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Problem solution · SQL

Manager of the Largest Department

Manager of the Largest Department: 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
21 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 Manager of the Largest Department, 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

  • 21 lines of SQL from the credited upstream file 2988.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 codeManager of the Largest Department · SQLSQL
Use this to learn the idea, then write your own version.
WITH  RankedDepartments AS (    SELECT      dep_id,      DENSE_RANK() OVER(        ORDER BY COUNT(*) DESC      ) AS `rank`    FROM Employees    GROUP BY 1  )SELECT  Employees.emp_name AS manager_name,  Employees.dep_idFROM EmployeesINNER JOIN RankedDepartments  USING (dep_id)WHERE  Employees.position = 'Manager'  AND RankedDepartments.`rank` = 1ORDER BY dep_id;

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