HardSQL challenge #373WINDOW FUNCTIONPATTERNGROUP BY

Max Growth Streak per User

For each user, find their longest run of consecutive orders in which each order total is higher than the one before. Return the number of increases in that run.

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 user, find their longest run of consecutive orders in which each order total is higher than the one before. Return the number of increases in that run.

A correct answer returns 35 rows with 2 columns: user_id, streak.

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
user_id	streak
77	2
61	2
148	1
146	1
137	1

First 5 of 35 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
Grouping rows with GROUP BY to get one result row per group
Grouping by more than one column to get a row per combination
Querying the result of another query with a subquery in FROM
Turning conditions into values with CASE WHEN
Reading the previous row's value with LAG()
Accumulating a value row by row with an ordered window
Restarting a window calculation for each group with PARTITION BY
Sorting the result with ORDER BY

Hint and approach

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

Show the hint

Flag increases with LAG per user, group them with a running SUM of the non-increases, then take the longest group per user

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. Build an inner query first (a derived table):
  1. Build an inner query first (a derived table):
    1. Read from the orders table.
    2. Fetch total from the previous row with LAG(), ordered by order_date, restarting for each user_id. Where there is no such row the value is NULL.
    3. Label each row with a CASE expression (inc): the conditions are tested in order and the first one that is true decides the value.
  2. Compute SUM(CASE WHEN inc = 0 THEN 1 ELSE 0 END) as a running window function: it covers the rows from the start up to the current one, ordered by order_date, restarting for each user_id.
2. Filter in WHERE: keep the rows where inc equals 1. 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. Group the rows by user_id and grp, so the result has one row for each combination.
4. For each group, compute COUNT(*).
Group the rows by user_id, so the result has one row for each value.
For each group, compute MAX(streak).
Sort the result by streak descending.

Model solution

Checking your session…

What to watch for

Running totals. With ORDER BY and no frame, rows that tie on the sort column are added together in one step. Use ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW for a strict row-by-row total.
LAG(). LAG() returns NULL for the first row of each partition, and any arithmetic with that NULL is NULL too. Supply a default as the third argument if you need a number.
PARTITION BY. PARTITION BY is not GROUP BY: it keeps every row and only decides which rows a window function looks at.