HardSQL challenge #334SELF JOINGROUP BYAGGREGATION

Count Direct Reports per Manager

Return each manager id with the number of employees reporting to them, largest team first.

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.

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The task

Find managers with highest number of direct reports.

A correct answer returns 10 rows with 2 columns: manager_id, subordinates.

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
manager_idsubordinates
914
1013
713
213
812
411
311
111
510
62

The complete result: 10 rows. This is the output of the model solution on the practice database.

What you practise

  • Filtering rows with a WHERE condition
  • Testing for missing values with IS NULL and IS NOT NULL
  • 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
  • Sorting the result with ORDER BY

Hint and approach

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

Show the hint

Self join GROUP BY manager_id

Show the step-by-step approach

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

  1. Read from the employees table.
  2. Filter in WHERE: keep the rows where manager_id has a value (it is not NULL).
  3. Group the rows by manager_id, so the result has one row for each value.
  4. For each group, compute COUNT(*).
  5. Sort the result by subordinates descending.

Model solution

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What to watch for

No join is needed: grouping on manager_id counts the reports. A self join is only required to show the manager's name. The WHERE clause removes the people at the top, whose manager_id is NULL.

  • GROUP BY. Every selected column must either be in GROUP BY or sit inside an aggregate. SQLite accepts other columns and returns an arbitrary value for them; most databases raise an error.
  • IS NULL. Comparing with = NULL or <> NULL never matches any row. NULL has to be tested with IS NULL or IS NOT NULL.
  • ORDER BY. Without ORDER BY the row order is not guaranteed, even if it looks stable. Rows that tie on the sort columns can come back in any order.
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Same technique, other tables

More practice on this pattern with different columns:

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