feat: add solutions to lc problems: No.0512,1571

* No.0512.Game Play Analysis II
* No.1571.Warehouse Manager

Author: yanglbme

solution/0500-0599/0511.Game Play Analysis I/README.md

@@ -60,10 +60,12 @@ Result 表：
 
 ```sql
 SELECT
-    player_id, MIN(event_date) first_login
+    player_id,
+    MIN(event_date) first_login
 FROM
     Activity
-GROUP BY player_id;
+GROUP BY
+    player_id;
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0511.Game Play Analysis I/README_EN.md

@@ -61,10 +61,12 @@ Activity table:
 
 ```sql
 SELECT
-    player_id, MIN(event_date) first_login
+    player_id,
+    MIN(event_date) first_login
 FROM
     Activity
-GROUP BY player_id;
+GROUP BY
+    player_id;
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0511.Game Play Analysis I/Solution.sql

@@ -1,5 +1,7 @@
-SELECT 
-    player_id, MIN(event_date) first_login
+SELECT
+    player_id,
+    MIN(event_date) first_login
 FROM
     Activity
-GROUP BY player_id;
+GROUP BY
+    player_id;

solution/0500-0599/0512.Game Play Analysis II/README.md

@@ -56,7 +56,22 @@ Result table:
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+select
+    player_id,
+    device_id
+from
+    Activity
+where
+    (player_id, event_date) in (
+        select
+            player_id,
+            min(event_date) event_date
+        from
+            Activity
+        group by
+            player_id
+    )
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0512.Game Play Analysis II/README_EN.md

@@ -60,7 +60,22 @@ Activity table:
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+select
+    player_id,
+    device_id
+from
+    Activity
+where
+    (player_id, event_date) in (
+        select
+            player_id,
+            min(event_date) event_date
+        from
+            Activity
+        group by
+            player_id
+    )
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0512.Game Play Analysis II/Solution.sql

@@ -0,0 +1,16 @@
+# Write your MySQL query statement below
+select
+    player_id,
+    device_id
+from
+    Activity
+where
+    (player_id, event_date) in (
+        select
+            player_id,
+            min(event_date) event_date
+        from
+            Activity
+        group by
+            player_id
+    )

solution/0500-0599/0586.Customer Placing the Largest Number of Orders/README.md

@@ -66,13 +66,17 @@ customer_number 为 '3' 的顾客有两个订单，比顾客 '1' 或者 '2' 都
 ### **SQL**
 
 ```sql
+# Write your MySQL query statement below
 SELECT
     customer_number
 FROM
-    Orders
-GROUP BY customer_number
-ORDER BY COUNT(customer_number) DESC
-LIMIT 1;
+    orders
+GROUP BY
+    customer_number
+ORDER BY
+    count(1) DESC
+LIMIT
+    1;
 ```
 
 SQL Server

solution/0500-0599/0586.Customer Placing the Largest Number of Orders/README_EN.md

@@ -60,13 +60,17 @@ So the result is customer_number 3.
 ### **SQL**
 
 ```sql
+# Write your MySQL query statement below
 SELECT
     customer_number
 FROM
-    Orders
-GROUP BY customer_number
-ORDER BY COUNT(customer_number) DESC
-LIMIT 1;
+    orders
+GROUP BY
+    customer_number
+ORDER BY
+    count(1) DESC
+LIMIT
+    1;
 ```
 
 SQL Server

solution/0500-0599/0586.Customer Placing the Largest Number of Orders/Solution.sql

@@ -1,7 +1,11 @@
-SELECT 
+# Write your MySQL query statement below
+SELECT
     customer_number
 FROM
-    Orders
-GROUP BY customer_number
-ORDER BY COUNT(customer_number) DESC
-LIMIT 1;
+    orders
+GROUP BY
+    customer_number
+ORDER BY
+    count(1) DESC
+LIMIT
+    1;

solution/1500-1599/1571.Warehouse Manager/README.md

@@ -98,26 +98,18 @@ Id为4的商品(LC-T-Shirt)的存货量为 4x10x20 = 800
 
 <!-- tabs:start -->
 
-### **Python3**
-
-<!-- 这里可写当前语言的特殊实现逻辑 -->
-
-```python
-
-```
-
-### **Java**
-
-<!-- 这里可写当前语言的特殊实现逻辑 -->
-
-```java
-
-```
-
-### **...**
-
-```
-
+### **SQL**
+
+```sql
+# Write your MySQL query statement below
+SELECT
+    name AS warehouse_name,
+    SUM(units * Width * Length * Height) AS volume
+FROM
+    Warehouse w
+    JOIN Products p ON w.product_id = p.product_id
+GROUP BY
+    name
 ```
 
 <!-- tabs:end -->

solution/1500-1599/1571.Warehouse Manager/README_EN.md

@@ -94,22 +94,18 @@ LCHouse3: 1 unit of LC-T-Shirt.
 
 <!-- tabs:start -->
 
-### **Python3**
-
-```python
-
-```
-
-### **Java**
-
-```java
-
-```
-
-### **...**
-
-```
-
+### **SQL**
+
+```sql
+# Write your MySQL query statement below
+SELECT
+    name AS warehouse_name,
+    SUM(units * Width * Length * Height) AS volume
+FROM
+    Warehouse w
+    JOIN Products p ON w.product_id = p.product_id
+GROUP BY
+    name
 ```
 
 <!-- tabs:end -->

solution/1500-1599/1571.Warehouse Manager/Solution.sql

@@ -0,0 +1,9 @@
+# Write your MySQL query statement below
+SELECT
+    name AS warehouse_name,
+    SUM(units * Width * Length * Height) AS volume
+FROM
+    Warehouse w
+    JOIN Products p ON w.product_id = p.product_id
+GROUP BY
+    name

solution/2600-2699/2611.Mice and Cheese/README.md

@@ -57,11 +57,13 @@
 
 **方法一：贪心 + 排序**
 
-我们可以先将所有奶酪分给第二只老鼠，接下来，考虑将其中 $k$ 块奶酪分给第一只老鼠，那么我们应该如何选择这 $k$ 块奶酪呢？显然，将第 $i$ 块奶酪从第二只老鼠分给第一只老鼠，得分的变化量为 $reward1[i] - reward2[i]$，我们希望这个变化量尽可能大，这样才能使得总得分最大。
+我们可以先将所有奶酪分给第二只老鼠，因此初始得分为 $\sum_{i=0}^{n-1} reward2[i]$。
 
-因此，我们将奶酪按照 `reward1[i] - reward2[i]` 从大到小排序，前 $k$ 块奶酪由第一只老鼠吃掉，剩下的奶酪由第二只老鼠吃掉，即可得到最大得分。
+接下来，考虑将其中 $k$ 块奶酪分给第一只老鼠，那么我们应该如何选择这 $k$ 块奶酪呢？显然，将第 $i$ 块奶酪从第二只老鼠分给第一只老鼠，得分的变化量为 $reward1[i] - reward2[i]$，我们希望这个变化量尽可能大，这样才能使得总得分最大。
 
-时间复杂度 $O(n \times \log n)$，空间复杂度 $O(n)$。其中 $n$ 为奶酪的数量。
+因此，我们将奶酪按照 $reward1[i] - reward2[i]$ 从大到小排序，前 $k$ 块奶酪由第一只老鼠吃掉，剩下的奶酪由第二只老鼠吃掉，即可得到最大得分。也即是说，我们将初始得分加上 $\sum_{i=0}^{k-1} (reward1[i] - reward2[i])$ 即可。
+
+时间复杂度 $O(n \times \log n)$，空间复杂度 $O(\log n)$。其中 $n$ 为奶酪的数量。
 
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