feat: add sql solution to lc problem: No.1919

solution/1900-1999/1919.Leetcodify Similar Friends/README.md

@@ -106,7 +106,15 @@ Friendship table:
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT DISTINCT user1_id, user2_id
+FROM
+    Friendship AS f
+    LEFT JOIN Listens AS l1 ON user1_id = l1.user_id
+    LEFT JOIN Listens AS l2 ON user2_id = l2.user_id
+WHERE l1.song_id = l2.song_id AND l1.day = l2.day
+GROUP BY 1, 2, l1.day
+HAVING count(DISTINCT l1.song_id) >= 3;
 ```
 
 <!-- tabs:end -->

solution/1900-1999/1919.Leetcodify Similar Friends/README_EN.md

@@ -100,7 +100,15 @@ Users 2 and 5 are friends and listened to songs 10, 11, and 12, but they did not
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT DISTINCT user1_id, user2_id
+FROM
+    Friendship AS f
+    LEFT JOIN Listens AS l1 ON user1_id = l1.user_id
+    LEFT JOIN Listens AS l2 ON user2_id = l2.user_id
+WHERE l1.song_id = l2.song_id AND l1.day = l2.day
+GROUP BY 1, 2, l1.day
+HAVING count(DISTINCT l1.song_id) >= 3;
 ```
 
 <!-- tabs:end -->

solution/1900-1999/1919.Leetcodify Similar Friends/Solution.sql

@@ -0,0 +1,9 @@
+# Write your MySQL query statement below
+SELECT DISTINCT user1_id, user2_id
+FROM
+    Friendship AS f
+    LEFT JOIN Listens AS l1 ON user1_id = l1.user_id
+    LEFT JOIN Listens AS l2 ON user2_id = l2.user_id
+WHERE l1.song_id = l2.song_id AND l1.day = l2.day
+GROUP BY 1, 2, l1.day
+HAVING count(DISTINCT l1.song_id) >= 3;