feat: add solutions to lc problems: No.3058,3059 (doocs#2384)

Author: yanglbme

solution/3000-3099/3058.Friends With No Mutual Friends/README.md

@@ -62,12 +62,32 @@ Output table is ordered by user_id1 in ascending order.</pre>
 
 ## 解法
 
-### 方法一
+### 方法一：子查询
+
+我们先把所有的朋友关系都列出来，记录在 `T` 表中。然后再找出没有共同朋友的朋友对。
+
+接下来，我们可以使用子查询来找出没有共同朋友的朋友对，即这个朋友对不属于其他某个人的朋友。
 
 <!-- tabs:start -->
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT user_id1, user_id2 FROM Friends
+        UNION ALL
+        SELECT user_id2, user_id1 FROM Friends
+    )
+SELECT user_id1, user_id2
+FROM Friends
+WHERE
+    (user_id1, user_id2) NOT IN (
+        SELECT t1.user_id1, t2.user_id1
+        FROM
+            T AS t1
+            JOIN T AS t2 ON t1.user_id2 = t2.user_id2
+    )
+ORDER BY 1, 2;
 ```
 
 <!-- tabs:end -->

solution/3000-3099/3058.Friends With No Mutual Friends/README_EN.md

@@ -60,12 +60,32 @@ Output table is ordered by user_id1 in ascending order.</pre>
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Subquery
+
+First, we list all the friend relationships and record them in table `T`. Then we find the pairs of friends who do not have common friends.
+
+Next, we can use a subquery to find pairs of friends who do not have common friends, i.e., this pair of friends does not belong to any other person's friends.
 
 <!-- tabs:start -->
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT user_id1, user_id2 FROM Friends
+        UNION ALL
+        SELECT user_id2, user_id1 FROM Friends
+    )
+SELECT user_id1, user_id2
+FROM Friends
+WHERE
+    (user_id1, user_id2) NOT IN (
+        SELECT t1.user_id1, t2.user_id1
+        FROM
+            T AS t1
+            JOIN T AS t2 ON t1.user_id2 = t2.user_id2
+    )
+ORDER BY 1, 2;
 ```
 
 <!-- tabs:end -->

solution/3000-3099/3058.Friends With No Mutual Friends/Solution.sql

@@ -0,0 +1,17 @@
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT user_id1, user_id2 FROM Friends
+        UNION ALL
+        SELECT user_id2, user_id1 FROM Friends
+    )
+SELECT user_id1, user_id2
+FROM Friends
+WHERE
+    (user_id1, user_id2) NOT IN (
+        SELECT t1.user_id1, t2.user_id1
+        FROM
+            T AS t1
+            JOIN T AS t2 ON t1.user_id2 = t2.user_id2
+    )
+ORDER BY 1, 2;

solution/3000-3099/3059.Find All Unique Email Domains/README.md

@@ -57,12 +57,19 @@ Output table is ordered by email_domains in ascending order.
 
 ## 解法
 
-### 方法一
+### 方法一：使用 `SUBSTRING_INDEX` 函数 + 分组统计
+
+我们先筛选出所有以 `.com` 结尾的邮箱，然后使用 `SUBSTRING_INDEX` 函数提取出邮箱的域名，最后使用 `GROUP BY` 统计每个域名的个数。
 
 <!-- tabs:start -->
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT SUBSTRING_INDEX(email, '@', -1) AS email_domain, COUNT(1) AS count
+FROM Emails
+WHERE email LIKE '%.com'
+GROUP BY 1
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

solution/3000-3099/3059.Find All Unique Email Domains/README_EN.md

@@ -55,12 +55,19 @@ Output table is ordered by email_domains in ascending order.
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Using `SUBSTRING_INDEX` Function + Grouping Statistics
+
+First, we filter out all emails ending with `.com`, then use the `SUBSTRING_INDEX` function to extract the domain name of the email. Finally, we use `GROUP BY` to count the number of each domain.
 
 <!-- tabs:start -->
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT SUBSTRING_INDEX(email, '@', -1) AS email_domain, COUNT(1) AS count
+FROM Emails
+WHERE email LIKE '%.com'
+GROUP BY 1
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

solution/3000-3099/3059.Find All Unique Email Domains/Solution.sql

@@ -0,0 +1,6 @@
+# Write your MySQL query statement below
+SELECT SUBSTRING_INDEX(email, '@', -1) AS email_domain, COUNT(1) AS count
+FROM Emails
+WHERE email LIKE '%.com'
+GROUP BY 1
+ORDER BY 1;