LeetCode 1075. 1076 Project Employees

drop table Project
drop table Employees

Create table Project(project_id int,employee_id int);

insert into Project values(1,1);
insert into Project values(1,2);
insert into Project values(1,3);
insert into Project values(2,1);
insert into Project values(2,4);

Create table Employees(employee_id int,name varchar(20),experience_years int);
insert into Employees values(1,'Khaled',3);
insert into Employees values(2,'Ali',2);
insert into Employees values(3,'John',1);
insert into Employees values(4,'Doe',2);

select * from Employees
select * from Project

 

1075. Project Employees I

Write an SQL query that reports the average experience years of all the employees for each project, rounded to 2 digits.

select project_id,ROUND(AVG(CAST(e.experience_years AS FLOAT)), 2) as average_years
from Project p
join Employees e
on p.employee_id=e.employee_id
group by p.project_id

 

1076. Project Employees II

Write an SQL query that reports all the projects that have the most employees.

select project_id
from Project
group by project_id
having count(*) in
(
select TOP 1 count(employee_id) as cnt
from Project
group by project_id
order by cnt desc
)