HardSQL challenge #345WINDOW FUNCTIONLAGAGGREGATION

Max Daily Sales Growth

Find products with highest growth between consecutive days.

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

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The task

Find products with highest growth between consecutive days.

A correct answer returns 492 rows with 3 columns: product_id, day, growth.

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
iduser_idtotaldiscountitem_countstatusorder_dateshipped_dateshipping_cityshipping_country
1432638.110.253cancelled2024-04-24NULLLos AngelesUSA
2829470.195delivered2024-03-152023-04-09ChicagoUSA
3136683.240.082delivered2023-02-152023-07-10Los AngelesUSA

order_items 496 rows

id INTEGER (PK) · order_id INTEGER · product_id INTEGER · quantity INTEGER · unit_price REAL · subtotal REAL

First 3 rows of the order_items table
idorder_idproduct_idquantityunit_pricesubtotal
11765192.21961.05
21823107.1321.3
316515.395.39

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
product_iddaygrowth
72024-05-034
102022-10-234
142023-09-144
172023-11-194
322024-01-094

First 5 of 492 rows. This is the output of the model solution on the practice database.

What you practise

  • Summarising many rows into one value with COUNT, SUM, AVG, MIN or MAX
  • Grouping by more than one column to get a row per combination
  • Combining rows from two tables on a matching key with an inner join
  • Querying the result of another query with a subquery in FROM
  • Reading the previous row's value with LAG()
  • Restarting a window calculation for each group with PARTITION BY
  • Sorting the result with ORDER 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 daily sales

Show the step-by-step approach

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

  1. Build an inner query first (a derived table):
    1. Read from the order_items table.
    2. Join orders on order_items.order_id = orders.id.
    3. Group the rows by order_items.product_id and day, so the result has one row for each combination.
    4. For each group, compute SUM(order_items.quantity).
  2. Fetch sales from the previous row with LAG(), ordered by day, restarting for each product_id. Where there is no such row the value is NULL.
  3. Combine those values into growth: sales - LAG(sales) OVER (PARTITION BY product_id ORDER BY day).
  4. Sort the result by growth descending.

Model solution

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What to watch for

  • 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.
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