MediumSQL challenge #164SUBQUERY

Salary Above Dept Max

Find employees whose salary is equal to max in their department.

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 employees whose salary is equal to max in their department.

A correct answer returns 10 rows with 12 columns: id, first_name, last_name, email, department, department_id, title, salary, hire_date, manager_id, city, status.

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
idfirst_namelast_nameemaildepartmentdepartment_idtitlesalaryhire_datemanager_idcitystatus
6AmyNguyenanguyen90@corp.ioDesign10Product Designer1945432021-01-16NULLSeattleon_leave
11LauraRiveralrivera10@corp.ioSales3Sales Manager1957592022-11-191San Diegoactive
48StevenNguyensnguyen35@corp.ioMarketing2Content Lead2044492015-01-145Phoenixactive
50PaulRobinsonprobinson12@corp.ioFinance5Finance Manager1960322019-05-054Jacksonvilleactive
51StevenAdamssadams85@corp.ioProduct9Group PM1882732024-09-167Charlotteactive
55KennethCampbellkcampbell32@corp.ioLegal8Legal Counsel1922202016-03-139New Yorkactive
63RonaldRamirezrramirez6@corp.ioHR4HR Manager1910372021-05-012Phoenixactive
100NancyThompsonnthompson43@corp.ioCustomer Success7Senior CS Manager2039992023-01-2310New Yorkactive
106DonnaJohnsondjohnson40@corp.ioEngineering1VP Engineering1990292018-06-212Jacksonvilleactive
111JohnLeejlee7@corp.ioOperations6Operations Analyst2034672015-03-199Seattleactive

The complete result: 10 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

  • Returning every column of a table with SELECT *
  • Filtering rows with a WHERE condition
  • Summarising many rows into one value with COUNT, SUM, AVG, MIN or MAX
  • Writing a subquery that refers to the current row of the outer query

Hint and approach

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

Show the hint

Subquery MAX

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 salary equals the value returned by a subquery.
  3. The subquery is correlated: it refers to the current outer row, so it is evaluated for each row. It works like this:
    1. Read from the employees table.
    2. Filter in WHERE: keep the rows where department equals the outer row's department.
    3. Aggregate over all the rows that are left: compute MAX(salary). With no GROUP BY, the result is a single row.
  4. Return every column with SELECT *.

Model solution

Checking your session…

What to watch for

  • Correlated subquery. A correlated subquery is evaluated per outer row, which gets slow on large tables. A join to a grouped subquery or a window function usually does the same work in one pass.
  • Aggregate functions. Aggregates skip NULLs. AVG divides by the number of non-NULL values and COUNT(column) can be smaller than COUNT(*).
  • WHERE. WHERE is evaluated before the SELECT list, so it cannot use a column alias defined there or the result of an aggregate.
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