select distinct *
from   t t1
where  5 >= (
  select count ( distinct t2.sal )
  from   t t2
  where  t2.deptno   = t1.deptno
  and    t2.sal >= t1.sal
);

Why 5? I dunno - you'll have to ask the original author for that one!

As to what this query is doing...

It's returning a list of all the rows where:

- For the same deptno
- There are five or less different values for sal that are the same or more than the sal for the current row

Plugging in employees, the question it's answering is:

Which employees have five or less salaries greater than or equal to their salary in their department?

Or to put it another way:

Which employees earn one of the top five salaries in their department?

Which gives this query:

select employee_id, salary, department_id
from   hr.employees t1
where  5 >= (
  select count( distinct t2.salary )
  from   hr.employees t2
  where  t2.department_id = t1.department_id
  and    t2.salary >= t1.salary
);

EMPLOYEE_ID   SALARY   DEPARTMENT_ID   
          100    24000              90 
          101    17000              90 
          102    17000              90 
          103     9000              60 
          104     6000              60 
          105     4800              60 
          106     4800              60 
          107     4200              60 
          108    12008             100 
          109     9000             100 
          110     8200             100 
          111     7700             100 
          112     7800             100 
          114    11000              30 
          115     3100              30 
          116     2900              30 
          117     2800              30 
          118     2600              30 
          120     8000              50 
          121     8200              50 
          122     7900              50 
          123     6500              50 
          124     5800              50 
          145    14000              80 
          146    13500              80 
          147    12000              80 
          148    11000              80 
          168    11500              80 
          174    11000              80 
          178     7000          <null> 
          200     4400              10 
          201    13000              20 
          202     6000              20 
          203     6500              40 
          204    10000              70 
          205    12008             110 
          206     8300             110

There's no need for the second query against employees. With dense_rank you can find the top five salaries per department.

Which gives a query like:

select employee_id, salary, department_id 
from (
  select employee_id, salary, department_id ,
         dense_rank () over (
           partition by department_id 
           order by salary desc
         ) dr
  from   hr.employees 
)
where  dr <= 5;