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Medium/Project Employees 3.sql

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+-- Question 85
+-- 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 most experienced employees in each project. 
+-- In case of a tie, report all employees with the maximum number of experience years.
+
+-- 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   | 3                |
+-- | 4           | Doe    | 2                |
+-- +-------------+--------+------------------+
+
+-- Result table:
+-- +-------------+---------------+
+-- | project_id  | employee_id   |
+-- +-------------+---------------+
+-- | 1           | 1             |
+-- | 1           | 3             |
+-- | 2           | 1             |
+-- +-------------+---------------+
+-- Both employees with id 1 and 3 have the 
+-- most experience among the employees of the first project. For the second project, the employee with id 1 has the most experience.
+
+-- Solution
+with t1 as(
+select p.project_id, p.employee_id, e.experience_years,
+rank() over(partition by project_id order by experience_years desc) as rk
+from project p
+join employee e
+on p.employee_id = e.employee_id)
+
+select t1.project_id, t1.employee_id
+from t1
+where t1.rk = 1