add 615

Author: fishercoder1534

README.md

@@ -625,6 +625,7 @@ Your ideas/fixes/algorithms are more than welcome!
 |620|[Not Boring Movies](https://leetcode.com/problems/not-boring-movies/)|[Solution](../master/database/_620.sql) | | | Easy |
 |619|[Biggest Single Number](https://leetcode.com/problems/biggest-single-number/)|[Solution](../master/database/_619.sql) | | | Easy |
 |618|[Students Report By Geography](https://leetcode.com/problems/students-report-by-geography/)|[Solution](../master/database/_618.sql) | | | Hard | Session Variables
+|615|[Average Salary: Departments VS Company](https://leetcode.com/problems/average-salary-departments-vs-company/)|[Solution](../master/database/_615.sql) | | | Hard| 
 |614|[Second Degree Follower](https://leetcode.com/problems/second-degree-follower/)|[Solution](../master/database/_614.sql) | | | Medium | Inner Join
 |613|[Shortest Distance in a Line](https://leetcode.com/problems/shortest-distance-in-a-line/)|[Solution](../master/database/_613.sql) | || Easy|
 |612|[Shortest Distance in a Plane](https://leetcode.com/problems/shortest-distance-in-a-plane/)|[Solution](../master/database/_612.sql) | || Medium|

database/_615.sql

@@ -0,0 +1,53 @@
+--615. Average Salary: Departments VS Company
+--Given two tables as below, write a query to display the comparison result (higher/lower/same) of the average salary of employees in a department to the company's average salary.
+--Table: salary
+--
+--| 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 |
+--
+--The employee_id column refers to the employee_id in the following table employee.
+--
+--| employee_id | department_id |
+--|-------------|---------------|
+--| 1           | 1             |
+--| 2           | 2             |
+--| 3           | 2             |
+--
+--So for the sample data above, the result is:
+--
+--| pay_month | department_id | comparison  |
+--|-----------|---------------|-------------|
+--| 2017-03   | 1             | higher      |
+--| 2017-03   | 2             | lower       |
+--| 2017-02   | 1             | same        |
+--| 2017-02   | 2             | same        |
+--
+--Explanation
+--
+--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 he 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.
+
+SELECT d1.pay_month, d1.department_id,
+CASE WHEN d1.department_avg > c1.company_avg THEN 'higher'
+     WHEN d1.department_avg < c1.company_avg THEN 'lower'
+     ELSE 'same'
+END AS 'comparison'
+FROM ((SELECT LEFT(s1.pay_date, 7) pay_month, e1.department_id, AVG(s1.amount) department_avg
+FROM salary s1
+JOIN employee e1 ON s1.employee_id = e1.employee_id
+GROUP BY pay_month, e1.department_id) d1
+LEFT JOIN (SELECT LEFT(pay_date, 7) pay_month, AVG(amount) company_avg
+FROM salary
+GROUP BY pay_month) c1 ON d1.pay_month = c1.pay_month)
+ORDER BY pay_month DESC, department_id;