We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
**from Company c, Lead_Manager l, Senior_Manager s, Manager m, Employee e **
Actually this line is type of cross-join so for each record of each table are cross-joined.
There can be a better solution.
/*
Enter your query here.
*/
select company.company_code,company.founder,lead_table.leadm,senior_table.senior,manager_table.manager,emp_table.emp
from
(select company_code,count(distinct employee_code) as emp
from employee
group by company_code)as emp_table
inner join
(select company_code,count(distinct manager_code) as manager
from manager
group by company_code) as manager_table
on(emp_table.company_code=manager_table.company_code)
inner join
(select company_code,count(distinct senior_manager_code) as senior
from senior_manager
group by company_code)senior_table
on(senior_table.company_code=manager_table.company_code)
inner join
(select company_code,count(distinct lead_manager_code) as leadm
from lead_manager
group by company_code)lead_table
on(senior_table.company_code=lead_table.company_code)
inner join
Company
on(company.company_code=lead_table.company_code)
order by company.company_code;
Cookie support is required to access HackerRank
Seems like cookies are disabled on this browser, please enable them to open this website
New Companies
You are viewing a single comment's thread. Return to all comments →
**from Company c, Lead_Manager l, Senior_Manager s, Manager m, Employee e **
Actually this line is type of cross-join so for each record of each table are cross-joined. There can be a better solution.
/* Enter your query here. */ select company.company_code,company.founder,lead_table.leadm,senior_table.senior,manager_table.manager,emp_table.emp from (select company_code,count(distinct employee_code) as emp from employee group by company_code)as emp_table inner join (select company_code,count(distinct manager_code) as manager from manager group by company_code) as manager_table on(emp_table.company_code=manager_table.company_code) inner join (select company_code,count(distinct senior_manager_code) as senior from senior_manager group by company_code)senior_table on(senior_table.company_code=manager_table.company_code) inner join (select company_code,count(distinct lead_manager_code) as leadm from lead_manager group by company_code)lead_table on(senior_table.company_code=lead_table.company_code) inner join Company on(company.company_code=lead_table.company_code) order by company.company_code;