Employees above their department average
Return employee ID, name, and salary for employees earning strictly more than the average non-NULL salary in their non-NULL department. Return only the requested columns and follow the specified ordering. This question probes aggregate department averages once and join them back to employees.
Input
Setup script defining employees(id INT, name VARCHAR(30), department_id INT, manager_id INT, salary INT).
Output
id, name, salary; ascending id.
Constraints
0–10000 rows per table. IDs are unique and non-NULL unless stated otherwise; foreign keys may be NULL. Numeric amounts are integers from 0 to 1000000. Dates are valid YYYY-MM-DD text. Text uses case-sensitive comparison; test data has consistent case. Duplicate business values and NULLs are allowed unless excluded by the task.
Example
Tables and sample rows
CREATE TABLE employees(id INT, name VARCHAR(30), department_id INT, manager_id INT, salary INT); INSERT INTO employees VALUES (1,'Ana',10,NULL,100),(2,'Ben',10,1,200),(3,'Cara',20,1,200),(4,'Dan',20,3,300),(5,'Eve',NULL,NULL,NULL);
Expected result
id | name | salary 2 | Ben | 200 4 | Dan | 300
Another dataset (1)
CREATE TABLE employees(id INT, name VARCHAR(30), department_id INT, manager_id INT, salary INT); INSERT INTO employees VALUES (1,'Low',10,NULL,50),(2,'High',10,NULL,100),(3,'Null',10,NULL,NULL);
Expected result
id | name | salary 2 | High | 100
Covers missing matches, ties, or NULL values without relying on insertion order.
Hints (2)
- Build one average per department.
- Compare each employee against their own department average.
Target complexity: time O(n) · space O(n)