HardSQL challenge #325WINDOW FUNCTIONPARTITIONSTATISTICS

Calculate the Median Salary per Department

Return the middle salary of each department, averaging the two middle values when the count is even.

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.

Solve This SQL Challenge

The task

Find median salary per department.

A correct answer returns 10 rows with 2 columns: department, AVG(salary).

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
department	AVG(salary)
Customer Success	137737
Design	150673
Engineering	123267
Finance	122626
HR	120155
Legal	101862
Marketing	103819
Operations	125723
Product	121472
Sales	123218

The complete result: 10 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

Filtering rows with a WHERE condition
Summarising many rows into one value with COUNT, SUM, AVG, MIN or MAX
Grouping rows with GROUP BY to get one result row per group
Querying the result of another query with a subquery in FROM
Numbering rows with ROW_NUMBER() to pick one or a few per group
Attaching a total or average to every row with an aggregate OVER ()
Restarting a window calculation for each group with PARTITION BY

Hint and approach

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

Show the hint

ROW_NUMBER + COUNT partition

Show the step-by-step approach

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

Build an inner query first (a derived table):
1. Read from the employees table.
2. Number the rows with ROW_NUMBER(), ordered by salary, restarting for each department.
3. Compute COUNT(*) as a window function over each department, so the group's value is repeated on every one of its rows.
Filter in WHERE: keep the rows where rn is one of (cnt+1)/2 or (cnt+2)/2. The filter sits in the outer query because a window function cannot be tested in the WHERE clause of the query that computes it.
Group the rows by department, so the result has one row for each value.
For each group, compute AVG(salary).

Model solution

Checking your session…

What to watch for

Both window functions use the same partition: one numbers the salaries inside the department and the other counts them. The middle one or two rows of each department are then averaged.

ROW_NUMBER(). ROW_NUMBER() breaks ties arbitrarily. When the ORDER BY inside OVER does not identify one row, the row that gets number 1 can change between runs.
Window aggregates. A window aggregate keeps every row, while a plain aggregate collapses them. Mixing the two in one SELECT without GROUP BY is an error.
PARTITION BY. PARTITION BY is not GROUP BY: it keeps every row and only decides which rows a window function looks at.