Set each employee bonus to 10% of the department average salary.
Problem Statement
Examples
Input: employees table: +----+---------+------------+--------+-------+ | id | name | department | salary | bonus | +----+---------+------------+--------+-------+ | 1 | Alice | IT | 90000 | 0 | | 2 | Bob | IT | 80000 | 0 | | 3 | Charlie | HR | 60000 | 0 | +----+---------+------------+--------+-------+
Output: +----+---------+------------+--------+-------+ | id | name | department | salary | bonus | +----+---------+------------+--------+-------+ | 1 | Alice | IT | 90000 | 8500 | | 2 | Bob | IT | 80000 | 8500 | | 3 | Charlie | HR | 60000 | 6000 | +----+---------+------------+--------+-------+
Explanation: The query retrieves the requested records satisfying all problem requirements.
Complexity
Time Complexity: -
Space Complexity: -
Hints
Editorial & Approach
Problem Overview & Intuition
To solve "UPDATE with Subquery", we query the relational database engine using declarative SQL. The goal is to set each employee bonus to 10% of the department average salary. 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)
UPDATE employees SET bonus = (SELECT AVG(salary) * 0.1 FROM employees e2 WHERE e2.department = employees.department); SELECT * FROM employees ORDER BY id;
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.