Merge pull request #87 from remy727/1075-project-employees-i

1075. Project Employees I

Author: remy727

README.md

@@ -384,6 +384,7 @@
 | 595 | Big Countries | [MySQL](./database/0595-big-countries.sql) | Easy |
 | 620 | Not Boring Movies | [MySQL](./database/0620-not-boring-movies.sql) | Easy |
 | 1068 | Product Sales Analysis I | [Ruby](./database/1068-product-sales-analysis-i.sql) | Easy |
+| 1075 | Project Employees I | [Ruby](./database/1075-project-employees-i.sql) | Easy |
 | 1148 | Article Views I | [MySQL](./database/1148-article-views-i.sql) | Easy |
 | 1251 | Average Selling Price | [MySQL](./database/1251-average-selling-price.sql) | Easy |
 | 1280 | Students and Examinations | [MySQL](./database/1280-students-and-examinations.sql) | Easy |

database/1075-project-employees-i.sql

@@ -0,0 +1,71 @@
+-- 1075. Project Employees I
+-- Easy
+-- https://leetcode.com/problems/project-employees-i
+
+/*
+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.
+Each row of this table indicates that the employee with employee_id is working on the project with project_id.
+
+Table: Employee
++------------------+---------+
+| Column Name      | Type    |
++------------------+---------+
+| employee_id      | int     |
+| name             | varchar |
+| experience_years | int     |
++------------------+---------+
+employee_id is the primary key of this table. It's guaranteed that experience_years is not NULL.
+Each row of this table contains information about one employee.
+
+Write an SQL query that reports the average experience years of all the employees for each project, rounded to 2 digits.
+Return the result table in any order.
+The query result format is in the following example.
+
+Example 1:
+Input: 
+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                |
++-------------+--------+------------------+
+Output: 
++-------------+---------------+
+| project_id  | average_years |
++-------------+---------------+
+| 1           | 2.00          |
+| 2           | 2.50          |
++-------------+---------------+
+Explanation: 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
+*/
+
+SELECT
+	p.project_id,
+ROUND(AVG(e.experience_years), 2) as average_years
+FROM
+	Project p
+JOIN
+	Employee e ON p.employee_id = e.employee_id
+GROUP BY
+	p.project_id;