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2004. 职员招聘人数 🔒

题目描述

表: Candidates

+-------------+------+
| Column Name | Type |
+-------------+------+
| employee_id | int  |
| experience  | enum |
| salary      | int  |
+-------------+------+
employee_id是此表的主键列。
经验是包含一个值(“高级”、“初级”)的枚举类型。
此表的每一行都显示候选人的id、月薪和经验。

 

一家公司想雇佣新员工。公司的工资预算是 70000 美元。公司的招聘标准是:

  1. 雇佣最多的高级员工。
  2. 在雇佣最多的高级员工后,使用剩余预算雇佣最多的初级员工。

编写一个SQL查询,查找根据上述标准雇佣的高级员工和初级员工的数量。
任意顺序 返回结果表。
查询结果格式如下例所示。

 

示例 1:

输入: 
Candidates table:
+-------------+------------+--------+
| employee_id | experience | salary |
+-------------+------------+--------+
| 1           | Junior     | 10000  |
| 9           | Junior     | 10000  |
| 2           | Senior     | 20000  |
| 11          | Senior     | 20000  |
| 13          | Senior     | 50000  |
| 4           | Junior     | 40000  |
+-------------+------------+--------+
输出: 
+------------+---------------------+
| experience | accepted_candidates |
+------------+---------------------+
| Senior     | 2                   |
| Junior     | 2                   |
+------------+---------------------+
说明:
我们可以雇佣2名ID为(2,11)的高级员工。由于预算是7万美元,他们的工资总额是4万美元,我们还有3万美元,但他们不足以雇佣ID为13的高级员工。
我们可以雇佣2名ID为(1,9)的初级员工。由于剩下的预算是3万美元,他们的工资总额是2万美元,我们还有1万美元,但他们不足以雇佣ID为4的初级员工。

示例 2:

输入: 
Candidates table:
+-------------+------------+--------+
| employee_id | experience | salary |
+-------------+------------+--------+
| 1           | Junior     | 10000  |
| 9           | Junior     | 10000  |
| 2           | Senior     | 80000  |
| 11          | Senior     | 80000  |
| 13          | Senior     | 80000  |
| 4           | Junior     | 40000  |
+-------------+------------+--------+
输出: 
+------------+---------------------+
| experience | accepted_candidates |
+------------+---------------------+
| Senior     | 0                   |
| Junior     | 3                   |
+------------+---------------------+
解释:
我们不能用目前的预算雇佣任何高级员工,因为我们需要至少80000美元来雇佣一名高级员工。
我们可以用剩下的预算雇佣三名初级员工。

解法

方法一:窗口函数

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# Write your MySQL query statement below
WITH
    s AS (
        SELECT
            employee_id,
            SUM(salary) OVER (ORDER BY salary) AS cur
        FROM Candidates
        WHERE experience = 'Senior'
    ),
    j AS (
        SELECT
            employee_id,
            IFNULL(
                SELECT
                    MAX(cur)
                FROM s
                WHERE cur <= 70000,
                0
            ) + SUM(salary) OVER (ORDER BY salary) AS cur
        FROM Candidates
        WHERE experience = 'Junior'
    )
SELECT
    'Senior' AS experience,
    COUNT(employee_id) AS accepted_candidates
FROM s
WHERE cur <= 70000
UNION ALL
SELECT
    'Junior' AS experience,
    COUNT(employee_id) AS accepted_candidates
FROM j
WHERE cur <= 70000;

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