task #184

README.md

@@ -2,6 +2,7 @@
 
 [![MIT licensed](https://img.shields.io/badge/License-MIT-yellow.svg)](./LICENSE)
 [![CI](https://github.com/ivanbyone/leetcode-sql/actions/workflows/ci.yml/badge.svg)](https://github.com/Ivanbyone/leetcode-sql//actions)
+![Postgres](https://img.shields.io/badge/postgres-%23316192.svg?style=for-the-badge&logo=postgresql&logoColor=white)
 
 ## Description
 
@@ -44,6 +45,7 @@ Have a good contributing!
     - [596. Classes With at Least 5 Students](./leetcode/easy/596.%20Classes%20With%20at%20Least%205%20Students.sql)
 2. [Medium](./leetcode/medium/)
     - [176. Second Highest Salary](./leetcode/medium/176.%20Second%20Highest%20Salary.sql)
+    - [184. Department Highest Salary](./leetcode/medium/184.%20Department%20Highest%20Salary.sql)
 
 ## License
 

leetcode/medium/184. Department Highest Salary.sql

@@ -0,0 +1,49 @@
+/*
+Question 184. Department Highest Salary
+Link: https://leetcode.com/problems/department-highest-salary/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 columns) 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. It is guaranteed that department name is not NULL.
+Each row of this table indicates the ID of a department and its name.
+
+
+Write a solution to find employees who have the highest salary 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.salary = (
+    SELECT MAX(em.salary) 
+    FROM Employee AS em
+    WHERE em.departmentId = e.departmentId
+)