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feat: add sql solution to lc problem: No.3156 (#2859)
No.3156.Employee Task Duration and Concurrent Tasks
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‎solution/3100-3199/3156.Employee Task Duration and Concurrent Tasks/README.md

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---
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comments: true
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difficulty: 困难
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3156.Employee%20Task%20Duration%20and%20Concurrent%20Tasks/README.md
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---
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<!-- problem:start -->
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# [3156. 员工任务持续时间和并发任务 🔒](https://leetcode.cn/problems/employee-task-duration-and-concurrent-tasks)
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[English Version](/solution/3100-3199/3156.Employee%20Task%20Duration%20and%20Concurrent%20Tasks/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>表：<code>Tasks</code></p>
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<pre>
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+---------------+----------+
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| Column Name | Type |
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+---------------+----------+
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| task_id | int |
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| employee_id | int |
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| start_time | datetime |
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| end_time | datetime |
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+---------------+----------+
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(task_id, employee_id) 是这张表的主键。
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这张表的每一行包含任务标识，员工标识和每个任务的开始和结束时间。
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</pre>
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<p>编写一个解决方案来查找 <strong>每个</strong> 员工的任务 <strong>总持续时间</strong> 以及员工在任何时间点处理的 <strong>最大并发任务数</strong>。总时长应该 <strong>舍入</strong> 到最近的 <strong>整小时</strong>。</p>
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<p>返回结果表以&nbsp;<code>employee_id</code><strong> <em>升序</em></strong><em>&nbsp;排序。</em></p>
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<p>结果格式如下所示。</p>
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<p>&nbsp;</p>
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<p><strong class="example">示例：</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong></p>
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<p>Tasks 表：</p>
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<pre class="example-io">
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+---------+-------------+---------------------+---------------------+
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| task_id | employee_id | start_time | end_time |
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+---------+-------------+---------------------+---------------------+
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| 1 | 1001 | 2023-05-01 08:00:00 | 2023-05-01 09:00:00 |
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| 2 | 1001 | 2023-05-01 08:30:00 | 2023-05-01 10:30:00 |
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| 3 | 1001 | 2023-05-01 11:00:00 | 2023-05-01 12:00:00 |
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| 7 | 1001 | 2023-05-01 13:00:00 | 2023-05-01 15:30:00 |
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| 4 | 1002 | 2023-05-01 09:00:00 | 2023-05-01 10:00:00 |
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| 5 | 1002 | 2023-05-01 09:30:00 | 2023-05-01 11:30:00 |
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| 6 | 1003 | 2023-05-01 14:00:00 | 2023-05-01 16:00:00 |
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+---------+-------------+---------------------+---------------------+
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</pre>
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<p><strong>输出：</strong></p>
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<pre class="example-io">
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+-------------+------------------+----------------------+
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| employee_id | total_task_hours | max_concurrent_tasks |
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+-------------+------------------+----------------------+
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| 1001 | 6 | 2 |
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| 1002 | 2 | 2 |
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| 1003 | 2 | 1 |
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+-------------+------------------+----------------------+
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</pre>
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<p><strong>解释：</strong></p>
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<ul>
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<li>对于员工 ID 1001：
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<ul>
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<li>任务 1 和任务 2 从 08:30 到&nbsp;09:00 重叠（30 分钟）。</li>
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<li>任务 7 持续时间为 150 分钟（2 小时 30 分钟）。</li>
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<li>总工作小时：60（任务 1）+ 120（任务 2）+ 60（任务&nbsp;3）+ 150（任务 7）- 30（重叠）= 360 分钟 = 6 小时。</li>
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<li>最大并发任务：2 （重叠期间）。</li>
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</ul>
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</li>
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<li>对于员工 ID 1002：
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<ul>
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<li>任务 4 和任务 5 从 09:30 到&nbsp;10:00 重叠（30 分钟）。</li>
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<li>总工作时间：60 （任务&nbsp;4）+ 120（任务 5）- 30（重叠）= 150 分钟 = 2 小时 30 分钟。</li>
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<li>总工作小时：（舍入后）：2 小时。</li>
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<li>最大并发任务：2 （重叠期间）。</li>
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</ul>
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</li>
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<li>对于员工 ID 1003：
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<ul>
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<li>没有重叠的工作。</li>
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<li>总工作时间：120 分钟 = 2 小时。</li>
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<li>最大并发任务：1。</li>
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</ul>
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</li>
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</ul>
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<p><b>注意：</b>输出表以 employee_id 升序排序。</p>
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</div>
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<!-- description:end -->
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## 解法
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<!-- solution:start -->
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### 方法一
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<!-- tabs:start -->
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#### MySQL
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```sql
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# Write your MySQL query statement below
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WITH
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T AS (
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SELECT DISTINCT employee_id, start_time AS st
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FROM Tasks
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UNION DISTINCT
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SELECT DISTINCT employee_id, end_time AS st
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FROM Tasks
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),
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P AS (
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SELECT
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*,
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LEAD(st) OVER (
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PARTITION BY employee_id
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ORDER BY st
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) AS ed
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FROM T
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),
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S AS (
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SELECT
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P.*,
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COUNT(1) AS concurrent_count
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FROM
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P
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INNER JOIN Tasks USING (employee_id)
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WHERE P.st >= Tasks.start_time AND P.ed <= Tasks.end_time
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GROUP BY 1, 2, 3
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)
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SELECT
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employee_id,
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FLOOR(SUM(TIME_TO_SEC(TIMEDIFF(ed, st)) / 3600)) AS total_task_hours,
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MAX(concurrent_count) AS max_concurrent_tasks
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FROM S
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GROUP BY 1
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ORDER BY 1;
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

‎solution/3100-3199/3156.Employee Task Duration and Concurrent Tasks/README_EN.md

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---
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comments: true
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difficulty: Hard
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3156.Employee%20Task%20Duration%20and%20Concurrent%20Tasks/README_EN.md
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---
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<!-- problem:start -->
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# [3156. Employee Task Duration and Concurrent Tasks 🔒](https://leetcode.com/problems/employee-task-duration-and-concurrent-tasks)
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[中文文档](/solution/3100-3199/3156.Employee%20Task%20Duration%20and%20Concurrent%20Tasks/README.md)
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## Description
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<!-- description:start -->
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<p>Table: <code>Tasks</code></p>
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<pre>
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+---------------+----------+
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| Column Name | Type |
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+---------------+----------+
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| task_id | int |
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| employee_id | int |
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| start_time | datetime |
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| end_time | datetime |
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+---------------+----------+
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(task_id, employee_id) is the primary key for this table.
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Each row in this table contains the task identifier, the employee identifier, and the start and end times of each task.
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</pre>
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<p>Write a solution to find the <strong>total duration</strong> of tasks for <strong>each</strong> employee and the <strong>maximum number of concurrent tasks</strong> an employee handled at <strong>any point in time</strong>. The total duration should be <strong>rounded down</strong> to the nearest number of <strong>full hours</strong>.</p>
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<p>Return <em>the result table ordered by</em>&nbsp;<code>employee_id</code><strong> <em>ascending</em></strong><em> order</em>.</p>
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<p>The result format is in the following example.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong></p>
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<p>Tasks table:</p>
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<pre class="example-io">
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+---------+-------------+---------------------+---------------------+
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| task_id | employee_id | start_time | end_time |
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+---------+-------------+---------------------+---------------------+
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| 1 | 1001 | 2023-05-01 08:00:00 | 2023-05-01 09:00:00 |
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| 2 | 1001 | 2023-05-01 08:30:00 | 2023-05-01 10:30:00 |
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| 3 | 1001 | 2023-05-01 11:00:00 | 2023-05-01 12:00:00 |
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| 7 | 1001 | 2023-05-01 13:00:00 | 2023-05-01 15:30:00 |
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| 4 | 1002 | 2023-05-01 09:00:00 | 2023-05-01 10:00:00 |
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| 5 | 1002 | 2023-05-01 09:30:00 | 2023-05-01 11:30:00 |
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| 6 | 1003 | 2023-05-01 14:00:00 | 2023-05-01 16:00:00 |
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+---------+-------------+---------------------+---------------------+
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</pre>
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<p><strong>Output:</strong></p>
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<pre class="example-io">
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+-------------+------------------+----------------------+
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| employee_id | total_task_hours | max_concurrent_tasks |
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+-------------+------------------+----------------------+
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| 1001 | 6 | 2 |
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| 1002 | 2 | 2 |
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| 1003 | 2 | 1 |
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+-------------+------------------+----------------------+
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</pre>
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<p><strong>Explanation:</strong></p>
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<ul>
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<li>For employee ID 1001:
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<ul>
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<li>Task 1 and Task 2 overlap from 08:30 to 09:00 (30 minutes).</li>
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<li>Task 7 has a duration of 150 minutes (2 hours and 30 minutes).</li>
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<li>Total task time: 60 (Task 1) + 120 (Task 2) + 60 (Task 3) + 150 (Task 7) - 30 (overlap) = 360 minutes = 6 hours.</li>
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<li>Maximum concurrent tasks: 2 (during the overlap period).</li>
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</ul>
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</li>
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<li>For employee ID 1002:
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<ul>
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<li>Task 4 and Task 5 overlap from 09:30 to 10:00 (30 minutes).</li>
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<li>Total task time: 60 (Task 4) + 120 (Task 5) - 30 (overlap) = 150 minutes = 2 hours and 30 minutes.</li>
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<li>Total task hours (rounded down): 2 hours.</li>
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<li>Maximum concurrent tasks: 2 (during the overlap period).</li>
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</ul>
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</li>
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<li>For employee ID 1003:
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<ul>
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<li>No overlapping tasks.</li>
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<li>Total task time: 120 minutes = 2 hours.</li>
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<li>Maximum concurrent tasks: 1.</li>
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</ul>
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</li>
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</ul>
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<p><b>Note:</b> Output table is ordered by employee_id in ascending order.</p>
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</div>
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<!-- description:end -->
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## Solutions
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<!-- solution:start -->
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### Solution 1
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<!-- tabs:start -->
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#### MySQL
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```sql
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# Write your MySQL query statement below
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WITH
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T AS (
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SELECT DISTINCT employee_id, start_time AS st
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FROM Tasks
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UNION DISTINCT
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SELECT DISTINCT employee_id, end_time AS st
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FROM Tasks
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),
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P AS (
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SELECT
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*,
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LEAD(st) OVER (
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PARTITION BY employee_id
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ORDER BY st
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) AS ed
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FROM T
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),
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S AS (
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SELECT
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P.*,
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COUNT(1) AS concurrent_count
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FROM
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P
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INNER JOIN Tasks USING (employee_id)
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WHERE P.st >= Tasks.start_time AND P.ed <= Tasks.end_time
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GROUP BY 1, 2, 3
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)
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SELECT
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employee_id,
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FLOOR(SUM(TIME_TO_SEC(TIMEDIFF(ed, st)) / 3600)) AS total_task_hours,
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MAX(concurrent_count) AS max_concurrent_tasks
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FROM S
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GROUP BY 1
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ORDER BY 1;
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

‎solution/3100-3199/3156.Employee Task Duration and Concurrent Tasks/Solution.sql

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# Write your MySQL query statement below
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WITH
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T AS (
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SELECT DISTINCT employee_id, start_time AS st
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FROM Tasks
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UNION DISTINCT
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SELECT DISTINCT employee_id, end_time AS st
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FROM Tasks
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),
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P AS (
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SELECT
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*,
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LEAD(st) OVER (
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PARTITION BY employee_id
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ORDER BY st
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) AS ed
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FROM T
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),
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S AS (
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SELECT
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P.*,
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COUNT(1) AS concurrent_count
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FROM
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P
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INNER JOIN Tasks USING (employee_id)
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WHERE P.st >= Tasks.start_time AND P.ed <= Tasks.end_time
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GROUP BY 1, 2, 3
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)
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SELECT
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employee_id,
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FLOOR(SUM(TIME_TO_SEC(TIMEDIFF(ed, st)) / 3600)) AS total_task_hours,
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MAX(concurrent_count) AS max_concurrent_tasks
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FROM S
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GROUP BY 1
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ORDER BY 1;

‎solution/DATABASE_README.md

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| 3126 | [服务器利用时间](/solution/3100-3199/3126.Server%20Utilization%20Time/README.md) | `数据库` | 中等 | 🔒 |
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| 3140 | [连续空余座位 II](/solution/3100-3199/3140.Consecutive%20Available%20Seats%20II/README.md) | `数据库` | 中等 | 🔒 |
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| 3150 | [无效的推文 II](/solution/3100-3199/3150.Invalid%20Tweets%20II/README.md) | `数据库` | 简单 | 🔒 |
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| 3156 | [员工任务持续时间和并发任务](/solution/3100-3199/3156.Employee%20Task%20Duration%20and%20Concurrent%20Tasks/README.md) | | 困难 | 🔒 |
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## 版权
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‎solution/DATABASE_README_EN.md

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@@ -277,6 +277,7 @@ Press <kbd>Control</kbd> + <kbd>F</kbd>(or <kbd>Command</kbd> + <kbd>F</kbd> on
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| 3126 | [Server Utilization Time](/solution/3100-3199/3126.Server%20Utilization%20Time/README_EN.md) | `Database` | Medium | 🔒 |
278278
| 3140 | [Consecutive Available Seats II](/solution/3100-3199/3140.Consecutive%20Available%20Seats%20II/README_EN.md) | `Database` | Medium | 🔒 |
279279
| 3150 | [Invalid Tweets II](/solution/3100-3199/3150.Invalid%20Tweets%20II/README_EN.md) | `Database` | Easy | 🔒 |
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| 3156 | [Employee Task Duration and Concurrent Tasks](/solution/3100-3199/3156.Employee%20Task%20Duration%20and%20Concurrent%20Tasks/README_EN.md) | | Hard | 🔒 |
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## Copyright
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‎solution/README.md

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| 3153 | [所有数对中数位不同之和](/solution/3100-3199/3153.Sum%20of%20Digit%20Differences%20of%20All%20Pairs/README.md) | | 中等 | 第 398 场周赛 |
31673167
| 3154 | [到达第 K 级台阶的方案数](/solution/3100-3199/3154.Find%20Number%20of%20Ways%20to%20Reach%20the%20K-th%20Stair/README.md) | | 困难 | 第 398 场周赛 |
31683168
| 3155 | [Maximum Number of Upgradable Servers](/solution/3100-3199/3155.Maximum%20Number%20of%20Upgradable%20Servers/README.md) | | 中等 | 🔒 |
3169+
| 3156 | [员工任务持续时间和并发任务](/solution/3100-3199/3156.Employee%20Task%20Duration%20and%20Concurrent%20Tasks/README.md) | | 困难 | 🔒 |
31693170

31703171
## 版权
31713172

‎solution/README_EN.md

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@@ -3164,6 +3164,7 @@ Press <kbd>Control</kbd> + <kbd>F</kbd>(or <kbd>Command</kbd> + <kbd>F</kbd> on
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| 3153 | [Sum of Digit Differences of All Pairs](/solution/3100-3199/3153.Sum%20of%20Digit%20Differences%20of%20All%20Pairs/README_EN.md) | | Medium | Weekly Contest 398 |
31653165
| 3154 | [Find Number of Ways to Reach the K-th Stair](/solution/3100-3199/3154.Find%20Number%20of%20Ways%20to%20Reach%20the%20K-th%20Stair/README_EN.md) | | Hard | Weekly Contest 398 |
31663166
| 3155 | [Maximum Number of Upgradable Servers](/solution/3100-3199/3155.Maximum%20Number%20of%20Upgradable%20Servers/README_EN.md) | | Medium | 🔒 |
3167+
| 3156 | [Employee Task Duration and Concurrent Tasks](/solution/3100-3199/3156.Employee%20Task%20Duration%20and%20Concurrent%20Tasks/README_EN.md) | | Hard | 🔒 |
31673168

31683169
## Copyright
31693170

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