HardSQL challenge #229WINDOW FUNCTIONLAGAGGREGATION

Find the Longest Gap Between Orders for Each Customer

Return each user with the largest number of days between two consecutive orders.

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

Find max gap in days between orders per user.

A correct answer returns 114 rows with 2 columns: user_id, MAX(gap).

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	MAX(gap)
1	NULL
2	NULL
4	607
5	NULL
6	361

First 5 of 114 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

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
Working with dates in SQLite: strftime, julianday and DATE
Reading the previous row's value with LAG()
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

LAG + MAX

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 orders table.
2. Fetch order_date 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. Combine those values into gap: julianday(order_date) - julianday(LAG(order_date) OVER (PARTITION BY user_id ORDER BY order_date)).
Group the rows by user_id, so the result has one row for each value.
For each group, compute MAX(gap).

Model solution

Checking your session…

What to watch for

Two levels are needed: the inner query calculates every gap with LAG(), and the outer query takes the maximum per user. Customers with a single order have no gap, so their result is NULL.

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

Solve This SQL Challenge

Calculate the Average Days Between Orders per CustomerSimilar levelHard
Find Customers Who Never Went More Than 7 Days Between OrdersSimilar levelHard
Count Orders per Month with strftimeEasierMedium
Calculate Days Between Consecutive Orders per Customer with LAG()Similar levelHard
Find the Longest Gap Between Consecutive HiresSimilar levelHard

Same technique, other tables

More practice on this pattern with different columns:

Keep learning

Lesson: LAG & LEAD
Lesson: Window Functions
Lesson: SQL Subqueries
Lesson: GROUP BY
All SQL practice challenges, by difficulty and topic
SQL interview questions with answers
SQL Playground: run any query on the same database