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Commit cb593b9

Browse files
feat: add sql solutions to lc problems: No.2993~2995 (doocs#2182)
* No.2993.Friday Purchases I * No.2994.Friday Purchases II * No.2995.Viewers Turned Streamers
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‎.prettierrc‎

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"solution/2600-2699/2686.Immediate Food Delivery III/Solution.sql",
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"solution/2400-2499/2494.Merge Overlapping Events in the Same Hall/Solution.sql",
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"solution/2700-2799/2752.Customers with Maximum Number of Transactions on Consecutive Days/Solution.sql",
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"solution/2700-2799/2793.Status of Flight Tickets/Solution.sql"
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"solution/2700-2799/2793.Status of Flight Tickets/Solution.sql",
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"solution/2900-2999/2994.Friday Purchases II/Solution.sql"
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],
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"options": {
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"parser": "bigquery"

‎package-lock.json‎

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Some generated files are not rendered by default. Learn more about customizing how changed files appear on GitHub.

‎package.json‎

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"lint-staged": "^13.2.2",
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"prettier": "^2.8.8",
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"prettier-plugin-rust": "^0.1.9",
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"prettier-plugin-sql-cst": "^0.8.2"
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"prettier-plugin-sql-cst": "^0.10.0"
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},
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"lint-staged": {
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"*.{js,ts,php,sql,rs,md}": "prettier --write",

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

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@@ -63,14 +63,32 @@ Output table is ordered by week_of_month in ascending order.</pre>
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:日期函数**
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我们用到的日期函数有:
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- `DATE_FORMAT(date, format)`:将日期格式化为字符串
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- `DAYOFWEEK(date)`:返回日期对应的星期几,1 代表星期日,2 代表星期一,以此类推
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- `DAYOFMONTH(date)`:返回日期对应的月份中的第几天
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我们先用 `DATE_FORMAT` 函数将日期格式化为 `YYYYMM` 的形式,然后筛选出 2023 年 11 月且是星期五的记录,然后将记录按照 `purchase_date` 分组,计算出每个星期五的总消费金额。
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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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# Write your MySQL query statement below
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SELECT
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CEIL(DAYOFMONTH(purchase_date) / 7) AS week_of_month,
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purchase_date,
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SUM(amount_spend) AS total_amount
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FROM Purchases
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WHERE DATE_FORMAT(purchase_date, '%Y%m') = '202311' AND DAYOFWEEK(purchase_date) = 6
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GROUP BY 2
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ORDER BY 1;
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```
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<!-- tabs:end -->

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

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## Solutions
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**Solution 1: Date Functions**
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The date functions we use include:
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- `DATE_FORMAT(date, format)`: Formats a date as a string
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- `DAYOFWEEK(date)`: Returns the day of the week for a date, where 1 represents Sunday, 2 represents Monday, and so on
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- `DAYOFMONTH(date)`: Returns the day of the month for a date
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First, we use the `DATE_FORMAT` function to format the date in the form of `YYYYMM`, then filter out the records of November 2023 that fall on a Friday. Next, we group the records by `purchase_date` and calculate the total consumption amount for each Friday.
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<!-- tabs:start -->
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### **SQL**
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```sql
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# Write your MySQL query statement below
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SELECT
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CEIL(DAYOFMONTH(purchase_date) / 7) AS week_of_month,
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purchase_date,
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SUM(amount_spend) AS total_amount
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FROM Purchases
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WHERE DATE_FORMAT(purchase_date, '%Y%m') = '202311' AND DAYOFWEEK(purchase_date) = 6
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GROUP BY 2
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ORDER BY 1;
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```
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<!-- tabs:end -->
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# Write your MySQL query statement below
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SELECT
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CEIL(DAYOFMONTH(purchase_date) / 7) AS week_of_month,
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purchase_date,
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SUM(amount_spend) AS total_amount
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FROM Purchases
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WHERE DATE_FORMAT(purchase_date, '%Y%m') = '202311' AND DAYOFWEEK(purchase_date) = 6
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GROUP BY 2
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ORDER BY 1;

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

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:递归 + 左连接 + 日期函数**
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我们可以使用递归生成一个包含 2023 年 11 月所有日期的表 `T`,然后使用左连接将 `T` 与 `Purchases` 表按照日期进行连接,最后按照题目要求进行分组求和即可。
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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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WITH RECURSIVE
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T AS (
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SELECT '2023年11月01日' AS purchase_date
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UNION
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SELECT purchase_date + INTERVAL 1 DAY
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FROM T
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WHERE purchase_date < '2023年11月30日'
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)
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SELECT
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CEIL(DAYOFMONTH(purchase_date) / 7) AS week_of_month,
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purchase_date,
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IFNULL(SUM(amount_spend), 0) AS total_amount
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FROM
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T
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LEFT JOIN Purchases USING (purchase_date)
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WHERE DAYOFWEEK(purchase_date) = 6
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GROUP BY 2
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ORDER BY 1;
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```
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<!-- tabs:end -->

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

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## Solutions
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**Solution 1: Recursion + Left Join + Date Functions**
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We can generate a table `T` that contains all dates in November 2023 using recursion, then use a left join to connect `T` and the `Purchases` table by date. Finally, group and sum according to the requirements of the problem.
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### **SQL**
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```sql
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WITH RECURSIVE
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T AS (
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SELECT '2023年11月01日' AS purchase_date
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UNION
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SELECT purchase_date + INTERVAL 1 DAY
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FROM T
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WHERE purchase_date < '2023年11月30日'
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)
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SELECT
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CEIL(DAYOFMONTH(purchase_date) / 7) AS week_of_month,
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purchase_date,
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IFNULL(SUM(amount_spend), 0) AS total_amount
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FROM
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T
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LEFT JOIN Purchases USING (purchase_date)
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WHERE DAYOFWEEK(purchase_date) = 6
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GROUP BY 2
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ORDER BY 1;
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```
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<!-- tabs:end -->
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WITH RECURSIVE
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T AS (
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SELECT '2023年11月01日' AS purchase_date
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UNION
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SELECT purchase_date + INTERVAL 1 DAY
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FROM T
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WHERE purchase_date < '2023年11月30日'
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)
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SELECT
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CEIL(DAYOFMONTH(purchase_date) / 7) AS week_of_month,
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purchase_date,
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IFNULL(SUM(amount_spend), 0) AS total_amount
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FROM
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T
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LEFT JOIN Purchases USING (purchase_date)
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WHERE DAYOFWEEK(purchase_date) = 6
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GROUP BY 2
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ORDER BY 1;

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

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:窗口函数 + 等值连接**
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我们可以用窗口函数 `RANK()` 按照 `user_id` 维度,对每个会话进行排名,记录在表 `T` 中,然后再将 `T` 与 `Sessions` 表按照 `user_id` 进行等值连接,并且筛选出 `T` 中排名为 1 的记录,并且 `session_type` 为 `Viewer`,`Sessions` 表中 `session_type` 为 `Streamer` 的记录,最后按照 `user_id` 进行分组求和即可。
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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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# 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_type,
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RANK() OVER (
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PARTITION BY user_id
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ORDER BY session_start
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) AS rk
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FROM Sessions
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)
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SELECT user_id, COUNT(1) AS sessions_count
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FROM
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T AS t
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JOIN Sessions AS s USING (user_id)
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WHERE rk = 1 AND t.session_type = 'Viewer' AND s.session_type = 'Streamer'
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GROUP BY 1
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ORDER BY 2 DESC, 1 DESC;
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```
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<!-- tabs:end -->

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