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‎README.md

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| # | Title | Difficulty | Tag | Time, ms | Time, %
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|------|----------------|-------------|-------------|----------|--------
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| 3611 |[Find Overbooked Employees](src/main/kotlin/g3601_3700/s3611_find_overbooked_employees)| Medium | Database | 516 | 100.00
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| 3609 |[Minimum Moves to Reach Target in Grid](src/main/kotlin/g3601_3700/s3609_minimum_moves_to_reach_target_in_grid)| Hard | Math | 1 | 100.00
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| 3608 |[Minimum Time for K Connected Components](src/main/kotlin/g3601_3700/s3608_minimum_time_for_k_connected_components)| Medium | Sorting, Binary_Search, Graph, Union_Find | 31 | 100.00
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| 3607 |[Power Grid Maintenance](src/main/kotlin/g3601_3700/s3607_power_grid_maintenance)| Medium | Array, Hash_Table, Depth_First_Search, Breadth_First_Search, Heap_Priority_Queue, Graph, Union_Find, Ordered_Set | 91 | 100.00

‎src/main/kotlin/g3601_3700/s3611_find_overbooked_employees/readme.md

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[![](https://img.shields.io/github/stars/javadev/LeetCode-in-Kotlin?label=Stars&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin)
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[![](https://img.shields.io/github/forks/javadev/LeetCode-in-Kotlin?label=Fork%20me%20on%20GitHub%20&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin/fork)
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## 3611\. Find Overbooked Employees
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Medium
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Table: `employees`
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+---------------+---------+
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| Column Name | Type |
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+---------------+---------+
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| employee_id | int |
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| employee_name | varchar |
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| department | varchar |
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+---------------+---------+
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employee_id is the unique identifier for this table.
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Each row contains information about an employee and their department.
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Table: `meetings`
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+---------------+---------+
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| Column Name | Type |
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+---------------+---------+
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| meeting_id | int |
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| employee_id | int |
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| meeting_date | date |
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| meeting_type | varchar |
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| duration_hours| decimal |
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+---------------+---------+
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meeting_id is the unique identifier for this table.
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Each row represents a meeting attended by an employee. meeting_type can be 'Team', 'Client', or 'Training'.
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Write a solution to find employees who are **meeting-heavy** - employees who spend more than `50%` of their working time in meetings during any given week.
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* Assume a standard work week is `40` **hours**
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* Calculate **total meeting hours** per employee **per week** (**Monday to Sunday**)
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* An employee is meeting-heavy if their weekly meeting hours `>` `20` hours (`50%` of `40` hours)
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* Count how many weeks each employee was meeting-heavy
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* **Only include** employees who were meeting-heavy for **at least** `2` **weeks**
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Return _the result table ordered by the number of meeting-heavy weeks in **descending** order, then by employee name in **ascending** order_.
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The result format is in the following example.
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**Example:**
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**Input:**
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employees table:
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+-------------+----------------+-------------+
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| employee_id | employee_name | department |
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+-------------+----------------+-------------+
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| 1 | Alice Johnson | Engineering |
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| 2 | Bob Smith | Marketing |
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| 3 | Carol Davis | Sales |
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| 4 | David Wilson | Engineering |
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| 5 | Emma Brown | HR |
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+-------------+----------------+-------------+
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meetings table:
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+------------+-------------+--------------+--------------+----------------+
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| meeting_id | employee_id | meeting_date | meeting_type | duration_hours |
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+------------+-------------+--------------+--------------+----------------+
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| 1 | 1 | 2023-06-05 | Team | 8.0 |
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| 2 | 1 | 2023-06-06 | Client | 6.0 |
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| 3 | 1 | 2023-06-07 | Training | 7.0 |
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| 4 | 1 | 2023-06-12 | Team | 12.0 |
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| 5 | 1 | 2023-06-13 | Client | 9.0 |
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| 6 | 2 | 2023-06-05 | Team | 15.0 |
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| 7 | 2 | 2023-06-06 | Client | 8.0 |
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| 8 | 2 | 2023-06-12 | Training | 10.0 |
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| 9 | 3 | 2023-06-05 | Team | 4.0 |
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| 10 | 3 | 2023-06-06 | Client | 3.0 |
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| 11 | 4 | 2023-06-05 | Team | 25.0 |
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| 12 | 4 | 2023-06-19 | Client | 22.0 |
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| 13 | 5 | 2023-06-05 | Training | 2.0 |
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+------------+-------------+--------------+--------------+----------------+
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**Output:**
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+-------------+---------------+-------------+---------------------+
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| employee_id | employee_name | department | meeting_heavy_weeks |
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+-------------+---------------+-------------+---------------------+
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| 1 | Alice Johnson | Engineering | 2 |
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| 4 | David Wilson | Engineering | 2 |
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+-------------+---------------+-------------+---------------------+
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**Explanation:**
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* **Alice Johnson (employee\_id = 1):**
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* Week of June 5-11 (2023-06-05 to 2023-06-11): 8.0 + 6.0 + 7.0 = 21.0 hours (> 20 hours)
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* Week of June 12-18 (2023-06-12 to 2023-06-18): 12.0 + 9.0 = 21.0 hours (> 20 hours)
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* Meeting-heavy for 2 weeks
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* **David Wilson (employee\_id = 4):**
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* Week of June 5-11: 25.0 hours (> 20 hours)
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* Week of June 19-25: 22.0 hours (> 20 hours)
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* Meeting-heavy for 2 weeks
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* **Employees not included:**
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* Bob Smith (employee\_id = 2): Week of June 5-11: 15.0 + 8.0 = 23.0 hours (> 20), Week of June 12-18: 10.0 hours (< 20). Only 1 meeting-heavy week
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* Carol Davis (employee\_id = 3): Week of June 5-11: 4.0 + 3.0 = 7.0 hours (< 20). No meeting-heavy weeks
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* Emma Brown (employee\_id = 5): Week of June 5-11: 2.0 hours (< 20). No meeting-heavy weeks
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The result table is ordered by meeting\_heavy\_weeks in descending order, then by employee name in ascending order.
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## Solution
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```sql
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# Write your MySQL query statement below
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WITH process_1 AS (
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SELECT
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employee_id,
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SUM(duration_hours) AS duration_total
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FROM
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meetings
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GROUP BY
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employee_id,
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WEEKOFYEAR(meeting_date),
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YEAR(meeting_date)
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)
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SELECT
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p.employee_id,
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e.employee_name,
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e.department,
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COUNT(p.employee_id) AS meeting_heavy_weeks
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FROM
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process_1 p
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INNER JOIN employees e ON p.employee_id = e.employee_id
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WHERE
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duration_total > 20
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GROUP BY
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p.employee_id,
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e.employee_name,
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e.department
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HAVING
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COUNT(p.employee_id) > 1
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ORDER BY
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meeting_heavy_weeks DESC,
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employee_name ASC;
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```

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