570. 至少有5名直接下属的经理

题目描述

表: Employee

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| name        | varchar |
| department  | varchar |
| managerId   | int     |
+-------------+---------+
id 是此表的主键（具有唯一值的列）。
该表的每一行表示雇员的名字、他们的部门和他们的经理的id。
如果managerId为空，则该员工没有经理。
没有员工会成为自己的管理者。

 

编写一个解决方案，找出至少有五个直接下属的经理。

以 任意顺序 返回结果表。

查询结果格式如下所示。

 

示例 1:

输入: 
Employee 表:
+-----+-------+------------+-----------+
| id  | name  | department | managerId |
+-----+-------+------------+-----------+
| 101 | John  | A          | Null      |
| 102 | Dan   | A          | 101       |
| 103 | James | A          | 101       |
| 104 | Amy   | A          | 101       |
| 105 | Anne  | A          | 101       |
| 106 | Ron   | B          | 101       |
+-----+-------+------------+-----------+
输出: 
+------+
| name |
+------+
| John |
+------+

解法

方法一：分组统计 + 连接

我们可以先统计每个经理的直接下属人数，然后再连接 Employee 表，找出直接下属人数大于等于 \(5\) 的经理。

Python3MySQL

import pandas as pd


def find_managers(employee: pd.DataFrame) -> pd.DataFrame:
    # Group the employees by managerId and count the number of direct reports
    manager_report_count = (
        employee.groupby("managerId").size().reset_index(name="directReports")
    )

    # Filter managers with at least five direct reports
    result = manager_report_count[manager_report_count["directReports"] >= 5]

    # Merge with the Employee table to get the names of these managers
    result = result.merge(
        employee[["id", "name"]], left_on="managerId", right_on="id", how="inner"
    )

    # Select only the 'name' column and drop the 'id' and 'directReports' columns
    result = result[["name"]]

    return result

# Write your MySQL query statement below
SELECT name
FROM
    Employee
    JOIN (
        SELECT managerId AS id, COUNT(1) AS cnt
        FROM Employee
        GROUP BY 1
        HAVING cnt >= 5
    ) AS t
        USING (id);

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