Exercise 28：2010.The Number of Seniors and Juniors to Join the Company II
1.Description
Table: Candidates
+-------------+------+
| Column Name | Type |
+-------------+------+
| employee_id | int |
| experience | enum |
| salary | int |
+-------------+------+
employee_id is the primary key column for this table.
experience is an enum with one of the values ('Senior', 'Junior').
Each row of this table indicates the id of a candidate, their monthly salary, and their experience.
The salary of each candidate is guaranteed to be unique.
A company wants to hire new employees. The budget of the company for the salaries is $70000. The company's criteria for hiring are:
Keep hiring the senior with the smallest salary until you cannot hire any more seniors.
Use the remaining budget to hire the junior with the smallest salary.
Keep hiring the junior with the smallest salary until you cannot hire any more juniors.
Write an SQL query to find the ids of seniors and juniors hired under the mentioned criteria.
Return the result table in any order.
The query result format is in the following example.
Example 1:
Input:
Candidates table:
+-------------+------------+--------+
| employee_id | experience | salary |
+-------------+------------+--------+
| 1 | Junior | 10000 |
| 9 | Junior | 15000 |
| 2 | Senior | 20000 |
| 11 | Senior | 16000 |
| 13 | Senior | 50000 |
| 4 | Junior | 40000 |
+-------------+------------+--------+
Output:
+-------------+
| employee_id |
+-------------+
| 11 |
| 2 |
| 1 |
| 9 |
+-------------+
Explanation:
We can hire 2 seniors with IDs (11, 2). Since the budget is $70000 and the sum of their salaries is $36000, we still have $34000 but they are not enough to hire the senior candidate with ID 13.
We can hire 2 juniors with IDs (1, 9). Since the remaining budget is $34000 and the sum of their salaries is $25000, we still have $9000 but they are not enough to hire the junior candidate with ID 4.
Example 2:
Input:
Candidates table:
+-------------+------------+--------+
| employee_id | experience | salary |
+-------------+------------+--------+
| 1 | Junior | 25000 |
| 9 | Junior | 10000 |
| 2 | Senior | 85000 |
| 11 | Senior | 80000 |
| 13 | Senior | 90000 |
| 4 | Junior | 30000 |
+-------------+------------+--------+
Output:
+-------------+
| employee_id |
+-------------+
| 9 |
| 1 |
| 4 |
+-------------+
Explanation:
We cannot hire any seniors with the current budget as we need at least $80000 to hire one senior.
We can hire all three juniors with the remaining budget.
2.Create Table and insert into values
create table if not exists leetcode.ex_2010_candidates_1
(employee_id string ,
experience string,
salary int
) stored as orc ;
INSERT INTO table leetcode.ex_2010_candidates_1 VALUES
('1','Junior','10000'),
('9','Junior','15000'),
('2','Senior','20000'),
('11','Senior','16000'),
('13','Senior','50000'),
('4','Junior','40000') ;
create table if not exists leetcode.ex_2010_candidates_2
(employee_id string ,
experience string,
salary int
) stored as orc ;
INSERT INTO table leetcode.ex_2010_candidates_2 VALUES
('1','Junior','25000'),
('9','Junior','10000'),
('2','Senior','85000'),
('11','Senior','80000'),
('13','Senior','90000'),
('4','Junior','30000') ;
The difference between this question and question 27 is only in the final Query , other steps are the same.