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

Author: yanglbme

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

@@ -57,6 +57,14 @@ Cinema 表:
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：自连接**
+
+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。
+
+**方法二：窗口函数**
+
+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。
+
 <!-- tabs:start -->
 
 ### **SQL**
@@ -85,4 +93,22 @@ FROM T
 WHERE a = 2 OR b = 2;
 ```
 
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            *,
+            sum(free = 1) OVER (
+                ORDER BY seat_id
+                ROWS BETWEEN 1 PRECEDING AND 1 FOLLOWING
+            ) AS cnt
+        FROM Cinema
+    )
+SELECT seat_id
+FROM T
+WHERE free = 1 AND cnt > 1
+ORDER BY 1;
+```
+
 <!-- tabs:end -->

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

@@ -54,6 +54,14 @@ Cinema table:
 
 ## Solutions
 
+**Solution 1: Self-Join**
+
+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 $1$, and where both seats are empty.
+
+**Solution 2: Window Function**
+
+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.
+
 <!-- tabs:start -->
 
 ### **SQL**
@@ -82,4 +90,22 @@ FROM T
 WHERE a = 2 OR b = 2;
 ```
 
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            *,
+            sum(free = 1) OVER (
+                ORDER BY seat_id
+                ROWS BETWEEN 1 PRECEDING AND 1 FOLLOWING
+            ) AS cnt
+        FROM Cinema
+    )
+SELECT seat_id
+FROM T
+WHERE free = 1 AND cnt > 1
+ORDER BY 1;
+```
+
 <!-- tabs:end -->

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

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

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

@@ -55,17 +55,24 @@ Point 表:
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：自连接**
+
+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。
+
+**方法二：窗口函数**
+
+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。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    min(abs(p1.x - p2.x)) AS shortest
+SELECT min(p2.x - p1.x) AS shortest
 FROM
     Point AS p1
-    JOIN Point AS p2 ON p1.x != p2.x;
+    JOIN Point AS p2 ON p1.x < p2.x;
 ```
 
 ```sql

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

@@ -49,17 +49,24 @@ Point table:
 
 ## Solutions
 
+**Solution 1: Self-Join**
+
+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.
+
+**Solution 2: Window Function**
+
+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.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    min(abs(p1.x - p2.x)) AS shortest
+SELECT min(p2.x - p1.x) AS shortest
 FROM
     Point AS p1
-    JOIN Point AS p2 ON p1.x != p2.x;
+    JOIN Point AS p2 ON p1.x < p2.x;
 ```
 
 ```sql

solution/0600-0699/0613.Shortest Distance in a Line/Solution.sql

@@ -1,5 +1,5 @@
 # Write your MySQL query statement below
-SELECT x - lag(x) OVER (ORDER BY x) AS shortest
-FROM Point
-ORDER BY 1
-LIMIT 1, 1;
+SELECT min(p2.x - p1.x) AS shortest
+FROM
+    Point AS p1
+    JOIN Point AS p2 ON p1.x < p2.x;

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

@@ -122,34 +122,43 @@ Calls 表:
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：等值连接 + 分组 + 子查询**
+
+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。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
-with t as (
-    select
-        left(phone_number, 3) as country_code,
-        avg(duration) as duration
-    from
-        Person
-        join Calls on id in (caller_id, callee_id)
-    group by
-        country_code
-)
-select
-    c.name country
-from
-    Country c
-    join t on c.country_code = t.country_code
-where
-    t.duration > (
-        select
-            avg(duration)
-        from
-            Calls
+SELECT country
+FROM
+    (
+        SELECT c.name AS country, avg(duration) AS duration
+        FROM
+            Person
+            JOIN Calls ON id IN (caller_id, callee_id)
+            JOIN Country AS c ON left(phone_number, 3) = country_code
+        GROUP BY 1
+    ) AS t
+WHERE duration > (SELECT avg(duration) FROM Calls);
+```
+
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT c.name AS country, avg(duration) AS duration
+        FROM
+            Person
+            JOIN Calls ON id IN (caller_id, callee_id)
+            JOIN Country AS c ON left(phone_number, 3) = country_code
+        GROUP BY 1
     )
+SELECT country
+FROM T
+WHERE duration > (SELECT avg(duration) FROM Calls);
 ```
 
 <!-- tabs:end -->

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

@@ -117,34 +117,43 @@ Since Peru is the only country where the average call duration is greater than t
 
 ## Solutions
 
+**Solution 1: Equi-Join + Group By + Subquery**
+
+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.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
-with t as (
-    select
-        left(phone_number, 3) as country_code,
-        avg(duration) as duration
-    from
-        Person
-        join Calls on id in (caller_id, callee_id)
-    group by
-        country_code
-)
-select
-    c.name country
-from
-    Country c
-    join t on c.country_code = t.country_code
-where
-    t.duration > (
-        select
-            avg(duration)
-        from
-            Calls
+SELECT country
+FROM
+    (
+        SELECT c.name AS country, avg(duration) AS duration
+        FROM
+            Person
+            JOIN Calls ON id IN (caller_id, callee_id)
+            JOIN Country AS c ON left(phone_number, 3) = country_code
+        GROUP BY 1
+    ) AS t
+WHERE duration > (SELECT avg(duration) FROM Calls);
+```
+
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT c.name AS country, avg(duration) AS duration
+        FROM
+            Person
+            JOIN Calls ON id IN (caller_id, callee_id)
+            JOIN Country AS c ON left(phone_number, 3) = country_code
+        GROUP BY 1
     )
+SELECT country
+FROM T
+WHERE duration > (SELECT avg(duration) FROM Calls);
 ```
 
 <!-- tabs:end -->

solution/1500-1599/1501.Countries You Can Safely Invest In/Solution.sql

@@ -1,22 +1,13 @@
 # Write your MySQL query statement below
 WITH
-    t AS (
-        SELECT
-            left(phone_number, 3) AS country_code,
-            avg(duration) AS duration
+    T AS (
+        SELECT c.name AS country, avg(duration) AS duration
         FROM
             Person
             JOIN Calls ON id IN (caller_id, callee_id)
-        GROUP BY country_code
+            JOIN Country AS c ON left(phone_number, 3) = country_code
+        GROUP BY 1
     )
-SELECT
-    c.name AS country
-FROM
-    Country AS c
-    JOIN t ON c.country_code = t.country_code
-WHERE
-    t.duration > (
-        SELECT
-            avg(duration)
-        FROM Calls
-    );
+SELECT country
+FROM T
+WHERE duration > (SELECT avg(duration) FROM Calls);

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

@@ -61,57 +61,23 @@ Products table:
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：合并**
+
+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。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
-```sql
-SELECT
-    product_id,
-    'store1' AS store,
-    store1 AS price
-FROM Products
-WHERE store1 IS NOT NULL
-UNION
-SELECT
-    product_id,
-    'store2' AS store,
-    store2 AS price
-FROM Products
-WHERE store2 IS NOT NULL
-UNION
-SELECT
-    product_id,
-    'store3' AS store,
-    store3 AS price
-FROM Products
-WHERE store3 IS NOT NULL;
-```
-
 ```sql
 # Write your MySQL query statement below
-SELECT
-    product_id,
-    'store1' AS store,
-    store1 AS price
-FROM Products
-WHERE store1 > 0
+SELECT product_id, 'store1' AS store, store1 AS price FROM Products WHERE store1 IS NOT NULL
 UNION
-SELECT
-    product_id,
-    'store2' AS store,
-    store2 AS price
-FROM Products
-WHERE store2 > 0
+SELECT product_id, 'store2' AS store, store2 AS price FROM Products WHERE store2 IS NOT NULL
 UNION
-SELECT
-    product_id,
-    'store3' AS store,
-    store3 AS price
-FROM Products
-WHERE store3 > 0;
+SELECT product_id, 'store3' AS store, store3 AS price FROM Products WHERE store3 IS NOT NULL;
 ```
 
 <!-- tabs:end -->