MediumSQL challenge #118SUBQUERY

Find Employees Earning More Than Their Department Average

Return the employees whose salary is above the average of their own department.

The challenge runs on the SQLabHub practice database (SQLite) and uses the employees table. You can solve it in the browser, with nothing to install.

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The task

Find employees earning more than department average.

A correct answer returns 59 rows with 12 columns: id, first_name, last_name, email, department, department_id, title, salary, hire_date, manager_id, city, status.

Tables used in this challenge

`employees` 120 rows

id INTEGER (PK) · first_name TEXT · last_name TEXT · email TEXT · department TEXT · department_id INTEGER · title TEXT · salary REAL · hire_date TEXT · manager_id INTEGER · city TEXT · status TEXT

First 3 rows of the employees table
id	first_name	last_name	email	department	department_id	title	salary	hire_date	manager_id	city	status
1	William	Lewis	wlewis18@corp.io	Customer Success	7	Senior CS Manager	61774	2022-07-24	NULL	San Diego	active
2	John	Miller	jmiller51@corp.io	Legal	8	Paralegal	191694	2021-08-09	NULL	San Diego	active
3	Daniel	Rivera	drivera50@corp.io	Engineering	1	VP Engineering	135067	2016-06-06	NULL	Houston	active

The first rows of each table are shown. See how all the tables relate on the database schema page.

Expected output

Expected output of the challenge
id	first_name	last_name	email	department	department_id	title	salary	hire_date	manager_id	city	status
2	John	Miller	jmiller51@corp.io	Legal	8	Paralegal	191694	2021-08-09	NULL	San Diego	active
3	Daniel	Rivera	drivera50@corp.io	Engineering	1	VP Engineering	135067	2016-06-06	NULL	Houston	active
5	Jason	Mitchell	jmitchell78@corp.io	Design	10	Product Designer	156865	2019-10-17	NULL	Denver	active
6	Amy	Nguyen	anguyen90@corp.io	Design	10	Product Designer	194543	2021-01-16	NULL	Seattle	on_leave
10	Michelle	Miller	mmiller42@corp.io	Operations	6	COO	178286	2022-10-15	NULL	Fort Worth	active

First 5 of 59 rows. This is the output of the model solution on the practice database. The task does not fix a sort order, so your rows may come back in a different order.

What you practise

Returning every column of a table with SELECT *
Filtering rows with a WHERE condition
Summarising many rows into one value with COUNT, SUM, AVG, MIN or MAX
Writing a subquery that refers to the current row of the outer query

Hint and approach

Try the challenge first. Open these only when you need them.

Show the hint

Correlated subquery

Show the step-by-step approach

How the model solution is built, in the order the database evaluates it:

Read from the employees table.
Filter in WHERE: keep the rows where salary is greater than the value returned by a subquery.
The subquery is correlated: it refers to the current outer row, so it is evaluated for each row. It works like this:
1. Read from the employees table.
2. Filter in WHERE: keep the rows where department equals the outer row's department.
3. Aggregate over all the rows that are left: compute AVG(salary). With no GROUP BY, the result is a single row.
Return every column with SELECT *.

Model solution

Checking your session…

What to watch for

The subquery mentions the outer row's department, so it is correlated: conceptually it runs once per employee. A window function, AVG(salary) OVER (PARTITION BY department), gives the same comparison in one pass.

Correlated subquery. A correlated subquery is evaluated per outer row, which gets slow on large tables. A join to a grouped subquery or a window function usually does the same work in one pass.
Aggregate functions. Aggregates skip NULLs. AVG divides by the number of non-NULL values and COUNT(column) can be smaller than COUNT(*).
WHERE. WHERE is evaluated before the SELECT list, so it cannot use a column alias defined there or the result of an aggregate.

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