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feat: add new lc problems (doocs#2181)
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‎solution/2900-2999/2993.Friday Purchases I/README.md

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# [2993. Friday Purchases I](https://leetcode.cn/problems/friday-purchases-i)
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[English Version](/solution/2900-2999/2993.Friday%20Purchases%20I/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>Table: <code>Purchases</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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| user_id | int |
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| purchase_date | date |
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| amount_spend | int |
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+---------------+------+
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(user_id, purchase_date, amount_spend) is the primary key (combination of columns with unique values) for this table.
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purchase_date will range from November 1, 2023, to November 30, 2023, inclusive of both dates.
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Each row contains user id, purchase date, and amount spend.
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</pre>
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<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>
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<p>Return <em>the result table ordered by week of month</em><em> in <strong>ascending</strong></em><em><strong> </strong>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 1:</strong></p>
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<pre>
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<strong>Input:</strong>
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Purchases table:
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+---------+---------------+--------------+
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| user_id | purchase_date | amount_spend |
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+---------+---------------+--------------+
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| 11 | 2023-11-07 | 1126 |
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| 15 | 2023-11-30 | 7473 |
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| 17 | 2023-11-14 | 2414 |
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| 12 | 2023-11-24 | 9692 |
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| 8 | 2023-11-03 | 5117 |
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| 1 | 2023-11-16 | 5241 |
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| 10 | 2023-11-12 | 8266 |
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| 13 | 2023-11-24 | 12000 |
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+---------+---------------+--------------+
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<strong>Output:</strong>
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+---------------+---------------+--------------+
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| week_of_month | purchase_date | total_amount |
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+---------------+---------------+--------------+
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| 1 | 2023-11-03 | 5117 |
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| 4 | 2023-11-24 | 21692 |
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+---------------+---------------+--------------+
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<strong>Explanation:</strong>
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- During the first week of November 2023, transactions amounting to $5,117 occurred on Friday, 2023-11-03.
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- For the second week of November 2023, there were no transactions on Friday, 2023-11-10.
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- Similarly, during the third week of November 2023, there were no transactions on Friday, 2023-11-17.
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- 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.
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Output table is ordered by week_of_month in ascending order.</pre>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **SQL**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```sql
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```
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<!-- 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)
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[中文文档](/solution/2900-2999/2993.Friday%20Purchases%20I/README.md)
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## Description
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<p>Table: <code>Purchases</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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| user_id | int |
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| purchase_date | date |
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| amount_spend | int |
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+---------------+------+
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(user_id, purchase_date, amount_spend) is the primary key (combination of columns with unique values) for this table.
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purchase_date will range from November 1, 2023, to November 30, 2023, inclusive of both dates.
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Each row contains user id, purchase date, and amount spend.
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</pre>
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<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>
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<p>Return <em>the result table ordered by week of month</em><em> in <strong>ascending</strong></em><em><strong> </strong>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 1:</strong></p>
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<pre>
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<strong>Input:</strong>
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Purchases table:
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+---------+---------------+--------------+
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| user_id | purchase_date | amount_spend |
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+---------+---------------+--------------+
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| 11 | 2023-11-07 | 1126 |
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| 15 | 2023-11-30 | 7473 |
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| 17 | 2023-11-14 | 2414 |
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| 12 | 2023-11-24 | 9692 |
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| 8 | 2023-11-03 | 5117 |
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| 1 | 2023-11-16 | 5241 |
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| 10 | 2023-11-12 | 8266 |
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| 13 | 2023-11-24 | 12000 |
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+---------+---------------+--------------+
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<strong>Output:</strong>
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+---------------+---------------+--------------+
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| week_of_month | purchase_date | total_amount |
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+---------------+---------------+--------------+
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| 1 | 2023-11-03 | 5117 |
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| 4 | 2023-11-24 | 21692 |
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+---------------+---------------+--------------+
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<strong>Explanation:</strong>
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- During the first week of November 2023, transactions amounting to $5,117 occurred on Friday, 2023-11-03.
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- For the second week of November 2023, there were no transactions on Friday, 2023-11-10.
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- Similarly, during the third week of November 2023, there were no transactions on Friday, 2023-11-17.
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- 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.
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Output table is ordered by week_of_month in ascending order.</pre>
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## Solutions
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<!-- tabs:start -->
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### **SQL**
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```sql
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```
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<!-- 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)
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[English Version](/solution/2900-2999/2994.Friday%20Purchases%20II/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>Table: <code>Purchases</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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| user_id | int |
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| purchase_date | date |
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| amount_spend | int |
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+---------------+------+
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(user_id, purchase_date, amount_spend) is the primary key (combination of columns with unique values) for this table.
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purchase_date will range from November 1, 2023, to November 30, 2023, inclusive of both dates.
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Each row contains user id, purchase date, and amount spend.
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</pre>
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<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>
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<p>Return <em>the result table ordered by week of month</em><em> in <strong>ascending</strong></em><em><strong> </strong>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 1:</strong></p>
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<pre>
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<strong>Input:</strong>
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Purchases table:
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+---------+---------------+--------------+
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| user_id | purchase_date | amount_spend |
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+---------+---------------+--------------+
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| 11 | 2023-11-07 | 1126 |
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| 15 | 2023-11-30 | 7473 |
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| 17 | 2023-11-14 | 2414 |
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| 12 | 2023-11-24 | 9692 |
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| 8 | 2023-11-03 | 5117 |
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| 1 | 2023-11-16 | 5241 |
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| 10 | 2023-11-12 | 8266 |
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| 13 | 2023-11-24 | 12000 |
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+---------+---------------+--------------+
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<strong>Output:</strong>
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+---------------+---------------+--------------+
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| week_of_month | purchase_date | total_amount |
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+---------------+---------------+--------------+
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| 1 | 2023-11-03 | 5117 |
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| 2 | 2023-11-10 | 0 |
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| 3 | 2023-11-17 | 0 |
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| 4 | 2023-11-24 | 21692 |
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+---------------+---------------+--------------+
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<strong>Explanation:</strong>
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- During the first week of November 2023, transactions amounting to $5,117 occurred on Friday, 2023-11-03.
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- 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.
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- 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.
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- 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.
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Output table is ordered by week_of_month in ascending order.</pre>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **SQL**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```sql
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```
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<!-- 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)
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[中文文档](/solution/2900-2999/2994.Friday%20Purchases%20II/README.md)
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## Description
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<p>Table: <code>Purchases</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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| user_id | int |
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| purchase_date | date |
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| amount_spend | int |
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+---------------+------+
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(user_id, purchase_date, amount_spend) is the primary key (combination of columns with unique values) for this table.
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purchase_date will range from November 1, 2023, to November 30, 2023, inclusive of both dates.
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Each row contains user id, purchase date, and amount spend.
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</pre>
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<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>
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<p>Return <em>the result table ordered by week of month</em><em> in <strong>ascending</strong></em><em><strong> </strong>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 1:</strong></p>
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<pre>
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<strong>Input:</strong>
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Purchases table:
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+---------+---------------+--------------+
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| user_id | purchase_date | amount_spend |
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+---------+---------------+--------------+
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| 11 | 2023-11-07 | 1126 |
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| 15 | 2023-11-30 | 7473 |
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| 17 | 2023-11-14 | 2414 |
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| 12 | 2023-11-24 | 9692 |
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| 8 | 2023-11-03 | 5117 |
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| 1 | 2023-11-16 | 5241 |
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| 10 | 2023-11-12 | 8266 |
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| 13 | 2023-11-24 | 12000 |
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+---------+---------------+--------------+
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<strong>Output:</strong>
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+---------------+---------------+--------------+
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| week_of_month | purchase_date | total_amount |
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+---------------+---------------+--------------+
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| 1 | 2023-11-03 | 5117 |
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| 2 | 2023-11-10 | 0 |
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| 3 | 2023-11-17 | 0 |
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| 4 | 2023-11-24 | 21692 |
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+---------------+---------------+--------------+
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<strong>Explanation:</strong>
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- During the first week of November 2023, transactions amounting to $5,117 occurred on Friday, 2023-11-03.
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- 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.
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- 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.
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- 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.
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Output table is ordered by week_of_month in ascending order.</pre>
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## Solutions
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<!-- tabs:start -->
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### **SQL**
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```sql
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```
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<!-- 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)
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[English Version](/solution/2900-2999/2995.Viewers%20Turned%20Streamers/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>表：&nbsp;<code>Sessions</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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| user_id | int |
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| session_start | datetime |
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| session_end | datetime |
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| session_id | int |
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| session_type | enum |
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+---------------+----------+
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session_id 是这张表具有唯一值的列。
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session_type 是一个 ENUM (枚举) 类型，包含(Viewer, Streamer)两个类别。
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这张表包含 user id, session start, session end, session id 和 session type。
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</pre>
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<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>
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<p>按照会话数量和 <code>user_id</code> <strong>降序</strong> 排序返回结果表。</p>
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<p>结果格式如下例所示。</p>
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<p>&nbsp;</p>
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<p><b>示例 1：</b></p>
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<pre>
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<b>输入：</b>
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Sessions table:
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+---------+---------------------+---------------------+------------+--------------+
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| user_id | session_start | session_end | session_id | session_type |
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+---------+---------------------+---------------------+------------+--------------+
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| 101 | 2023-11-06 13:53:42 | 2023-11-06 14:05:42 | 375 | Viewer |
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| 101 | 2023-11-22 16:45:21 | 2023-11-22 20:39:21 | 594 | Streamer |
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| 102 | 2023-11-16 13:23:09 | 2023-11-16 16:10:09 | 777 | Streamer |
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| 102 | 2023-11-17 13:23:09 | 2023-11-17 16:10:09 | 778 | Streamer |
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| 101 | 2023-11-20 07:16:06 | 2023-11-20 08:33:06 | 315 | Streamer |
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| 104 | 2023-11-27 03:10:49 | 2023-11-27 03:30:49 | 797 | Viewer |
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| 103 | 2023-11-27 03:10:49 | 2023-11-27 03:30:49 | 798 | Streamer |
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+---------+---------------------+---------------------+------------+--------------+
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<b>输出：</b>
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+---------+----------------+
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| user_id | sessions_count |
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+---------+----------------+
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| 101 | 2 |
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+---------+----------------+
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<b>解释</b>
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- 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。
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- 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。
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- user_id 103 只参与了一次会话，即作为主播，因此不会考虑在内。
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- 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。
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输出表按照会话数量和 user_id 降序排序。
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</pre>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **SQL**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```sql
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```
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<!-- tabs:end -->

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