Find the top 1 highest-paid employee in each department.
Problem Statement
Examples
Input: employees table: +----+---------+------------+--------+ | id | name | department | salary | +----+---------+------------+--------+ | 1 | Alice | IT | 90000 | | 2 | Bob | IT | 120000 | | 3 | Charlie | HR | 60000 | | 4 | Dana | HR | 75000 | | 5 | Eve | Sales | 80000 | +----+---------+------------+--------+
Output: +------+------------+--------+ | name | department | salary | +------+------------+--------+ | Dana | HR | 75000 | | Bob | IT | 120000 | | Eve | Sales | 80000 | +------+------------+--------+
Explanation: The query retrieves the requested records satisfying all problem requirements.
Complexity
Time Complexity: -
Space Complexity: -
Hints
Editorial & Approach
Problem Overview & Intuition
To solve "Top N Per Group (ROW_NUMBER + PARTITION)", we query the relational database engine using declarative SQL. The goal is to find the top 1 highest-paid employee in each department. By formulating an optimal execution plan with appropriate projection and filtering, the database engine executes the query with minimal overhead.
Step-by-Step Approach
- Analyze Schema: Identify the target tables, necessary foreign keys, and expected output columns.
- Construct Filtering & Logic: Apply WHERE filtering to isolate the requested data.
- Format & Order: Sort the resulting records according to specified order criteria.
Optimal Implementation (SQL)
SELECT name, department, salary FROM (SELECT name, department, salary, ROW_NUMBER() OVER (PARTITION BY department ORDER BY salary DESC) AS rn FROM employees) sub WHERE rn = 1;
Complexity Analysis
Key Considerations & Edge Cases
- Empty Tables: The query executes safely returning zero rows without syntax error.
- NULL Values: Columns containing NULL values are properly handled by standard ANSI SQL semantics.
- Case Sensitivity: String comparisons and keywords adhere to PostgreSQL/standard SQL rules.