feat: add sql solution to lc problem: No.1951

No.1951.All the Pairs With the Maximum Number of Common Followers

Author: yanglbme

solution/1900-1999/1951.All the Pairs With the Maximum Number of Common Followers/README.md

@@ -73,7 +73,30 @@ Result 表:
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+# Write your MySQL query statement below
+with t as (
+    select
+        r1.user_id user1_id,
+        r2.user_id user2_id,
+        rank() over(
+            order by
+                count(1) desc
+        ) rk
+    from
+        Relations r1
+        join Relations r2 on r1.follower_id = r2.follower_id
+        and r1.user_id < r2.user_id
+    group by
+        r1.user_id,
+        r2.user_id
+)
+select
+    user1_id,
+    user2_id
+from
+    t
+where
+    rk = 1;
 ```
 
 <!-- tabs:end -->

solution/1900-1999/1951.All the Pairs With the Maximum Number of Common Followers/README_EN.md

@@ -67,7 +67,30 @@ Note that we do not have any information about the users that follow users 3, 4,
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+with t as (
+    select
+        r1.user_id user1_id,
+        r2.user_id user2_id,
+        rank() over(
+            order by
+                count(1) desc
+        ) rk
+    from
+        Relations r1
+        join Relations r2 on r1.follower_id = r2.follower_id
+        and r1.user_id < r2.user_id
+    group by
+        r1.user_id,
+        r2.user_id
+)
+select
+    user1_id,
+    user2_id
+from
+    t
+where
+    rk = 1;
 ```
 
 <!-- tabs:end -->

solution/1900-1999/1951.All the Pairs With the Maximum Number of Common Followers/Solution.sql

@@ -0,0 +1,24 @@
+# Write your MySQL query statement below
+with t as (
+    select
+        r1.user_id user1_id,
+        r2.user_id user2_id,
+        rank() over(
+            order by
+                count(1) desc
+        ) rk
+    from
+        Relations r1
+        join Relations r2 on r1.follower_id = r2.follower_id
+        and r1.user_id < r2.user_id
+    group by
+        r1.user_id,
+        r2.user_id
+)
+select
+    user1_id,
+    user2_id
+from
+    t
+where
+    rk = 1;