feat: update solutions to lc problems: No.0180,0610,1164 (#1850)

* No.0180.Consecutive Numbers
* No.0610.Triangle Judgement
* No.1164.Product Price at a Given Date

Author: yanglbme

.prettierrc

@@ -14,6 +14,7 @@
     "overrides": [
         {
             "files": [
+                "solution/0100-0199/0180.Consecutive Numbers/Solution.sql",
                 "solution/0500-0599/0550.Game Play Analysis IV/Solution.sql",
                 "solution/0500-0599/0578.Get Highest Answer Rate Question/Solution.sql",
                 "solution/0600-0699/0610.Triangle Judgement/Solution.sql",
@@ -55,4 +56,4 @@
             }
         }
     ]
-}
+}

solution/0100-0199/0180.Consecutive Numbers/README.md

@@ -57,26 +57,62 @@ Result 表：
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：两次连接**
+
+我们可以使用两次连接来解决这个问题。
+
+我们首先进行一次自连接，连接条件是 `l1.num = l2.num` 并且 `l1.id = l2.id - 1`，这样我们就可以找出所有至少连续出现两次的数字。然后，我们再进行一次自连接，连接条件是 `l2.num = l3.num` 并且 `l2.id = l3.id - 1`，这样我们就可以找出所有至少连续出现三次的数字。最后，我们只需要筛选出去重的 `l2.num` 即可。
+
+**方法二：窗口函数**
+
+我们可以使用窗口函数 `LAG` 和 `LEAD` 来获取上一行的 `num` 和下一行的 `num`，记录在字段 $a$ 和 $b$ 中。最后，我们只需要筛选出 $a =num$ 并且 $b = num$ 的行，这些行就是至少连续出现三次的数字。注意，我们需要使用 `DISTINCT` 关键字来对结果去重。
+
+我们也可以对数字进行分组，具体做法是使用 `IF` 函数来判断当前行与前一行的 `num` 是否相等，如果相等则记为 $0$，否则记为 $1$，然后使用窗口函数 `SUM` 来计算前缀和，这样计算出的前缀和就是分组的标识。最后，我们只需要按照分组标识进行分组，然后筛选出每组中的行数大于等于 $3$ 的数字即可。同样，我们需要使用 `DISTINCT` 关键字来对结果去重。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
+```sql
+# Write your MySQL query statement below
+SELECT DISTINCT l2.num AS ConsecutiveNums
+FROM
+    Logs AS l1
+    JOIN Logs AS l2 ON l1.id = l2.id - 1 AND l1.num = l2.num
+    JOIN Logs AS l3 ON l2.id = l3.id - 1 AND l2.num = l3.num;
+```
+
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            *,
+            LAG(num) OVER () AS a,
+            LEAD(num) OVER () AS b
+        FROM Logs
+    )
+SELECT DISTINCT num AS ConsecutiveNums
+FROM T
+WHERE a = num AND b = num;
+```
+
 ```sql
 # Write your MySQL query statement below
 WITH
-    t AS (
+    T AS (
         SELECT
             *,
-            CASE
-                WHEN (LAG(num) OVER (ORDER BY id)) = num THEN 0
-                ELSE 1
-            END AS mark
+            IF(num = (LAG(num) OVER ()), 0, 1) AS st
         FROM Logs
     ),
-    p AS (SELECT num, SUM(mark) OVER (ORDER BY id) AS gid FROM t)
+    S AS (
+        SELECT *, SUM(st) OVER (ORDER BY id) AS p
+        FROM T
+    )
 SELECT DISTINCT num AS ConsecutiveNums
-FROM p
-GROUP BY gid
+FROM S
+GROUP BY p
 HAVING COUNT(1) >= 3;
 ```
 

solution/0100-0199/0180.Consecutive Numbers/README_EN.md

@@ -53,26 +53,62 @@ Logs table:
 
 ## Solutions
 
+**Solution 1: Two Joins**
+
+We can use two joins to solve this problem.
+
+First, we perform a self-join with the condition `l1.num = l2.num` and `l1.id = l2.id - 1`, so that we can find all numbers that appear at least twice in a row. Then, we perform another self-join with the condition `l2.num = l3.num` and `l2.id = l3.id - 1`, so that we can find all numbers that appear at least three times in a row. Finally, we only need to select the distinct `l2.num`.
+
+**Solution 2: Window Function**
+
+We can use the window functions `LAG` and `LEAD` to obtain the `num` of the previous row and the next row of the current row, and record them in the fields $a$ and $b$, respectively. Finally, we only need to filter out the rows where $a = num$ and $b = num$, which are the numbers that appear at least three times in a row. Note that we need to use the `DISTINCT` keyword to remove duplicates from the results.
+
+We can also group the numbers by using the `IF` function to determine whether the `num` of the current row is equal to the `num` of the previous row. If they are equal, we set it to $0$, otherwise we set it to $1$. Then, we use the window function `SUM` to calculate the prefix sum, which is the grouping identifier. Finally, we only need to group by the grouping identifier and filter out the numbers with a row count greater than or equal to $3$ in each group. Similarly, we need to use the `DISTINCT` keyword to remove duplicates from the results.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
+```sql
+# Write your MySQL query statement below
+SELECT DISTINCT l2.num AS ConsecutiveNums
+FROM
+    Logs AS l1
+    JOIN Logs AS l2 ON l1.id = l2.id - 1 AND l1.num = l2.num
+    JOIN Logs AS l3 ON l2.id = l3.id - 1 AND l2.num = l3.num;
+```
+
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            *,
+            LAG(num) OVER () AS a,
+            LEAD(num) OVER () AS b
+        FROM Logs
+    )
+SELECT DISTINCT num AS ConsecutiveNums
+FROM T
+WHERE a = num AND b = num;
+```
+
 ```sql
 # Write your MySQL query statement below
 WITH
-    t AS (
+    T AS (
         SELECT
             *,
-            CASE
-                WHEN (LAG(num) OVER (ORDER BY id)) = num THEN 0
-                ELSE 1
-            END AS mark
+            IF(num = (LAG(num) OVER ()), 0, 1) AS st
         FROM Logs
     ),
-    p AS (SELECT num, SUM(mark) OVER (ORDER BY id) AS gid FROM t)
+    S AS (
+        SELECT *, SUM(st) OVER (ORDER BY id) AS p
+        FROM T
+    )
 SELECT DISTINCT num AS ConsecutiveNums
-FROM p
-GROUP BY gid
+FROM S
+GROUP BY p
 HAVING COUNT(1) >= 3;
 ```
 

solution/0100-0199/0180.Consecutive Numbers/Solution.sql

@@ -1,16 +1,16 @@
 # Write your MySQL query statement below
 WITH
-    t AS (
+    T AS (
         SELECT
             *,
-            CASE
-                WHEN (LAG(num) OVER (ORDER BY id)) = num THEN 0
-                ELSE 1
-            END AS mark
+            IF(num = (LAG(num) OVER ()), 0, 1) AS st
         FROM Logs
     ),
-    p AS (SELECT num, SUM(mark) OVER (ORDER BY id) AS gid FROM t)
+    S AS (
+        SELECT *, SUM(st) OVER (ORDER BY id) AS p
+        FROM T
+    )
 SELECT DISTINCT num AS ConsecutiveNums
-FROM p
-GROUP BY gid
+FROM S
+GROUP BY p
 HAVING COUNT(1) >= 3;

solution/0600-0699/0610.Triangle Judgement/README.md

@@ -53,9 +53,9 @@ Triangle 表:
 
 <!-- 这里可写通用的实现逻辑 -->
 
-**方法一：if 语句 + 三角形判断条件**
+**方法一：IF 语句 + 三角形判断条件**
 
-三条边能否构成三角形的条件是：任意两边之和大于第三边。因此，我们可以使用 `if` 语句来判断是否满足这个条件，如果满足则返回 `Yes`，否则返回 `No`。
+三条边能否构成三角形的条件是：任意两边之和大于第三边。因此，我们可以使用 `IF` 语句来判断是否满足这个条件，如果满足则返回 `Yes`，否则返回 `No`。
 
 <!-- tabs:start -->
 

solution/0600-0699/0610.Triangle Judgement/README_EN.md

@@ -49,6 +49,10 @@ Triangle table:
 
 ## Solutions
 
+**Solution 1: IF Statement + Triangle Inequality**
+
+The condition for whether three sides can form a triangle is that the sum of any two sides is greater than the third side. Therefore, we can use an `IF` statement to determine whether this condition is satisfied. If it is satisfied, we return `Yes`, otherwise we return `No`.
+
 <!-- tabs:start -->
 
 ### **SQL**

solution/1100-1199/1164.Product Price at a Given Date/README.md

@@ -57,38 +57,33 @@ Products 表:
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：子查询 + 连接**
+
+我们可以使用子查询，找出每个产品在给定日期之前最后一次价格变更的价格，记录在 `P` 表中。然后，我们再找出所有产品的 `product_id`，记录在 `T` 表中。最后，我们将 `T` 表和 `P` 表按照 `product_id` 进行左连接，即可得到最终结果。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    p1.product_id AS product_id,
-    IFNULL(p2.price, 10) AS price
-FROM
-    (
-        SELECT DISTINCT
-            (product_id) AS product_id
+WITH
+    T AS (SELECT DISTINCT product_id FROM Products),
+    P AS (
+        SELECT product_id, new_price AS price
         FROM Products
-    ) AS p1
-    LEFT JOIN (
-        SELECT
-            t1.product_id,
-            t1.new_price AS price
-        FROM
-            Products AS t1
-            JOIN (
-                SELECT
-                    product_id,
-                    MAX(change_date) AS change_date
+        WHERE
+            (product_id, change_date) IN (
+                SELECT product_id, MAX(change_date) AS change_date
                 FROM Products
                 WHERE change_date <= '2019-08-16'
-                GROUP BY product_id
-            ) AS t2
-                ON t1.product_id = t2.product_id AND t1.change_date = t2.change_date
-    ) AS p2
-        ON p1.product_id = p2.product_id;
+                GROUP BY 1
+            )
+    )
+SELECT product_id, IFNULL(price, 10) AS price
+FROM
+    T
+    LEFT JOIN P USING (product_id);
 ```
 
 ```sql

solution/1100-1199/1164.Product Price at a Given Date/README_EN.md

@@ -53,38 +53,33 @@ Products table:
 
 ## Solutions
 
+**Solution 1: Subquery + Join**
+
+We can use a subquery to find the price of the last price change for each product before the given date, and record it in the `P` table. Then, we can find all `product_id`s in the `T` table. Finally, we can left join the `T` table with the `P` table on `product_id` to get the final result.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    p1.product_id AS product_id,
-    IFNULL(p2.price, 10) AS price
-FROM
-    (
-        SELECT DISTINCT
-            (product_id) AS product_id
+WITH
+    T AS (SELECT DISTINCT product_id FROM Products),
+    P AS (
+        SELECT product_id, new_price AS price
         FROM Products
-    ) AS p1
-    LEFT JOIN (
-        SELECT
-            t1.product_id,
-            t1.new_price AS price
-        FROM
-            Products AS t1
-            JOIN (
-                SELECT
-                    product_id,
-                    MAX(change_date) AS change_date
+        WHERE
+            (product_id, change_date) IN (
+                SELECT product_id, MAX(change_date) AS change_date
                 FROM Products
                 WHERE change_date <= '2019-08-16'
-                GROUP BY product_id
-            ) AS t2
-                ON t1.product_id = t2.product_id AND t1.change_date = t2.change_date
-    ) AS p2
-        ON p1.product_id = p2.product_id;
+                GROUP BY 1
+            )
+    )
+SELECT product_id, IFNULL(price, 10) AS price
+FROM
+    T
+    LEFT JOIN P USING (product_id);
 ```
 
 ```sql

solution/1100-1199/1164.Product Price at a Given Date/Solution.sql

@@ -1,24 +1,18 @@
 # Write your MySQL query statement below
 WITH
+    T AS (SELECT DISTINCT product_id FROM Products),
     P AS (
-        SELECT p1.product_id, new_price, change_date
-        FROM
-            (
-                SELECT DISTINCT product_id
+        SELECT product_id, new_price AS price
+        FROM Products
+        WHERE
+            (product_id, change_date) IN (
+                SELECT product_id, MAX(change_date) AS change_date
                 FROM Products
-            ) AS p1
-            LEFT JOIN Products AS p2
-                ON p1.product_id = p2.product_id AND p2.change_date <= '2019-08-16'
-    ),
-    T AS (
-        SELECT
-            *,
-            RANK() OVER (
-                PARTITION BY product_id
-                ORDER BY change_date DESC
-            ) AS rk
-        FROM P
+                WHERE change_date <= '2019-08-16'
+                GROUP BY 1
+            )
     )
-SELECT product_id, IFNULL(new_price, 10) AS price
-FROM T
-WHERE rk = 1;
+SELECT product_id, IFNULL(price, 10) AS price
+FROM
+    T
+    LEFT JOIN P USING (product_id);