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feat: add solutions to lc problems: No.3052,3057,3060 (#2385)
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‎solution/3000-3099/3052.Maximize Items/README.md

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@@ -72,12 +72,37 @@ Output table is ordered by item count in descending order.</pre>
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## 解法
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### 方法一
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### 方法一：连接查询 + 合并
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我们先计算出所有 prime_eligible 类型的物品的总面积，记录在 `T` 表的 `s` 字段中。
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接下来，我们分别计算 prime_eligible 和 not_prime 类型的物品的数量。对于 prime_eligible 类型的物品，我们可以存储的份数是 $\lfloor \frac{500000}{s} \rfloor$，对于 not_prime 类型的物品，我们可以存储的份数是 $\lfloor \frac{500000 \mod s}{\sum \text{s1}} \rfloor$。其中 $\sum \text{s1}$ 是所有 not_prime 类型的物品的总面积。再分别乘上 prime_eligible 和 not_prime 类型的物品的数量，就是我们的结果。
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<!-- tabs:start -->
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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 SUM(square_footage) AS s
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FROM Inventory
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WHERE item_type = 'prime_eligible'
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)
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SELECT
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'prime_eligible' AS item_type,
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COUNT(1) * FLOOR(500000 / s) AS item_count
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FROM
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Inventory
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JOIN T
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WHERE item_type = 'prime_eligible'
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UNION ALL
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SELECT
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'not_prime',
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IFNULL(COUNT(1) * FLOOR(IF(s = 0, 500000, 500000 % s) / SUM(square_footage)), 0)
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FROM
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Inventory
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JOIN T
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WHERE item_type = 'not_prime';
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```
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<!-- tabs:end -->

‎solution/3000-3099/3052.Maximize Items/README_EN.md

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## Solutions
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### Solution 1
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### Solution 1: Join Query + Union All
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First, we calculate the total area of all items of type `prime_eligible` and record it in the `s` field of table `T`.
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Next, we calculate the number of items of type `prime_eligible` and `not_prime` respectively. For items of type `prime_eligible`, the number of portions we can store is $\lfloor \frac{500000}{s} \rfloor$. For items of type `not_prime`, the number of portions we can store is $\lfloor \frac{500000 \mod s}{\sum \text{s1}} \rfloor$. Where $\sum \text{s1}$ is the total area of all items of type `not_prime`. Multiplying by the number of items of type `prime_eligible` and `not_prime` respectively gives us our result.
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<!-- tabs:start -->
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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 SUM(square_footage) AS s
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FROM Inventory
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WHERE item_type = 'prime_eligible'
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)
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SELECT
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'prime_eligible' AS item_type,
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COUNT(1) * FLOOR(500000 / s) AS item_count
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FROM
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Inventory
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JOIN T
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WHERE item_type = 'prime_eligible'
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UNION ALL
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SELECT
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'not_prime',
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IFNULL(COUNT(1) * FLOOR(IF(s = 0, 500000, 500000 % s) / SUM(square_footage)), 0)
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FROM
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Inventory
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JOIN T
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WHERE item_type = 'not_prime';
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```
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<!-- tabs:end -->

‎solution/3000-3099/3052.Maximize Items/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 SUM(square_footage) AS s
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FROM Inventory
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WHERE item_type = 'prime_eligible'
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)
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SELECT
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'prime_eligible' AS item_type,
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COUNT(1) * FLOOR(500000 / s) AS item_count
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FROM
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Inventory
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JOIN T
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WHERE item_type = 'prime_eligible'
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UNION ALL
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SELECT
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'not_prime',
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IFNULL(COUNT(1) * FLOOR(IF(s = 0, 500000, 500000 % s) / SUM(square_footage)), 0)
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FROM
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Inventory
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JOIN T
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WHERE item_type = 'not_prime';

‎solution/3000-3099/3057.Employees Project Allocation/README.md

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@@ -83,12 +83,35 @@ Result table orderd by employee_id, project_id in ascending order.
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## 解法
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### 方法一
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### 方法一：分组统计 + 等值连接
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我们先根据 `employee_id` 连接 `Project` 表和 `Employees` 表，然后再根据 `team` 分组统计每个团队的平均工作量，记录在临时表 `T` 中。
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然后，我们再次连接 `Project` 表和 `Employees` 表，同时连接 `T` 表，找出工作量大于团队平均工作量的员工，并且按照 `employee_id` 和 `project_id` 排序。
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<!-- tabs:start -->
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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 team, AVG(workload) AS avg_workload
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FROM
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Project
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JOIN Employees USING (employee_id)
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GROUP BY 1
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)
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SELECT
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employee_id,
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project_id,
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name AS employee_name,
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workload AS project_workload
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FROM
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Project
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JOIN Employees USING (employee_id)
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JOIN T USING (team)
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WHERE workload > avg_workload
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ORDER BY 1, 2;
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```
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<!-- tabs:end -->

‎solution/3000-3099/3057.Employees Project Allocation/README_EN.md

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@@ -81,12 +81,35 @@ Result table orderd by employee_id, project_id in ascending order.
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## Solutions
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### Solution 1
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### Solution 1: Grouping Statistics + Equi-Join
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First, we join the `Project` table and the `Employees` table based on `employee_id`, then group by `team` to calculate the average workload of each team, and record it in the temporary table `T`.
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Then, we join the `Project` table and the `Employees` table again, and also join the `T` table, to find employees whose workload is greater than the average workload of the team. Finally, we sort by `employee_id` and `project_id`.
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<!-- tabs:start -->
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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 team, AVG(workload) AS avg_workload
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FROM
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Project
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JOIN Employees USING (employee_id)
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GROUP BY 1
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)
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SELECT
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employee_id,
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project_id,
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name AS employee_name,
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workload AS project_workload
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FROM
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Project
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JOIN Employees USING (employee_id)
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JOIN T USING (team)
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WHERE workload > avg_workload
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ORDER BY 1, 2;
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```
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<!-- tabs:end -->

‎solution/3000-3099/3057.Employees Project Allocation/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 team, AVG(workload) AS avg_workload
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FROM
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Project
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JOIN Employees USING (employee_id)
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GROUP BY 1
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)
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SELECT
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employee_id,
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project_id,
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name AS employee_name,
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workload AS project_workload
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FROM
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Project
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JOIN Employees USING (employee_id)
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JOIN T USING (team)
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WHERE workload > avg_workload
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ORDER BY 1, 2;

‎solution/3000-3099/3060.User Activities within Time Bounds/README.md

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@@ -68,12 +68,29 @@ Output table is ordered by user_id in increasing order.
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## 解法
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### 方法一
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### 方法一：窗口函数 + 时间函数
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我们先使用 `LAG` 窗口函数，找到每个用户相同类型的会话的上一个会话的结束时间，记为 `prev_session_end`。然后我们使用 `TIMESTAMPDIFF` 函数计算当前会话的开始时间与上一个会话的结束时间的时间差，如果时间差小于等于 12 小时，那么这个用户就符合题目要求。
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<!-- tabs:start -->
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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
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user_id,
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session_start,
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LAG(session_end) OVER (
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PARTITION BY user_id, session_type
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ORDER BY session_end
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) AS prev_session_end
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FROM Sessions
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)
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SELECT DISTINCT
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user_id
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FROM T
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WHERE TIMESTAMPDIFF(HOUR, prev_session_end, session_start) <= 12;
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```
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<!-- tabs:end -->

‎solution/3000-3099/3060.User Activities within Time Bounds/README_EN.md

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## Solutions
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### Solution 1
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### Solution 1: Window Function + Time Function
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First, we use the `LAG` window function to find the end time of the previous session of the same type for each user, denoted as `prev_session_end`. Then we use the `TIMESTAMPDIFF` function to calculate the time difference between the start time of the current session and the end time of the previous session. If the time difference is less than or equal to 12 hours, then this user meets the requirements of the problem.
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<!-- tabs:start -->
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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
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user_id,
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session_start,
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LAG(session_end) OVER (
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PARTITION BY user_id, session_type
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ORDER BY session_end
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) AS prev_session_end
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FROM Sessions
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)
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SELECT DISTINCT
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user_id
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FROM T
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WHERE TIMESTAMPDIFF(HOUR, prev_session_end, session_start) <= 12;
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```
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<!-- tabs:end -->

‎solution/3000-3099/3060.User Activities within Time Bounds/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
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user_id,
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session_start,
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LAG(session_end) OVER (
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PARTITION BY user_id, session_type
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ORDER BY session_end
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) AS prev_session_end
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FROM Sessions
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)
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SELECT DISTINCT
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user_id
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FROM T
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WHERE TIMESTAMPDIFF(HOUR, prev_session_end, session_start) <= 12;

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