HardSQL challenge #295WINDOW FUNCTIONSTATISTICSSUBQUERY

Calculate a Running Median of Order Totals

Return each order with the median total of all orders placed up to and including its date.

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

Solve This SQL Challenge

The task

For each order, calculate the running median: the median total of all orders placed up to and including that order date.

A correct answer returns 200 rows with 4 columns: id, order_date, total, running_median.

Tables used in this challenge

`orders` 200 rows

id INTEGER (PK) · user_id INTEGER · total REAL · discount REAL · item_count INTEGER · status TEXT · order_date TEXT · shipped_date TEXT · shipping_city TEXT · shipping_country TEXT

First 3 rows of the orders table
id	user_id	total	discount	item_count	status	order_date	shipped_date	shipping_city	shipping_country
1	43	2638.11	0.25	3	cancelled	2024-04-24	NULL	Los Angeles	USA
2	8	2947	0.19	5	delivered	2024-03-15	2023-04-09	Chicago	USA
3	136	683.24	0.08	2	delivered	2023-02-15	2023-07-10	Los Angeles	USA

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	order_date	total	running_median
58	2022-01-06	1843.49	1843.49
4	2022-01-10	3920.44	2881.965
89	2022-01-15	1152.13	1843.49
101	2022-01-16	3391.66	2617.575
8	2022-01-17	4163.77	3391.66

First 5 of 200 rows. This is the output of the model solution on the practice database.

What you practise

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
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 ()
Sorting the result with ORDER BY

Hint and approach

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

Show the hint

Use a correlated subquery that computes the median of the earlier orders with ROW_NUMBER and COUNT

Show the step-by-step approach

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

Read from the orders table.
Number the rows with ROW_NUMBER(), ordered by total.
Compute COUNT(*) with an empty OVER (), so the figure for the whole result is repeated on every row.
A subquery in the SELECT list is evaluated for each output row, using values from that row:
1. Build an inner query first (a derived table):
  1. Read from the orders table.
  2. Filter in WHERE: keep the rows where order_date is at most the outer row's order_date.
  3. Number the rows with ROW_NUMBER(), ordered by total.
  4. 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(total). With no GROUP BY, the result is a single row.
Sort the result by order_date.

Model solution

Checking your session…

What to watch for

SQLite has no MEDIAN and no window frame that can produce one, so each row gets a correlated subquery that ranks the earlier orders and averages the middle one or two. That is quadratic work: fine for 200 orders, slow for a large table. Orders on the same date share one median, because the cutoff is the date.

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.
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.