feat: add sql solution to lc problem: No.1811 (doocs#1102)

No.1811.Find Interview Candidates

Author: yanglbme

solution/1800-1899/1811.Find Interview Candidates/README.md

@@ -114,7 +114,45 @@ Quarz在连续5场竞赛中赢得了奖牌(190, 191, 192, 193, and 194), 所以
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    S AS (
+        SELECT contest_id, gold_medal AS user_id, 1 AS type
+        FROM Contests
+        UNION
+        SELECT contest_id, silver_medal AS user_id, 2 AS type
+        FROM Contests
+        UNION
+        SELECT contest_id, bronze_medal AS user_id, 3 AS type
+        FROM Contests
+    ),
+    T AS (
+        SELECT
+            user_id,
+            (
+                contest_id - row_number() OVER (
+                    PARTITION BY user_id
+                    ORDER BY contest_id
+                )
+            ) AS diff
+        FROM S
+    ),
+    P AS (
+        SELECT user_id
+        FROM S
+        WHERE type = 1
+        GROUP BY user_id
+        HAVING count(1) >= 3
+        UNION
+        SELECT DISTINCT user_id
+        FROM T
+        GROUP BY user_id, diff
+        HAVING count(1) >= 3
+    )
+SELECT name, mail
+FROM
+    P AS p
+    LEFT JOIN Users AS u ON p.user_id = u.user_id;
 ```
 
 <!-- tabs:end -->

solution/1800-1899/1811.Find Interview Candidates/README_EN.md

@@ -107,7 +107,45 @@ Quarz won a medal in 5 consecutive contests (190, 191, 192, 193, and 194), so we
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    S AS (
+        SELECT contest_id, gold_medal AS user_id, 1 AS type
+        FROM Contests
+        UNION
+        SELECT contest_id, silver_medal AS user_id, 2 AS type
+        FROM Contests
+        UNION
+        SELECT contest_id, bronze_medal AS user_id, 3 AS type
+        FROM Contests
+    ),
+    T AS (
+        SELECT
+            user_id,
+            (
+                contest_id - row_number() OVER (
+                    PARTITION BY user_id
+                    ORDER BY contest_id
+                )
+            ) AS diff
+        FROM S
+    ),
+    P AS (
+        SELECT user_id
+        FROM S
+        WHERE type = 1
+        GROUP BY user_id
+        HAVING count(1) >= 3
+        UNION
+        SELECT DISTINCT user_id
+        FROM T
+        GROUP BY user_id, diff
+        HAVING count(1) >= 3
+    )
+SELECT name, mail
+FROM
+    P AS p
+    LEFT JOIN Users AS u ON p.user_id = u.user_id;
 ```
 
 <!-- tabs:end -->

solution/1800-1899/1811.Find Interview Candidates/Solution.sql

@@ -0,0 +1,39 @@
+# Write your MySQL query statement below
+WITH
+    S AS (
+        SELECT contest_id, gold_medal AS user_id, 1 AS type
+        FROM Contests
+        UNION
+        SELECT contest_id, silver_medal AS user_id, 2 AS type
+        FROM Contests
+        UNION
+        SELECT contest_id, bronze_medal AS user_id, 3 AS type
+        FROM Contests
+    ),
+    T AS (
+        SELECT
+            user_id,
+            (
+                contest_id - row_number() OVER (
+                    PARTITION BY user_id
+                    ORDER BY contest_id
+                )
+            ) AS diff
+        FROM S
+    ),
+    P AS (
+        SELECT user_id
+        FROM S
+        WHERE type = 1
+        GROUP BY user_id
+        HAVING count(1) >= 3
+        UNION
+        SELECT DISTINCT user_id
+        FROM T
+        GROUP BY user_id, diff
+        HAVING count(1) >= 3
+    )
+SELECT name, mail
+FROM
+    P AS p
+    LEFT JOIN Users AS u ON p.user_id = u.user_id;