[HR] New Companies

• Advanced Select > Join+Groupby+count
  https://www.hackerrank.com/challenges/the-company/problem?isFullScreen=true

select c.company_code, c.founder
      ,count(distinct l.lead_manager_code)
      ,count(distinct s.senior_manager_code)
      ,count(distinct m.manager_code)
      ,count(distinct e.employee_code)
from company c
left join lead_manager l on c.company_code = l.company_code
left join senior_manager s on l.lead_manager_code = s.lead_manager_code
left join manager m on s.senior_manager_code = m.senior_manager_code
left join employee e on m.manager_code = e.manager_code
group by c.company_code, c.founder
order by c.company_code