LC hard problem ~ DepartmentTop3Salaries.sql

Author: absognety

LeetCode/SQL50/DepartmentTop3Salaries.sql

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+-- Table: Employee
+
+-- +--------------+---------+
+-- | Column Name  | Type    |
+-- +--------------+---------+
+-- | id           | int     |
+-- | name         | varchar |
+-- | salary       | int     |
+-- | departmentId | int     |
+-- +--------------+---------+
+-- id is the primary key (column with unique values) for this table.
+-- departmentId is a foreign key (reference column) of the ID from the Department table.
+-- Each row of this table indicates the ID, name, and salary of an employee. It also contains the ID of their department.
+ 
+-- Table: Department
+
+-- +-------------+---------+
+-- | Column Name | Type    |
+-- +-------------+---------+
+-- | id          | int     |
+-- | name        | varchar |
+-- +-------------+---------+
+-- id is the primary key (column with unique values) for this table.
+-- Each row of this table indicates the ID of a department and its name.
+ 
+-- A company's executives are interested in seeing who earns the most money in each of the company's departments. A high earner in a department is an employee who has a salary in the top three unique salaries for that department.
+
+-- Write a solution to find the employees who are high earners in each of the departments.
+
+-- Return the result table in any order.
+
+-- The result format is in the following example.
+
+-- Example 1:
+
+-- Input: 
+-- Employee table:
+-- +----+-------+--------+--------------+
+-- | id | name  | salary | departmentId |
+-- +----+-------+--------+--------------+
+-- | 1  | Joe   | 85000  | 1            |
+-- | 2  | Henry | 80000  | 2            |
+-- | 3  | Sam   | 60000  | 2            |
+-- | 4  | Max   | 90000  | 1            |
+-- | 5  | Janet | 69000  | 1            |
+-- | 6  | Randy | 85000  | 1            |
+-- | 7  | Will  | 70000  | 1            |
+-- +----+-------+--------+--------------+
+-- Department table:
+-- +----+-------+
+-- | id | name  |
+-- +----+-------+
+-- | 1  | IT    |
+-- | 2  | Sales |
+-- +----+-------+
+-- Output: 
+-- +------------+----------+--------+
+-- | Department | Employee | Salary |
+-- +------------+----------+--------+
+-- | IT         | Max      | 90000  |
+-- | IT         | Joe      | 85000  |
+-- | IT         | Randy    | 85000  |
+-- | IT         | Will     | 70000  |
+-- | Sales      | Henry    | 80000  |
+-- | Sales      | Sam      | 60000  |
+-- +------------+----------+--------+
+-- Explanation: 
+-- In the IT department:
+-- - Max earns the highest unique salary
+-- - Both Randy and Joe earn the second-highest unique salary
+-- - Will earns the third-highest unique salary
+
+-- In the Sales department:
+-- - Henry earns the highest salary
+-- - Sam earns the second-highest salary
+-- - There is no third-highest salary as there are only two employees
+
+-- Write your PostgreSQL query statement below
+-- Solution
+with rank_salaries_each_dep as (
+    select employee.id as emp_id,
+           employee.name as emp_name,
+           department.id as dep_id,
+           department.name as dep_name,
+           employee.salary as employee_salary,
+           dense_rank() over (partition by department.id,department.name 
+           order by employee.salary desc) rank
+    from department inner join employee 
+    on department.id = employee.departmentId
+)
+select dep_name as Department,
+       emp_name as Employee,
+       employee_salary as Salary
+from rank_salaries_each_dep where rank in (1,2,3);