HardSQL challenge #368WINDOW FUNCTIONSTATISTICSSUBQUERY

Find Employees Earning Above the Median Salary

Return the employees whose salary is higher than the median salary.

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 employees earning above median salary.

A correct answer returns 60 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 60 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
Using the result of one query inside another
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 ()

Hint and approach

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

Show the hint

Median logic with ROW_NUMBER

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 runs once and returns a single value. It works like this:
1. Build an inner query first (a derived table):
  1. Read from the employees table.
  2. Number the rows with ROW_NUMBER(), ordered by salary.
  3. Compute COUNT(*) with an empty OVER (), so the figure for the whole result is repeated on every row.
2. 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.
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 median is calculated in a nested subquery and used as one number in the outer WHERE. Unlike the average, the median is not pulled up by a few very high salaries, so this returns exactly half of the employees when no salaries tie.

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.
Derived table. Columns computed in the inner query are only visible outside under their alias, so every calculated column there needs a name.