Added a new question- Average Salary: Departments Vs Company

Author: Indrajeet Mishra

Average Salary: Departments VS Company/README.md

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+# [Average Salary: Departments VS Company](https://leetcode.com/problems/average-salary-departments-vs-company/)
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+
+---
+
+## 🛠️ Problem Statement
+
+  <b>Table Name : Salary</b>
+
+|  Column Name  |Type |
+| ------------- | ------------- |
+| id          | int  |
+| employee_id | int  |
+| amount      | int  |
+| pay_date    | date |
+
+* id is the primary key column for this table. Each row of this table indicates the salary of an employee in one month.
+* employee_id is a foreign key from the Employee table.
+</br>
+
+  <b>Table Name : Employee</b>
+
+|  Column Name  |Type |
+| ------------- | ------------- |
+| employee_id   | int  |
+| department_id | int  |
+
+* employee_id is the primary key column for this table. Each row of this table indicates the department of an employee.
+
+</br>
+Write an SQL query to report the comparison result (higher/lower/same) of the average salary of employees
+in a department to the company’s average salary. Return the result table in any order.
+
+The query result format is in the following example.  
+</br>
+</br>
+ <details>
+<summary>
+Input
+</summary>
+</br>
+
+<b> Table Name: Salary </b></br>
+
+| id | employee_id | amount | pay_date   |
+| --------- | ------------- | ------ | --------- |
+| 1  | 1           | 9000   | 2017/03/31 |
+| 2  | 2           | 6000   | 2017/03/31 |
+| 3  | 3           | 10000  | 2017/03/31 |
+| 4  | 1           | 7000   | 2017/02/28 |
+| 5  | 2           | 6000   | 2017/02/28 |
+| 6  | 3           | 8000   | 2017/02/28 |
+
+<b> Table Name: Employee </b></br>
+
+| employee_id | department_id |
+|----|-------|
+| 1           | 1             |
+| 2           | 2             |
+| 3           | 2             |
+
+</details>
+
+<details>
+<summary>
+Output
+</summary>
+</br>
+
+| pay_month | department_id | comparison |
+| --- |--- | --- |
+| 2017-02   | 1             | same       |
+| 2017-03   | 1             | higher     |
+| 2017-02   | 2             | same       |
+| 2017-03   | 2             | lower      | 
+</details>
+
+<br>
+<b> Explanation: </b> 
+
+- In March, the company's average salary is (9000+6000+10000)/3 = 8333.33...
+- The average salary for department '1' is 9000, which is the salary of employee_id '1' since there is only
+one employee in this department. So the comparison result is 'higher' since 9000 > 8333.33 obviously.
+- The average salary of department '2' is (6000 + 10000)/2 = 8000, which is the average of employee_id '2' and '3'.
+So the comparison result is 'lower' since 8000 < 8333.33.
+
+With the same formula for the average salary comparison in February, the result is 'same' since both the department
+'1' and '2' have the same average salary with the company, which is 7000.
+
+---

Average Salary: Departments VS Company/schema.sql

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+-- Create the 'salary' table
+CREATE TABLE IF NOT EXISTS salary (
+    id INT,
+    employee_id INT,
+    amount INT,
+    pay_date DATE
+);
+
+-- Insert data into the 'salary' table
+INSERT INTO
+    salary (id, employee_id, amount, pay_date)
+VALUES
+    (1, 1, 9000, '2017-03-31'),
+    (2, 2, 6000, '2017-03-31'),
+    (3, 3, 10000, '2017-03-31'),
+    (4, 1, 7000, '2017-02-28'),
+    (5, 2, 6000, '2017-02-28'),
+    (6, 3, 8000, '2017-02-28');
+
+COMMIT;
+
+-- Create the 'employee' table
+CREATE TABLE IF NOT EXISTS employee (employee_id INT, department_id INT);
+
+-- Insert data into the 'employee' table
+INSERT INTO
+    employee (employee_id, department_id)
+VALUES
+    (1, 1),
+    (2, 2),
+    (3, 2);
+
+COMMIT;

Average Salary: Departments VS Company/solution.sql

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+WITH company_avg AS (
+    SELECT
+        to_char(pay_date, 'YYYY-MM') pay_month,
+        AVG(amount) company_avg
+    FROM
+        salary
+    GROUP BY
+        to_char(pay_date, 'YYYY-MM')
+),
+department_avg AS (
+    SELECT
+        to_char(a.pay_date, 'YYYY-MM') pay_month,
+        b.department_id,
+        AVG(a.amount) department_avg
+    FROM
+        salary a
+        INNER JOIN employee b ON a.employee_id = b.employee_id
+    GROUP BY
+        to_char(a.pay_date, 'YYYY-MM'),
+        b.department_id
+)
+SELECT
+    a.pay_month,
+    b.department_id,
+    CASE
+        WHEN b.department_avg > a.company_avg THEN 'higher'
+        WHEN b.department_avg < a.company_avg THEN 'lower'
+        ELSE 'same'
+    END AS comparison
+FROM
+    company_avg a
+    INNER JOIN department_avg b ON a.pay_month = b.pay_month
+ORDER BY
+    department_id,
+    pay_month;