SQL LeetCode: 569. Median Employee Salary

569. Median Employee Salary

Hard

The Employee table holds all employees. The employee table has three columns: Employee Id, Company Name, and Salary.

+-----+------------+--------+
|Id   | Company    | Salary |
+-----+------------+--------+
|1    | A          | 2341   |
|2    | A          | 341    |
|3    | A          | 15     |
|4    | A          | 15314  |
|5    | A          | 451    |
|6    | A          | 513    |
|7    | B          | 15     |
|8    | B          | 13     |
|9    | B          | 1154   |
|10   | B          | 1345   |
|11   | B          | 1221   |
|12   | B          | 234    |
|13   | C          | 2345   |
|14   | C          | 2645   |
|15   | C          | 2645   |
|16   | C          | 2652   |
|17   | C          | 65     |
+-----+------------+--------+

Write a SQL query to find the median salary of each company. Bonus points if you can solve it without using any built-in SQL functions.

+-----+------------+--------+
|Id   | Company    | Salary |
+-----+------------+--------+
|5    | A          | 451    |
|6    | A          | 513    |
|12   | B          | 234    |
|9    | B          | 1154   |
|14   | C          | 2645   |
+-----+------------+--------+

WITH CTE AS (
SELECT
Id, Company, salary,
row_number() OVER (partition by Company order by salary) as rank_id,
count(ID) OVER (partition by Company) as count_ID
FROM
Employee )

SELECT
ID,
Company,
Salary
FROM CTE
WHERE (CASE
WHEN count_ID%2!=0 THEN rank_id=(count_ID+1)/2
ELSE rank_id = count_ID/2 OR rank_id = (count_ID/2)+1
END)