task: #185

Author: Ivanbyone

README.md

@@ -60,6 +60,8 @@ Have a good contributing!
     - [1045. Customers Who Bought All Products](./leetcode/medium/1045.%20Customers%20Who%20Bought%20All%20Products.sql)
     - [1070. Product Sales Analysis III](./leetcode/medium/1070.%20Product%20Sales%20Analysis%203.sql)
     - [1158. Market Analysis 1](./leetcode/medium/1158.%20Market%20Analysis%201.sql)
+3. [Hard](./leetcode/hard/)
+    - [185. Department Top Three Salaries](./leetcode/hard/185.%20Department%20Top%20Three%20Salaries.sql)
 
 ## License
 

leetcode/hard/185. Department Top Three Salaries.sql

@@ -0,0 +1,56 @@
+/*
+Question 185. Department Top Three Salaries
+Link: https://leetcode.com/problems/department-top-three-salaries/description/
+
+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.
+*/
+
+SELECT 
+    d.name AS Department,
+    e.name AS Employee,
+    e.salary
+FROM Employee AS e
+LEFT JOIN 
+    Department AS d
+    ON e.departmentId = d.id
+WHERE (e.departmentId, e.salary) IN (
+    SELECT
+        e2.departmentId,
+        e2.salary 
+    FROM Employee AS e2
+    WHERE e2.departmentId = d.id 
+    GROUP BY e2.departmentId, e2.salary 
+    ORDER BY e2.salary DESC 
+    LIMIT 3
+)