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difficulty 中等
edit_url https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3611.Find%20Overbooked%20Employees/README.md
tags
数据库

3611. 查找超预订员工

English Version

题目描述

表：employees

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| employee_id   | int     |
| employee_name | varchar |
| department    | varchar |
+---------------+---------+
employee_id 是这张表的唯一主键。
每一行包含一个员工和他们部门的信息。

表：meetings

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| meeting_id    | int     |
| employee_id   | int     |
| meeting_date  | date    |
| meeting_type  | varchar |
| duration_hours| decimal |
+---------------+---------+
meeting_id 是这张表的唯一主键。
每一行表示一位员工参加的会议。meeting_type 可以是 'Team'，'Client' 或 'Training'。

编写一个解决方案来查找会议密集型的员工 -  在任何给定周内，花费超过 50% 工作时间在会议上的员工。

  • 假定一个标准工作周是 40 小时
  • 计算每位员工 每周（周一至周日）的 总会议小时数
  • 员工如果每周会议时间超过 20 小时（40 小时工作时间的 50%），则被视为会议密集型。
  • 统计每位员工有多少周是会议密集周
  • 仅查找 至少 2 周会议密集的员工

返回结果表按会议密集周的数量降序排列，然后按员工姓名升序排列。结果格式如下所示。

 

示例：

Input:

employees 表：

+-------------+----------------+-------------+
| employee_id | employee_name  | department  |
+-------------+----------------+-------------+
| 1           | Alice Johnson  | Engineering |
| 2           | Bob Smith      | Marketing   |
| 3           | Carol Davis    | Sales       |
| 4           | David Wilson   | Engineering |
| 5           | Emma Brown     | HR          |
+-------------+----------------+-------------+

meetings 表：

+------------+-------------+--------------+--------------+----------------+
| meeting_id | employee_id | meeting_date | meeting_type | duration_hours |
+------------+-------------+--------------+--------------+----------------+
| 1          | 1           | 2023-06-05   | Team         | 8.0            |
| 2          | 1           | 2023-06-06   | Client       | 6.0            |
| 3          | 1           | 2023-06-07   | Training     | 7.0            |
| 4          | 1           | 2023-06-12   | Team         | 12.0           |
| 5          | 1           | 2023-06-13   | Client       | 9.0            |
| 6          | 2           | 2023-06-05   | Team         | 15.0           |
| 7          | 2           | 2023-06-06   | Client       | 8.0            |
| 8          | 2           | 2023-06-12   | Training     | 10.0           |
| 9          | 3           | 2023-06-05   | Team         | 4.0            |
| 10         | 3           | 2023-06-06   | Client       | 3.0            |
| 11         | 4           | 2023-06-05   | Team         | 25.0           |
| 12         | 4           | 2023-06-19   | Client       | 22.0           |
| 13         | 5           | 2023-06-05   | Training     | 2.0            |
+------------+-------------+--------------+--------------+----------------+

输出：

+-------------+----------------+-------------+---------------------+
| employee_id | employee_name  | department  | meeting_heavy_weeks |
+-------------+----------------+-------------+---------------------+
| 1           | Alice Johnson  | Engineering | 2                   |
| 4           | David Wilson   | Engineering | 2                   |
+-------------+----------------+-------------+---------------------+

解释：

  • Alice Johnson (employee_id = 1): 
    <ul>
	<li>6 月 5 日至 11 日（2023-06-05 至 2023-06-11）：8.0 + 6.0 + 7.0 = 21.0 小时（&gt; 20 小时）</li>
	<li>6 月 12&nbsp;日至 18&nbsp;日（2023-06-12 至 2023-06-18）: 12.0 + 9.0 = 21.0 小时（&gt; 20 小时）</li>
	<li>2 周会议密集</li>
</ul>
</li>
<li><strong>David Wilson (employee_id = 4):</strong>
<ul>
	<li>6 月 5 日至 11 日：25.0 小时（&gt; 20 小时）</li>
	<li>6 月 19&nbsp;日至 25&nbsp;日：22.0 小时（&gt; 20 小时）</li>
	<li>2 周会议密集</li>
</ul>
</li>
<li><strong>未包含的员工：</strong>
<ul>
	<li>Bob Smith（employee_id = 2）：6 月 5 日至 11 日：15.0 + 8.0 = 23.0 小时（&gt; 20），6 月 12&nbsp;日至 18&nbsp;日：10.0 小时（&lt; 20）。只有 1 个会议密集周。</li>
	<li>Carol Davis（employee_id = 3）：6 月 5 日至 11 日：4.0 + 3.0 = 7.0 小时（&lt; 20）。没有会议密集周。</li>
	<li>Emma Brown（employee_id = 5）：6 月 5 日至 11 日：2.0 小时（&lt; 20）。没有会议密集周。</li>
</ul>
</li>

结果表按 meeting_heavy_weeks 降序排列，然后按员工姓名升序排列。

解法

方法一：分组聚合 + 连接查询

我们先将数据按照 employee_id、year 和 week 分组，计算每个员工每周的会议总时长。接着筛选出会议时长超过 20 小时的周数，并统计每个员工的会议密集周数。最后将结果与员工表连接，筛选出会议密集周数大于等于 2 的员工，并按要求排序。

MySQL

# Write your MySQL query statement below
WITH
    week_meeting_hours AS (
        SELECT
            employee_id,
            YEAR(meeting_date) AS year,
            WEEK(meeting_date, 1) AS week,
            SUM(duration_hours) hours
        FROM meetings
        GROUP BY 1, 2, 3
    ),
    intensive_weeks AS (
        SELECT
            employee_id,
            employee_name,
            department,
            count(1) AS meeting_heavy_weeks
        FROM
            week_meeting_hours
            JOIN employees USING (employee_id)
        WHERE hours >= 20
        GROUP BY 1
    )
SELECT employee_id, employee_name, department, meeting_heavy_weeks
FROM intensive_weeks
WHERE meeting_heavy_weeks >= 2
ORDER BY 4 DESC, 2;

Pandas

import pandas as pd


def find_overbooked_employees(
    employees: pd.DataFrame, meetings: pd.DataFrame
) -> pd.DataFrame:
    meetings["meeting_date"] = pd.to_datetime(meetings["meeting_date"])
    meetings["year"] = meetings["meeting_date"].dt.isocalendar().year
    meetings["week"] = meetings["meeting_date"].dt.isocalendar().week

    week_meeting_hours = (
        meetings.groupby(["employee_id", "year", "week"], as_index=False)[
            "duration_hours"
        ]
        .sum()
        .rename(columns={"duration_hours": "hours"})
    )

    intensive_weeks = week_meeting_hours[week_meeting_hours["hours"] >= 20]

    intensive_count = (
        intensive_weeks.groupby("employee_id")
        .size()
        .reset_index(name="meeting_heavy_weeks")
    )

    result = intensive_count.merge(employees, on="employee_id")

    result = result[result["meeting_heavy_weeks"] >= 2]

    result = result.sort_values(
        ["meeting_heavy_weeks", "employee_name"], ascending=[False, True]
    )

    return result[
        ["employee_id", "employee_name", "department", "meeting_heavy_weeks"]
    ].reset_index(drop=True)