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Commit 168ef90

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feat: update solutions to lc problems (#1790)
* No.0603.Consecutive Available Seats * No.0613.Shortest Distance in a Line * No.1501.Countries You Can Safely Invest In * No.1795.Rearrange Products Table
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‎solution/0600-0699/0603.Consecutive Available Seats/README.md‎

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@@ -57,6 +57,14 @@ Cinema 表:
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:自连接**
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我们可以使用自连接的方式,将相邻的两个座位连接起来,然后筛选出连续空余的座位并去重排序即可。
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**方法二:窗口函数**
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我们也可以使用 `LAG` 和 `LEAD` 函数(或者 `SUM() OVER(ROWS BETWEEN 1 PRECEDING AND 1 FOLLOWING)`)来获取相邻的座位信息,然后筛选出连续空余的座位并去重排序即可。
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<!-- tabs:start -->
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### **SQL**
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WHERE a = 2 OR b = 2;
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```
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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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*,
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sum(free = 1) OVER (
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ORDER BY seat_id
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ROWS BETWEEN 1 PRECEDING AND 1 FOLLOWING
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) AS cnt
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FROM Cinema
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)
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SELECT seat_id
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FROM T
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WHERE free = 1 AND cnt > 1
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ORDER BY 1;
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```
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<!-- tabs:end -->

‎solution/0600-0699/0603.Consecutive Available Seats/README_EN.md‎

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@@ -54,6 +54,14 @@ Cinema table:
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## Solutions
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**Solution 1: Self-Join**
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We can use a self-join to join the `Seat` table with itself, and then filter out the records where the `id` of the left seat is equal to the `id` of the right seat minus 1ドル,ドル and where both seats are empty.
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**Solution 2: Window Function**
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We can use the `LAG` and `LEAD` functions (or `SUM() OVER(ROWS BETWEEN 1 PRECEDING AND 1 FOLLOWING)`) to obtain the information of adjacent seats, and then filter out the consecutive empty seats and sort them in a unique way.
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<!-- tabs:start -->
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### **SQL**
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WHERE a = 2 OR b = 2;
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```
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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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*,
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sum(free = 1) OVER (
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ORDER BY seat_id
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ROWS BETWEEN 1 PRECEDING AND 1 FOLLOWING
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) AS cnt
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FROM Cinema
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)
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SELECT seat_id
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FROM T
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WHERE free = 1 AND cnt > 1
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ORDER BY 1;
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```
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<!-- tabs:end -->

‎solution/0600-0699/0603.Consecutive Available Seats/Solution.sql‎

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@@ -2,11 +2,14 @@
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WITH
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T AS (
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SELECT
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seat_id,
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(free + (lag(free) OVER (ORDER BY seat_id))) AS a,
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(free + (lead(free) OVER (ORDER BY seat_id))) AS b
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*,
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sum(free = 1) OVER (
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ORDER BY seat_id
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ROWS BETWEEN 1 PRECEDING AND 1 FOLLOWING
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) AS cnt
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FROM Cinema
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)
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SELECT seat_id
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FROM T
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WHERE a = 2 OR b = 2;
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WHERE free = 1 AND cnt > 1
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ORDER BY 1;

‎solution/0600-0699/0613.Shortest Distance in a Line/README.md‎

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@@ -55,17 +55,24 @@ Point 表:
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:自连接**
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我们可以使用自连接,将表中的每个点与其他更大的点进行连接,然后计算两点之间的距离,最后取最小值。
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**方法二:窗口函数**
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我们也可以使用窗口函数,将表中的点按照 $x$ 排序,然后计算相邻两点之间的距离,最后取最小值。
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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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min(abs(p1.x - p2.x)) AS shortest
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SELECT min(p2.x - p1.x) AS shortest
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FROM
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Point AS p1
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JOIN Point AS p2 ON p1.x != p2.x;
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JOIN Point AS p2 ON p1.x < p2.x;
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```
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```sql

‎solution/0600-0699/0613.Shortest Distance in a Line/README_EN.md‎

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@@ -49,17 +49,24 @@ Point table:
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## Solutions
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**Solution 1: Self-Join**
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We can use a self-join to join each point in the table with the larger points, and then calculate the distance between the two points. Finally, we can take the minimum distance.
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**Solution 2: Window Function**
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We can use a window function to sort the points in the table by their $x$ values, and then calculate the distance between adjacent points. Finally, we can take the minimum distance.
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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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min(abs(p1.x - p2.x)) AS shortest
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SELECT min(p2.x - p1.x) AS shortest
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FROM
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Point AS p1
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JOIN Point AS p2 ON p1.x != p2.x;
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JOIN Point AS p2 ON p1.x < p2.x;
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```
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```sql
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# Write your MySQL query statement below
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SELECT x - lag(x) OVER (ORDER BYx) AS shortest
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FROMPoint
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ORDER BY1
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LIMIT1, 1;
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SELECT min(p2.x - p1.x) AS shortest
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FROM
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PointAS p1
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JOINPointAS p2 ONp1.x<p2.x;

‎solution/1500-1599/1501.Countries You Can Safely Invest In/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:等值连接 + 分组 + 子查询**
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我们可以使用等值连接,将 `Person` 表和 `Calls` 表连接起来,连接的条件是 `Person.id = Calls.caller_id` 或者 `Person.id = Calls.callee_id`,然后再将连接后的表和 `Country` 表连接起来,连接的条件是 `left(phone_number, 3) = country_code`,最后按照国家分组,计算每个国家的平均通话时长,然后再使用子查询,找出平均通话时长大于全球平均通话时长的国家。
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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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with t as (
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select
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left(phone_number, 3) as country_code,
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avg(duration) as duration
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from
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Person
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join Calls on id in (caller_id, callee_id)
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group by
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country_code
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)
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select
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c.name country
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from
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Country c
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join t on c.country_code = t.country_code
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where
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t.duration > (
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select
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avg(duration)
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from
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Calls
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SELECT country
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FROM
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(
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SELECT c.name AS country, avg(duration) AS duration
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FROM
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Person
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JOIN Calls ON id IN (caller_id, callee_id)
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JOIN Country AS c ON left(phone_number, 3) = country_code
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GROUP BY 1
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) AS t
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WHERE duration > (SELECT avg(duration) FROM Calls);
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```
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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 c.name AS country, avg(duration) AS duration
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FROM
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Person
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JOIN Calls ON id IN (caller_id, callee_id)
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JOIN Country AS c ON left(phone_number, 3) = country_code
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GROUP BY 1
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)
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SELECT country
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FROM T
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WHERE duration > (SELECT avg(duration) FROM Calls);
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```
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<!-- tabs:end -->

‎solution/1500-1599/1501.Countries You Can Safely Invest In/README_EN.md‎

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## Solutions
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**Solution 1: Equi-Join + Group By + Subquery**
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We can use an equi-join to join the `Person` table and the `Calls` table on the condition of `Person.id = Calls.caller_id` or `Person.id = Calls.callee_id`, and then join the result with the `Country` table on the condition of `left(phone_number, 3) = country_code`. After that, we can group by country and calculate the average call duration for each country. Finally, we can use a subquery to find the countries whose average call duration is greater than the global average call duration.
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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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with t as (
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select
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left(phone_number, 3) as country_code,
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avg(duration) as duration
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from
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Person
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join Calls on id in (caller_id, callee_id)
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group by
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country_code
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)
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select
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c.name country
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from
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Country c
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join t on c.country_code = t.country_code
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where
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t.duration > (
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select
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avg(duration)
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from
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Calls
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SELECT country
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FROM
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(
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SELECT c.name AS country, avg(duration) AS duration
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FROM
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Person
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JOIN Calls ON id IN (caller_id, callee_id)
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JOIN Country AS c ON left(phone_number, 3) = country_code
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GROUP BY 1
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) AS t
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WHERE duration > (SELECT avg(duration) FROM Calls);
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```
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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 c.name AS country, avg(duration) AS duration
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FROM
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Person
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JOIN Calls ON id IN (caller_id, callee_id)
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JOIN Country AS c ON left(phone_number, 3) = country_code
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GROUP BY 1
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)
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SELECT country
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FROM T
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WHERE duration > (SELECT avg(duration) FROM Calls);
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```
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<!-- tabs:end -->
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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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left(phone_number, 3) AS country_code,
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avg(duration) AS duration
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T AS (
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SELECT c.name AS country, avg(duration) AS duration
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FROM
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Person
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JOIN Calls ON id IN (caller_id, callee_id)
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GROUP BY country_code
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JOIN Country AS c ON left(phone_number, 3) = country_code
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GROUP BY 1
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)
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SELECT
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c.name AS country
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FROM
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Country AS c
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JOIN t ON c.country_code = t.country_code
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WHERE
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t.duration > (
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SELECT
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avg(duration)
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FROM Calls
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);
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SELECT country
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FROM T
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WHERE duration > (SELECT avg(duration) FROM Calls);

‎solution/1700-1799/1795.Rearrange Products Table/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:合并**
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我们可以筛选出每个商店的产品和价格,然后使用 `UNION` 合并即可。
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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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SELECT
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product_id,
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'store1' AS store,
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store1 AS price
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FROM Products
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WHERE store1 IS NOT NULL
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UNION
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SELECT
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product_id,
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'store2' AS store,
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store2 AS price
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FROM Products
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WHERE store2 IS NOT NULL
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UNION
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SELECT
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product_id,
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'store3' AS store,
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store3 AS price
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FROM Products
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WHERE store3 IS NOT NULL;
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```
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```sql
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# Write your MySQL query statement below
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SELECT
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product_id,
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'store1' AS store,
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store1 AS price
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FROM Products
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WHERE store1 > 0
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SELECT product_id, 'store1' AS store, store1 AS price FROM Products WHERE store1 IS NOT NULL
10177
UNION
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SELECT
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product_id,
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'store2' AS store,
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store2 AS price
106-
FROM Products
107-
WHERE store2 > 0
78+
SELECT product_id, 'store2' AS store, store2 AS price FROM Products WHERE store2 IS NOT NULL
10879
UNION
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SELECT
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product_id,
111-
'store3' AS store,
112-
store3 AS price
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FROM Products
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WHERE store3 > 0;
80+
SELECT product_id, 'store3' AS store, store3 AS price FROM Products WHERE store3 IS NOT NULL;
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```
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<!-- tabs:end -->

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