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Problem solution · SQL

Department Top Three Salaries

Department Top Three Salaries: 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 Department Top Three Salaries, 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 185.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 codeDepartment Top Three Salaries · SQLSQL
Use this to learn the idea, then write your own version.
WITH  EmployeesWithRankInDepartment AS (    SELECT      Department.name AS department,      Employee.name AS employee,      Employee.salary,      DENSE_RANK() OVER(        PARTITION BY Employee.departmentId        ORDER BY Employee.salary DESC      ) AS `rank`    FROM Department    INNER JOIN Employee      ON (Department.id = Employee.departmentId )  )SELECT  department AS Department,  employee AS Employee,  salary AS SalaryFROM EmployeesWithRankInDepartmentWHERE `rank` <= 3;

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