2ORDER BY 1;

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗