feat: add sql solution to lc problem: No.2701 (#1143)

No.2701.Consecutive Transactions with Increasing Amounts

Author: yanglbme

solution/2700-2799/2701.Consecutive Transactions with Increasing Amounts/README.md

@@ -76,7 +76,32 @@ Transactions 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
+            t1.*,
+            sum(
+                CASE
+                    WHEN t2.customer_id IS NULL THEN 1
+                    ELSE 0
+                END
+            ) OVER (ORDER BY customer_id, transaction_date) AS s
+        FROM
+            Transactions AS t1
+            LEFT JOIN Transactions AS t2
+                ON t1.customer_id = t2.customer_id
+                AND t1.amount > t2.amount
+                AND datediff(t1.transaction_date, t2.transaction_date) = 1
+    )
+SELECT
+    customer_id,
+    min(transaction_date) AS consecutive_start,
+    max(transaction_date) AS consecutive_end
+FROM T
+GROUP BY customer_id, s
+HAVING count(1) >= 3
+ORDER BY customer_id;
 ```
 
 <!-- tabs:end -->

solution/2700-2799/2701.Consecutive Transactions with Increasing Amounts/README_EN.md

@@ -72,7 +72,32 @@ customer_id is sorted in ascending order.
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            t1.*,
+            sum(
+                CASE
+                    WHEN t2.customer_id IS NULL THEN 1
+                    ELSE 0
+                END
+            ) OVER (ORDER BY customer_id, transaction_date) AS s
+        FROM
+            Transactions AS t1
+            LEFT JOIN Transactions AS t2
+                ON t1.customer_id = t2.customer_id
+                AND t1.amount > t2.amount
+                AND datediff(t1.transaction_date, t2.transaction_date) = 1
+    )
+SELECT
+    customer_id,
+    min(transaction_date) AS consecutive_start,
+    max(transaction_date) AS consecutive_end
+FROM T
+GROUP BY customer_id, s
+HAVING count(1) >= 3
+ORDER BY customer_id;
 ```
 
 <!-- tabs:end -->

solution/2700-2799/2701.Consecutive Transactions with Increasing Amounts/Solution.sql

@@ -0,0 +1,26 @@
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            t1.*,
+            sum(
+                CASE
+                    WHEN t2.customer_id IS NULL THEN 1
+                    ELSE 0
+                END
+            ) OVER (ORDER BY customer_id, transaction_date) AS s
+        FROM
+            Transactions AS t1
+            LEFT JOIN Transactions AS t2
+                ON t1.customer_id = t2.customer_id
+                AND t1.amount > t2.amount
+                AND datediff(t1.transaction_date, t2.transaction_date) = 1
+    )
+SELECT
+    customer_id,
+    min(transaction_date) AS consecutive_start,
+    max(transaction_date) AS consecutive_end
+FROM T
+GROUP BY customer_id, s
+HAVING count(1) >= 3
+ORDER BY customer_id;