project-employees-i.sql

/*

Table: Project

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| project_id  | int     |
| employee_id | int     |
+-------------+---------+
(project_id, employee_id) is the primary key of this table.
employee_id is a foreign key to Employee table.
Table: Employee

+------------------+---------+
| Column Name      | Type    |
+------------------+---------+
| employee_id      | int     |
| name             | varchar |
| experience_years | int     |
+------------------+---------+
employee_id is the primary key of this table.
Write an SQL query that reports the average experience years of all the employees for each project, rounded to 2 digits.

The query result format is in the following example:

Project table:
+-------------+-------------+
| project_id  | employee_id |
+-------------+-------------+
| 1           | 1           |
| 1           | 2           |
| 1           | 3           |
| 2           | 1           |
| 2           | 4           |
+-------------+-------------+

Employee table:
+-------------+--------+------------------+
| employee_id | name   | experience_years |
+-------------+--------+------------------+
| 1           | Khaled | 3                |
| 2           | Ali    | 2                |
| 3           | John   | 1                |
| 4           | Doe    | 2                |
+-------------+--------+------------------+

Result table:
+-------------+---------------+
| project_id  | average_years |
+-------------+---------------+
| 1           | 2.00          |
| 2           | 2.50          |
+-------------+---------------+
The average experience years for the first project is (3 + 2 + 1) / 3 = 2.00

*/

# V0 
select project_id, round(avg(experience_years), 2) as average_years
from Project inner join Employee
on Project.employee_id = Employee.employee_id
group by project_id

# V1
# https://code.dennyzhang.com/project-employees-i
select project_id, round(avg(experience_years), 2) as average_years
from Project inner join Employee
on Project.employee_id = Employee.employee_id
group by project_id

# V2 
# Time:  O(m + n) 
# Space: O(m + n) 
SELECT project_id, 
       Round(Avg(experience_years), 2) AS average_years 
FROM   project AS p 
       INNER JOIN employee AS e 
               ON p.employee_id = e.employee_id 
GROUP  BY project_id 
ORDER  BY NULL