feat: add solutions to lc problems: No.1174,1193,1211,1633 (#1837)

* No.1174.Immediate Food Delivery II
* No.1193.Monthly Transactions I
* No.1211.Queries Quality and Percentage
* No.1633.Percentage of Users Attended a Contest

Author: yanglbme

solution/1100-1199/1174.Immediate Food Delivery II/README.md

@@ -67,35 +67,47 @@ Delivery 表：
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：子查询**
+
+我们可以使用子查询，先找到每个用户的首次订单，然后再计算即时订单的比例。
+
+**方法二：窗口函数**
+
+我们可以使用 `RANK()` 窗口函数，按照每个用户的订单日期升序排列，获取到每个用户的订单排名，然后我们筛选出排名为 $1$ 的订单，即为首次订单，再计算即时订单的比例。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
-select
-    ROUND(
-        SUM(
-            IF(
-                t1.order_date = t1.customer_pref_delivery_date,
-                1,
-                0
-            )
-        ) / COUNT(1) * 100,
-        2
-    ) as immediate_percentage
-from
-    Delivery t1
-    right join (
-        select
-            customer_id,
-            MIN(order_date) as order_date
-        from
-            Delivery
-        group by
-            customer_id
-    ) t2 on t1.customer_id = t2.customer_id
-    and t1.order_date = t2.order_date
+SELECT
+    ROUND(SUM(order_date = customer_pref_delivery_date) / COUNT(1) * 100, 2) AS immediate_percentage
+FROM Delivery
+WHERE
+    (customer_id, order_date) IN (
+        SELECT customer_id, MIN(order_date)
+        FROM Delivery
+        GROUP BY 1
+    );
+```
+
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            *,
+            RANK() OVER (
+                PARTITION BY customer_id
+                ORDER BY order_date
+            ) AS rk
+        FROM Delivery
+    )
+SELECT
+    ROUND(SUM(order_date = customer_pref_delivery_date) / COUNT(1) * 100, 2) AS immediate_percentage
+FROM T
+WHERE rk = 1;
 ```
 
 <!-- tabs:end -->

solution/1100-1199/1174.Immediate Food Delivery II/README_EN.md

@@ -62,35 +62,47 @@ Hence, half the customers have immediate first orders.
 
 ## Solutions
 
+**Solution 1: Subquery**
+
+We can use a subquery to first find the first order of each user, and then calculate the proportion of instant orders.
+
+**Solution 2: Window Function**
+
+We can use the `RANK()` window function to rank the orders of each user in ascending order by order date, and then filter out the orders with a rank of $1$, which are the first orders of each user. After that, we can calculate the proportion of instant orders.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
-select
-    ROUND(
-        SUM(
-            IF(
-                t1.order_date = t1.customer_pref_delivery_date,
-                1,
-                0
-            )
-        ) / COUNT(1) * 100,
-        2
-    ) as immediate_percentage
-from
-    Delivery t1
-    right join (
-        select
-            customer_id,
-            MIN(order_date) as order_date
-        from
-            Delivery
-        group by
-            customer_id
-    ) t2 on t1.customer_id = t2.customer_id
-    and t1.order_date = t2.order_date
+SELECT
+    ROUND(SUM(order_date = customer_pref_delivery_date) / COUNT(1) * 100, 2) AS immediate_percentage
+FROM Delivery
+WHERE
+    (customer_id, order_date) IN (
+        SELECT customer_id, MIN(order_date)
+        FROM Delivery
+        GROUP BY 1
+    );
+```
+
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            *,
+            RANK() OVER (
+                PARTITION BY customer_id
+                ORDER BY order_date
+            ) AS rk
+        FROM Delivery
+    )
+SELECT
+    ROUND(SUM(order_date = customer_pref_delivery_date) / COUNT(1) * 100, 2) AS immediate_percentage
+FROM T
+WHERE rk = 1;
 ```
 
 <!-- tabs:end -->

solution/1100-1199/1174.Immediate Food Delivery II/Solution.sql

@@ -1,16 +1,10 @@
 # Write your MySQL query statement below
 SELECT
-    ROUND(
-        SUM(IF(t1.order_date = t1.customer_pref_delivery_date, 1, 0)) / COUNT(1) * 100,
-        2
-    ) AS immediate_percentage
-FROM
-    Delivery AS t1
-    RIGHT JOIN (
-        SELECT
-            customer_id,
-            MIN(order_date) AS order_date
+    ROUND(SUM(order_date = customer_pref_delivery_date) / COUNT(1) * 100, 2) AS immediate_percentage
+FROM Delivery
+WHERE
+    (customer_id, order_date) IN (
+        SELECT customer_id, MIN(order_date)
         FROM Delivery
-        GROUP BY customer_id
-    ) AS t2
-        ON t1.customer_id = t2.customer_id AND t1.order_date = t2.order_date;
+        GROUP BY 1
+    );

solution/1100-1199/1193.Monthly Transactions I/README.md

@@ -59,6 +59,10 @@ Transactions</code> table:
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：分组求和**
+
+我们可以先按照月份和国家分组，然后利用 `COUNT` 和 `SUM` 函数分别求出每个分组的事务数、已批准的事务数、总金额和已批准的总金额。
+
 <!-- tabs:start -->
 
 ### **SQL**

solution/1100-1199/1193.Monthly Transactions I/README_EN.md

@@ -55,6 +55,10 @@ Transactions table:
 
 ## Solutions
 
+**Solution 1: Grouping and Aggregation**
+
+We can first group by month and country, and then use the `COUNT` and `SUM` functions to respectively calculate the number of transactions, the number of approved transactions, the total amount, and the total amount of approved transactions for each group.
+
 <!-- tabs:start -->
 
 ### **SQL**

solution/1200-1299/1211.Queries Quality and Percentage/README.md

@@ -81,6 +81,10 @@ Cat 查询结果的劣质查询百分比为 (1 / 3) * 100 = 33.33
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：分组统计**
+
+我们将查询结果按照 `query_name` 进行分组，然后利用 `AVG` 和 `ROUND` 函数计算 `quality` 和 `poor_query_percentage`。
+
 <!-- tabs:start -->
 
 ### **SQL**
@@ -90,9 +94,9 @@ Cat 查询结果的劣质查询百分比为 (1 / 3) * 100 = 33.33
 SELECT
     query_name,
     ROUND(AVG(rating / position), 2) AS quality,
-    ROUND(100 * AVG(rating < 3), 2) AS poor_query_percentage
+    ROUND(AVG(rating < 3) * 100, 2) AS poor_query_percentage
 FROM Queries
-GROUP BY query_name;
+GROUP BY 1;
 ```
 
 <!-- tabs:end -->

solution/1200-1299/1211.Queries Quality and Percentage/README_EN.md

@@ -76,6 +76,10 @@ Cat queries poor_ query_percentage is (1 / 3) * 100 = 33.33
 
 ## Solutions
 
+**Solution 1: Grouping and Aggregation**
+
+We can group the query results by `query_name`, and then use the `AVG` and `ROUND` functions to calculate `quality` and `poor_query_percentage`.
+
 <!-- tabs:start -->
 
 ### **SQL**
@@ -85,9 +89,9 @@ Cat queries poor_ query_percentage is (1 / 3) * 100 = 33.33
 SELECT
     query_name,
     ROUND(AVG(rating / position), 2) AS quality,
-    ROUND(100 * AVG(rating < 3), 2) AS poor_query_percentage
+    ROUND(AVG(rating < 3) * 100, 2) AS poor_query_percentage
 FROM Queries
-GROUP BY query_name;
+GROUP BY 1;
 ```
 
 <!-- tabs:end -->

solution/1200-1299/1211.Queries Quality and Percentage/Solution.sql

@@ -2,6 +2,6 @@
 SELECT
     query_name,
     ROUND(AVG(rating / position), 2) AS quality,
-    ROUND(100 * AVG(rating < 3), 2) AS poor_query_percentage
+    ROUND(AVG(rating < 3) * 100, 2) AS poor_query_percentage
 FROM Queries
-GROUP BY query_name;
+GROUP BY 1;

solution/1600-1699/1633.Percentage of Users Attended a Contest/README.md

@@ -91,6 +91,10 @@ Bob 注册了 207 赛事，注册率为 ((1/3) * 100) = 33.33%</pre>
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：分组统计 + 子查询**
+
+我们可以将 `Register` 表按照 `contest_id` 分组，统计每个赛事的注册人数，每个赛事的注册人数除以总注册人数即为该赛事的注册率。
+
 <!-- tabs:start -->
 
 ### **SQL**
@@ -101,8 +105,8 @@ SELECT
     contest_id,
     ROUND(COUNT(1) * 100 / (SELECT COUNT(1) FROM Users), 2) AS percentage
 FROM Register
-GROUP BY contest_id
-ORDER BY percentage DESC, contest_id;
+GROUP BY 1
+ORDER BY 2 DESC, 1;
 ```
 
 <!-- tabs:end -->

solution/1600-1699/1633.Percentage of Users Attended a Contest/README_EN.md

@@ -88,6 +88,10 @@ Bob registered in contest 207 and the percentage is ((1/3) * 100) = 33.33%
 
 ## Solutions
 
+**Solution 1: Grouping and Subquery**
+
+We can group the `Register` table by `contest_id` and count the number of registrations for each contest. The registration rate of each contest is the number of registrations divided by the total number of registrations.
+
 <!-- tabs:start -->
 
 ### **SQL**
@@ -98,8 +102,8 @@ SELECT
     contest_id,
     ROUND(COUNT(1) * 100 / (SELECT COUNT(1) FROM Users), 2) AS percentage
 FROM Register
-GROUP BY contest_id
-ORDER BY percentage DESC, contest_id;
+GROUP BY 1
+ORDER BY 2 DESC, 1;
 ```
 
 <!-- tabs:end -->

solution/1600-1699/1633.Percentage of Users Attended a Contest/Solution.sql

@@ -3,5 +3,5 @@ SELECT
     contest_id,
     ROUND(COUNT(1) * 100 / (SELECT COUNT(1) FROM Users), 2) AS percentage
 FROM Register
-GROUP BY contest_id
-ORDER BY percentage DESC, contest_id;
+GROUP BY 1
+ORDER BY 2 DESC, 1;