HardSQL challenge #382WINDOW FUNCTIONPARTITIONCALCULATION

Percent Difference of Each Salary from the Department Average

Return every employee with how far, in percent, their salary is from their department's average.

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

Calculate percentage difference of employee salary from department average.

A correct answer returns 120 rows with 13 columns: id, first_name, last_name, email, department, department_id, title, salary, hire_date, manager_id, city, status, pct_diff.

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	pct_diff
1	William	Lewis	wlewis18@corp.io	Customer Success	7	Senior CS Manager	61774	2022-07-24	NULL	San Diego	active	-53.948091023354216
15	Barbara	Baker	bbaker4@corp.io	Customer Success	7	Senior CS Manager	188280	2018-12-15	8	Philadelphia	on_leave	40.360886815211394
18	Robert	Green	rgreen61@corp.io	Customer Success	7	CS Manager	73118	2023-08-04	9	San Diego	active	-45.49125067901728
20	Edward	Ramirez	eramirez83@corp.io	Customer Success	7	Senior CS Manager	137737	2016-11-10	9	Indianapolis	active	2.6815777951283812
21	Charles	Allen	callen32@corp.io	Customer Success	7	CS Director	162901	2023-02-14	8	Houston	active	21.44109211326084

First 5 of 120 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

Attaching a total or average to every row with an aggregate OVER ()
Restarting a window calculation for each group with PARTITION BY
Deriving a new column from an expression and naming it with an alias

Hint and approach

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

Show the hint

salary - avg / avg

Show the step-by-step approach

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

Read from the employees table.
Compute AVG(salary) as a window function over each department, so the group's value is repeated on every one of its rows.
Combine those values into pct_diff: (salary - AVG(salary) OVER (PARTITION BY department)) * 100.0 / AVG(salary) OVER (PARTITION BY department).

Model solution

Checking your session…

What to watch for

The formula is (salary - department average) * 100.0 / department average, with the average supplied by a partitioned window. Positive means above the department average.

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.
Calculated columns. Dividing one integer by another truncates the result in SQLite. Multiply one side by 1.0 to get decimals.

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