feat: add new lc problems

solution/2900-2999/2993.Friday Purchases I/README.md

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+# [2993. Friday Purchases I](https://leetcode.cn/problems/friday-purchases-i)
+
+[English Version](/solution/2900-2999/2993.Friday%20Purchases%20I/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>Table: <code>Purchases</code></p>
+
+<pre>
++---------------+------+
+| Column Name   | Type |
++---------------+------+
+| user_id       | int  |
+| purchase_date | date |
+| amount_spend  | int  |
++---------------+------+
+(user_id, purchase_date, amount_spend) is the primary key (combination of columns with unique values) for this table.
+purchase_date will range from November 1, 2023, to November 30, 2023, inclusive of both dates.
+Each row contains user id, purchase date, and amount spend.
+</pre>
+
+<p>Write a solution to calculate the <strong>total spending</strong> by users on <strong>each Friday</strong> of <strong>every week</strong> in <strong>November 2023</strong>. Output only weeks that include <strong>at least one</strong> purchase on a <strong>Friday</strong>.</p>
+
+<p>Return <em>the result table ordered by week of month</em><em> in <strong>ascending</strong></em><em><strong> </strong>order.</em></p>
+
+<p>The result format is in the following example.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> 
+Purchases table:
++---------+---------------+--------------+
+| user_id | purchase_date | amount_spend |
++---------+---------------+--------------+
+| 11      | 2023-11-07    | 1126         |
+| 15      | 2023-11-30    | 7473         |
+| 17      | 2023-11-14    | 2414         |
+| 12      | 2023-11-24    | 9692         |
+| 8       | 2023-11-03    | 5117         |
+| 1       | 2023-11-16    | 5241         |
+| 10      | 2023-11-12    | 8266         |
+| 13      | 2023-11-24    | 12000        |
++---------+---------------+--------------+
+<strong>Output:</strong> 
++---------------+---------------+--------------+
+| week_of_month | purchase_date | total_amount |
++---------------+---------------+--------------+
+| 1             | 2023-11-03    | 5117         |
+| 4             | 2023-11-24    | 21692        |
++---------------+---------------+--------------+ 
+<strong>Explanation:</strong> 
+- During the first week of November 2023, transactions amounting to $5,117 occurred on Friday, 2023-11-03.
+- For the second week of November 2023, there were no transactions on Friday, 2023-11-10.
+- Similarly, during the third week of November 2023, there were no transactions on Friday, 2023-11-17.
+- In the fourth week of November 2023, two transactions took place on Friday, 2023-11-24, amounting to $12,000 and $9,692 respectively, summing up to a total of $21,692.
+Output table is ordered by week_of_month in ascending order.</pre>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **SQL**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```sql
+
+```
+
+<!-- tabs:end -->

solution/2900-2999/2993.Friday Purchases I/README_EN.md

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+# [2993. Friday Purchases I](https://leetcode.com/problems/friday-purchases-i)
+
+[中文文档](/solution/2900-2999/2993.Friday%20Purchases%20I/README.md)
+
+## Description
+
+<p>Table: <code>Purchases</code></p>
+
+<pre>
++---------------+------+
+| Column Name   | Type |
++---------------+------+
+| user_id       | int  |
+| purchase_date | date |
+| amount_spend  | int  |
++---------------+------+
+(user_id, purchase_date, amount_spend) is the primary key (combination of columns with unique values) for this table.
+purchase_date will range from November 1, 2023, to November 30, 2023, inclusive of both dates.
+Each row contains user id, purchase date, and amount spend.
+</pre>
+
+<p>Write a solution to calculate the <strong>total spending</strong> by users on <strong>each Friday</strong> of <strong>every week</strong> in <strong>November 2023</strong>. Output only weeks that include <strong>at least one</strong> purchase on a <strong>Friday</strong>.</p>
+
+<p>Return <em>the result table ordered by week of month</em><em> in <strong>ascending</strong></em><em><strong> </strong>order.</em></p>
+
+<p>The result format is in the following example.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> 
+Purchases table:
++---------+---------------+--------------+
+| user_id | purchase_date | amount_spend |
++---------+---------------+--------------+
+| 11      | 2023-11-07    | 1126         |
+| 15      | 2023-11-30    | 7473         |
+| 17      | 2023-11-14    | 2414         |
+| 12      | 2023-11-24    | 9692         |
+| 8       | 2023-11-03    | 5117         |
+| 1       | 2023-11-16    | 5241         |
+| 10      | 2023-11-12    | 8266         |
+| 13      | 2023-11-24    | 12000        |
++---------+---------------+--------------+
+<strong>Output:</strong> 
++---------------+---------------+--------------+
+| week_of_month | purchase_date | total_amount |
++---------------+---------------+--------------+
+| 1             | 2023-11-03    | 5117         |
+| 4             | 2023-11-24    | 21692        |
++---------------+---------------+--------------+ 
+<strong>Explanation:</strong> 
+- During the first week of November 2023, transactions amounting to $5,117 occurred on Friday, 2023-11-03.
+- For the second week of November 2023, there were no transactions on Friday, 2023-11-10.
+- Similarly, during the third week of November 2023, there were no transactions on Friday, 2023-11-17.
+- In the fourth week of November 2023, two transactions took place on Friday, 2023-11-24, amounting to $12,000 and $9,692 respectively, summing up to a total of $21,692.
+Output table is ordered by week_of_month in ascending order.</pre>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **SQL**
+
+```sql
+
+```
+
+<!-- tabs:end -->

solution/2900-2999/2994.Friday Purchases II/README.md

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+# [2994. Friday Purchases II](https://leetcode.cn/problems/friday-purchases-ii)
+
+[English Version](/solution/2900-2999/2994.Friday%20Purchases%20II/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>Table: <code>Purchases</code></p>
+
+<pre>
++---------------+------+
+| Column Name   | Type |
++---------------+------+
+| user_id       | int  |
+| purchase_date | date |
+| amount_spend  | int  |
++---------------+------+
+(user_id, purchase_date, amount_spend) is the primary key (combination of columns with unique values) for this table.
+purchase_date will range from November 1, 2023, to November 30, 2023, inclusive of both dates.
+Each row contains user id, purchase date, and amount spend.
+</pre>
+
+<p>Write a solution to calculate the <strong>total spending</strong> by users on <strong>each Friday</strong> of <strong>every week</strong> in <strong>November 2023</strong>. If there are <strong>no</strong> purchases on a particular <strong>Friday of a week</strong>, it will be considered as <code>0</code>.</p>
+
+<p>Return <em>the result table ordered by week of month</em><em> in <strong>ascending</strong></em><em><strong> </strong>order.</em></p>
+
+<p>The result format is in the following example.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> 
+Purchases table:
++---------+---------------+--------------+
+| user_id | purchase_date | amount_spend |
++---------+---------------+--------------+
+| 11      | 2023-11-07    | 1126         |
+| 15      | 2023-11-30    | 7473         |
+| 17      | 2023-11-14    | 2414         |
+| 12      | 2023-11-24    | 9692         |
+| 8       | 2023-11-03    | 5117         |
+| 1       | 2023-11-16    | 5241         |
+| 10      | 2023-11-12    | 8266         |
+| 13      | 2023-11-24    | 12000        |
++---------+---------------+--------------+
+<strong>Output:</strong> 
++---------------+---------------+--------------+
+| week_of_month | purchase_date | total_amount |
++---------------+---------------+--------------+
+| 1             | 2023-11-03    | 5117         |
+| 2             | 2023-11-10    | 0            |
+| 3             | 2023-11-17    | 0            |
+| 4             | 2023-11-24    | 21692        |
++---------------+---------------+--------------+ 
+<strong>Explanation:</strong> 
+- During the first week of November 2023, transactions amounting to $5,117 occurred on Friday, 2023-11-03.
+- For the second week of November 2023, there were no transactions on Friday, 2023-11-10, resulting in a value of 0 in the output table for that day.
+- Similarly, during the third week of November 2023, there were no transactions on Friday, 2023-11-17, reflected as 0 in the output table for that specific day.
+- In the fourth week of November 2023, two transactions took place on Friday, 2023-11-24, amounting to $12,000 and $9,692 respectively, summing up to a total of $21,692.
+Output table is ordered by week_of_month in ascending order.</pre>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **SQL**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```sql
+
+```
+
+<!-- tabs:end -->

solution/2900-2999/2994.Friday Purchases II/README_EN.md

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+# [2994. Friday Purchases II](https://leetcode.com/problems/friday-purchases-ii)
+
+[中文文档](/solution/2900-2999/2994.Friday%20Purchases%20II/README.md)
+
+## Description
+
+<p>Table: <code>Purchases</code></p>
+
+<pre>
++---------------+------+
+| Column Name   | Type |
++---------------+------+
+| user_id       | int  |
+| purchase_date | date |
+| amount_spend  | int  |
++---------------+------+
+(user_id, purchase_date, amount_spend) is the primary key (combination of columns with unique values) for this table.
+purchase_date will range from November 1, 2023, to November 30, 2023, inclusive of both dates.
+Each row contains user id, purchase date, and amount spend.
+</pre>
+
+<p>Write a solution to calculate the <strong>total spending</strong> by users on <strong>each Friday</strong> of <strong>every week</strong> in <strong>November 2023</strong>. If there are <strong>no</strong> purchases on a particular <strong>Friday of a week</strong>, it will be considered as <code>0</code>.</p>
+
+<p>Return <em>the result table ordered by week of month</em><em> in <strong>ascending</strong></em><em><strong> </strong>order.</em></p>
+
+<p>The result format is in the following example.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> 
+Purchases table:
++---------+---------------+--------------+
+| user_id | purchase_date | amount_spend |
++---------+---------------+--------------+
+| 11      | 2023-11-07    | 1126         |
+| 15      | 2023-11-30    | 7473         |
+| 17      | 2023-11-14    | 2414         |
+| 12      | 2023-11-24    | 9692         |
+| 8       | 2023-11-03    | 5117         |
+| 1       | 2023-11-16    | 5241         |
+| 10      | 2023-11-12    | 8266         |
+| 13      | 2023-11-24    | 12000        |
++---------+---------------+--------------+
+<strong>Output:</strong> 
++---------------+---------------+--------------+
+| week_of_month | purchase_date | total_amount |
++---------------+---------------+--------------+
+| 1             | 2023-11-03    | 5117         |
+| 2             | 2023-11-10    | 0            |
+| 3             | 2023-11-17    | 0            |
+| 4             | 2023-11-24    | 21692        |
++---------------+---------------+--------------+ 
+<strong>Explanation:</strong> 
+- During the first week of November 2023, transactions amounting to $5,117 occurred on Friday, 2023-11-03.
+- For the second week of November 2023, there were no transactions on Friday, 2023-11-10, resulting in a value of 0 in the output table for that day.
+- Similarly, during the third week of November 2023, there were no transactions on Friday, 2023-11-17, reflected as 0 in the output table for that specific day.
+- In the fourth week of November 2023, two transactions took place on Friday, 2023-11-24, amounting to $12,000 and $9,692 respectively, summing up to a total of $21,692.
+Output table is ordered by week_of_month in ascending order.</pre>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **SQL**
+
+```sql
+
+```
+
+<!-- tabs:end -->

solution/2900-2999/2995.Viewers Turned Streamers/README.md

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+# [2995. 观众变主播](https://leetcode.cn/problems/viewers-turned-streamers)
+
+[English Version](/solution/2900-2999/2995.Viewers%20Turned%20Streamers/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>表：&nbsp;<code>Sessions</code></p>
+
+<pre>
++---------------+----------+
+| Column Name   | Type     |
++---------------+----------+
+| user_id       | int      |
+| session_start | datetime |
+| session_end   | datetime |
+| session_id    | int      |
+| session_type  | enum     |
++---------------+----------+
+session_id 是这张表具有唯一值的列。
+session_type 是一个 ENUM (枚举) 类型，包含(Viewer, Streamer)两个类别。
+这张表包含 user id, session start, session end, session id 和 session type。
+</pre>
+
+<p>编写一个解决方案，找到&nbsp;<strong>首次会话&nbsp;</strong>为 <strong>观众</strong> 的用户的&nbsp;<strong>会话&nbsp;</strong>数量。</p>
+
+<p>按照会话数量和 <code>user_id</code> <strong>降序</strong> 排序返回结果表。</p>
+
+<p>结果格式如下例所示。</p>
+
+<p>&nbsp;</p>
+
+<p><b>示例 1：</b></p>
+
+<pre>
+<b>输入：</b>
+Sessions table:
++---------+---------------------+---------------------+------------+--------------+
+| user_id | session_start       | session_end         | session_id | session_type | 
++---------+---------------------+---------------------+------------+--------------+
+| 101     | 2023-11-06 13:53:42 | 2023-11-06 14:05:42 | 375        | Viewer       |  
+| 101     | 2023-11-22 16:45:21 | 2023-11-22 20:39:21 | 594        | Streamer     |   
+| 102     | 2023-11-16 13:23:09 | 2023-11-16 16:10:09 | 777        | Streamer     | 
+| 102     | 2023-11-17 13:23:09 | 2023-11-17 16:10:09 | 778        | Streamer     | 
+| 101     | 2023-11-20 07:16:06 | 2023-11-20 08:33:06 | 315        | Streamer     | 
+| 104     | 2023-11-27 03:10:49 | 2023-11-27 03:30:49 | 797        | Viewer       | 
+| 103     | 2023-11-27 03:10:49 | 2023-11-27 03:30:49 | 798        | Streamer     |  
++---------+---------------------+---------------------+------------+--------------+
+<b>输出：</b>
++---------+----------------+
+| user_id | sessions_count | 
++---------+----------------+
+| 101     | 2              | 
++---------+----------------+
+<b>解释</b>
+- user_id 101，在 2023-11-06 13:53:42 以观众身份开始了他们的初始会话，随后进行了两次主播会话，所以计数为 2。
+- user_id 102，尽管有两个会话，但初始会话是作为主播，因此将排除此用户。
+- user_id 103 只参与了一次会话，即作为主播，因此不会考虑在内。
+- User_id 104 以观众身份开始了他们的第一次会话，但没有后续会话，因此不会包括在最终计数中。
+输出表按照会话数量和 user_id 降序排序。
+</pre>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **SQL**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```sql
+
+```
+
+<!-- tabs:end -->