1075. Project Employees I

https://leetcode.com/problems/project-employees-i

Description

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 and for the second project is (3 + 2) / 2 = 2.50

ac