HardSQL challenge #297WINDOW FUNCTIONLAGDATE

Find the Longest Gap Between Consecutive Hires

Return the largest number of days between one hire and the next.

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.

Solve This SQL Challenge

The task

Find largest gap between consecutive hires.

A correct answer returns 1 row with one column: MAX(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
idfirst_namelast_nameemaildepartmentdepartment_idtitlesalaryhire_datemanager_idcitystatus
1WilliamLewiswlewis18@corp.ioCustomer Success7Senior CS Manager617742022-07-24NULLSan Diegoactive
2JohnMillerjmiller51@corp.ioLegal8Paralegal1916942021-08-09NULLSan Diegoactive
3DanielRiveradrivera50@corp.ioEngineering1VP Engineering1350672016-06-06NULLHoustonactive

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
MAX(diff)
133

The complete result: 1 row. 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
  • 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()
  • 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(hire_date)

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 employees table.
    2. Fetch hire_date from the previous row with LAG(), ordered by hire_date. Where there is no such row the value is NULL.
    3. Combine those values into diff: julianday(hire_date) - julianday(LAG(hire_date) OVER (ORDER BY hire_date)).
  2. Aggregate over all the rows that are left: compute MAX(diff). With no GROUP BY, the result is a single row.

Model solution

Checking your session…

What to watch for

A window function cannot be nested inside an aggregate, so MAX(LAG(...)) is not allowed. The gaps are calculated in a derived table and the maximum is taken outside it.

  • 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.
  • Derived table. Columns computed in the inner query are only visible outside under their alias, so every calculated column there needs a name.
  • Date functions. These date functions are SQLite syntax. MySQL uses DATE_FORMAT and DATEDIFF, PostgreSQL uses TO_CHAR, EXTRACT and date subtraction.
Solve This SQL Challenge

Keep learning