feat: update solutions to lc problems (#1784)

* No.0511.Game Play Analysis I
* No.0586.Customer Placing the Largest Number of Orders
* No.1212.Team Scores in Football Tournament
* No.1440.Evaluate Boolean Expression
* No.1571.Warehouse Manager
* No.1741.Find Total Time Spent by Each Employee
* No.1742.Maximum Number of Balls in a Box
* No.1890.The Latest Login in 2020
* No.2512.Reward Top K Students

Author: yanglbme

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

@@ -54,7 +54,7 @@ Result 表：
 
 <!-- 这里可写通用的实现逻辑 -->
 
-**方法一：GROUP BY**
+**方法一：分组求最小值**
 
 我们可以用 `GROUP BY` 对 `player_id` 进行分组，然后取每一组中最小的 `event_date` 作为玩家第一次登录平台的日期。
 
@@ -66,7 +66,7 @@ Result 表：
 # Write your MySQL query statement below
 SELECT player_id, min(event_date) AS first_login
 FROM Activity
-GROUP BY player_id;
+GROUP BY 1;
 ```
 
 <!-- tabs:end -->

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

@@ -55,6 +55,10 @@ Activity table:
 
 ## Solutions
 
+**Solution 1: Group By + Min Function**
+
+We can use `GROUP BY` to group the `player_id` and then take the minimum `event_date` in each group as the date when the player first logged into the platform.
+
 <!-- tabs:start -->
 
 ### **SQL**
@@ -63,7 +67,7 @@ Activity table:
 # Write your MySQL query statement below
 SELECT player_id, min(event_date) AS first_login
 FROM Activity
-GROUP BY player_id;
+GROUP BY 1;
 ```
 
 <!-- tabs:end -->

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

@@ -1,4 +1,4 @@
 # Write your MySQL query statement below
 SELECT player_id, min(event_date) AS first_login
 FROM Activity
-GROUP BY player_id;
+GROUP BY 1;

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

@@ -61,6 +61,10 @@ customer_number 为 '3' 的顾客有两个订单，比顾客 '1' 或者 '2' 都
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：分组 + 排序**
+
+我们可以使用 `GROUP BY` 将数据按照 `customer_number` 进行分组，然后按照 `count(1)` 进行降序排序，最后取第一条记录的 `customer_number` 即可。
+
 <!-- tabs:start -->
 
 ### **SQL**

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

@@ -55,6 +55,10 @@ So the result is customer_number 3.
 
 ## Solutions
 
+**Solution 1: Group By + Sorting**
+
+We can use `GROUP BY` to group the data by `customer_number`, and then sort the groups in descending order by `count(1)`. Finally, we can take the `customer_number` of the first record as the result.
+
 <!-- tabs:start -->
 
 ### **SQL**

solution/1200-1299/1212.Team Scores in Football Tournament/README.md

@@ -95,30 +95,42 @@ Teams </code>table:
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：左连接 + 分组 + CASE 表达式**
+
+我们可以通过左连接，将 `Teams` 表和 `Matches` 表连接起来，连接的条件为 `team_id = host_team OR team_id = guest_team`，这样就可以得到每个球队的所有比赛信息。
+
+接下来，我们按照 `team_id` 分组，然后使用 `CASE` 表达式计算每个球队的积分，计算规则如下：
+
+-   如果球队是主队，且主队进球数大于客队进球数，则积分加 $3$ 分；
+-   如果球队是客队，且客队进球数大于主队进球数，则积分加 $3$ 分；
+-   如果主队和客队进球数相同，则积分加 $1$ 分；
+
+最后，我们按照积分降序排序，如果积分相同，则按照 `team_id` 升序排序。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
 SELECT
-    t.team_id,
-    t.team_name,
-    SUM(
+    team_id,
+    team_name,
+    sum(
         CASE
-            WHEN t.team_id = m.host_team
-            AND m.host_goals > m.guest_goals THEN 3
-            WHEN m.host_goals = m.guest_goals THEN 1
-            WHEN t.team_id = m.guest_team
-            AND m.guest_goals > m.host_goals THEN 3
+            WHEN team_id = host_team
+            AND host_goals > guest_goals THEN 3
+            WHEN team_id = guest_team
+            AND guest_goals > host_goals THEN 3
+            WHEN host_goals = guest_goals THEN 1
             ELSE 0
         END
     ) AS num_points
 FROM
-    Teams AS t
-    LEFT JOIN Matches AS m ON t.team_id = m.host_team OR t.team_id = m.guest_team
-GROUP BY t.team_id
-ORDER BY num_points DESC, team_id;
+    Teams
+    LEFT JOIN Matches ON team_id = host_team OR team_id = guest_team
+GROUP BY 1
+ORDER BY 3 DESC, 1;
 ```
 
 <!-- tabs:end -->

solution/1200-1299/1212.Team Scores in Football Tournament/README_EN.md

@@ -90,30 +90,42 @@ Matches table:
 
 ## Solutions
 
+**Solution 1: Left Join + Group By + Case Expression**
+
+We can join the `Teams` table and the `Matches` table using a left join, where the join condition is `team_id = host_team OR team_id = guest_team`, to obtain all the match information for each team.
+
+Next, we group by `team_id` and use a `CASE` expression to calculate the points for each team according to the following rules:
+
+-   If the team is the host team and has more goals than the guest team, add $3$ points to the team's score.
+-   If the team is the guest team and has more goals than the host team, add $3$ points to the team's score.
+-   If the host team and the guest team have the same number of goals, add $1$ point to the team's score.
+
+Finally, we sort the result by points in descending order, and if the points are the same, we sort by `team_id` in ascending order.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
 SELECT
-    t.team_id,
-    t.team_name,
-    SUM(
+    team_id,
+    team_name,
+    sum(
         CASE
-            WHEN t.team_id = m.host_team
-            AND m.host_goals > m.guest_goals THEN 3
-            WHEN m.host_goals = m.guest_goals THEN 1
-            WHEN t.team_id = m.guest_team
-            AND m.guest_goals > m.host_goals THEN 3
+            WHEN team_id = host_team
+            AND host_goals > guest_goals THEN 3
+            WHEN team_id = guest_team
+            AND guest_goals > host_goals THEN 3
+            WHEN host_goals = guest_goals THEN 1
             ELSE 0
         END
     ) AS num_points
 FROM
-    Teams AS t
-    LEFT JOIN Matches AS m ON t.team_id = m.host_team OR t.team_id = m.guest_team
-GROUP BY t.team_id
-ORDER BY num_points DESC, team_id;
+    Teams
+    LEFT JOIN Matches ON team_id = host_team OR team_id = guest_team
+GROUP BY 1
+ORDER BY 3 DESC, 1;
 ```
 
 <!-- tabs:end -->

solution/1200-1299/1212.Team Scores in Football Tournament/Solution.sql

@@ -1,19 +1,19 @@
 # Write your MySQL query statement below
 SELECT
-    t.team_id,
-    t.team_name,
-    SUM(
+    team_id,
+    team_name,
+    sum(
         CASE
-            WHEN t.team_id = m.host_team
-            AND m.host_goals > m.guest_goals THEN 3
-            WHEN m.host_goals = m.guest_goals THEN 1
-            WHEN t.team_id = m.guest_team
-            AND m.guest_goals > m.host_goals THEN 3
+            WHEN team_id = host_team
+            AND host_goals > guest_goals THEN 3
+            WHEN team_id = guest_team
+            AND guest_goals > host_goals THEN 3
+            WHEN host_goals = guest_goals THEN 1
             ELSE 0
         END
     ) AS num_points
 FROM
-    Teams AS t
-    LEFT JOIN Matches AS m ON t.team_id = m.host_team OR t.team_id = m.guest_team
-GROUP BY t.team_id
-ORDER BY num_points DESC, team_id;
+    Teams
+    LEFT JOIN Matches ON team_id = host_team OR team_id = guest_team
+GROUP BY 1
+ORDER BY 3 DESC, 1;

solution/1400-1499/1440.Evaluate Boolean Expression/README.md

@@ -90,6 +90,10 @@ Expressions 表:
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：等值连接 + CASE 表达式**
+
+我们可以通过等值连接，将 `Expressions` 表中的每一行与 `Variables` 表中的两行进行关联，关联的条件是 `left_operand = name` 和 `right_operand = name`，然后通过 `CASE` 表达式来判断布尔表达式的值。如果 `operator` 为 `=`，则判断两个值是否相等；如果 `operator` 为 `>`，则判断左值是否大于右值；如果 `operator` 为 `<`，则判断左值是否小于右值。若是，那么布尔表达式的值为 `true`，否则为 `false`。
+
 <!-- tabs:start -->
 
 ### **SQL**
@@ -102,16 +106,16 @@ SELECT
     right_operand,
     CASE
         WHEN (
-            (e.operator = '=' AND v1.value = v2.value)
-            OR (e.operator = '>' AND v1.value > v2.value)
-            OR (e.operator = '<' AND v1.value < v2.value)
+            (operator = '=' AND v1.value = v2.value)
+            OR (operator = '>' AND v1.value > v2.value)
+            OR (operator = '<' AND v1.value < v2.value)
         ) THEN 'true'
         ELSE 'false'
     END AS value
 FROM
     Expressions AS e
-    LEFT JOIN Variables AS v1 ON e.left_operand = v1.name
-    LEFT JOIN Variables AS v2 ON e.right_operand = v2.name;
+    JOIN Variables AS v1 ON e.left_operand = v1.name
+    JOIN Variables AS v2 ON e.right_operand = v2.name;
 ```
 
 <!-- tabs:end -->

solution/1400-1499/1440.Evaluate Boolean Expression/README_EN.md

@@ -83,6 +83,10 @@ As shown, you need to find the value of each boolean expression in the table usi
 
 ## Solutions
 
+**Solution 1: Equi-Join + CASE Expression**
+
+We can associate each row in the `Expressions` table with two rows in the `Variables` table using an equi-join, where the conditions for the association are `left_operand = name` and `right_operand = name`. Then, we can use a `CASE` expression to determine the value of the boolean expression. If the `operator` is `=`, we check if the two values are equal. If the `operator` is `>`, we check if the left value is greater than the right value. If the `operator` is `<`, we check if the left value is less than the right value. If the condition is true, the boolean expression evaluates to `true`, otherwise it evaluates to `false`.
+
 <!-- tabs:start -->
 
 ### **SQL**
@@ -95,16 +99,16 @@ SELECT
     right_operand,
     CASE
         WHEN (
-            (e.operator = '=' AND v1.value = v2.value)
-            OR (e.operator = '>' AND v1.value > v2.value)
-            OR (e.operator = '<' AND v1.value < v2.value)
+            (operator = '=' AND v1.value = v2.value)
+            OR (operator = '>' AND v1.value > v2.value)
+            OR (operator = '<' AND v1.value < v2.value)
         ) THEN 'true'
         ELSE 'false'
     END AS value
 FROM
     Expressions AS e
-    LEFT JOIN Variables AS v1 ON e.left_operand = v1.name
-    LEFT JOIN Variables AS v2 ON e.right_operand = v2.name;
+    JOIN Variables AS v1 ON e.left_operand = v1.name
+    JOIN Variables AS v2 ON e.right_operand = v2.name;
 ```
 
 <!-- tabs:end -->

solution/1400-1499/1440.Evaluate Boolean Expression/Solution.sql

@@ -5,13 +5,13 @@ SELECT
     right_operand,
     CASE
         WHEN (
-            (e.operator = '=' AND v1.value = v2.value)
-            OR (e.operator = '>' AND v1.value > v2.value)
-            OR (e.operator = '<' AND v1.value < v2.value)
+            (operator = '=' AND v1.value = v2.value)
+            OR (operator = '>' AND v1.value > v2.value)
+            OR (operator = '<' AND v1.value < v2.value)
         ) THEN 'true'
         ELSE 'false'
     END AS value
 FROM
     Expressions AS e
-    LEFT JOIN Variables AS v1 ON e.left_operand = v1.name
-    LEFT JOIN Variables AS v2 ON e.right_operand = v2.name;
+    JOIN Variables AS v1 ON e.left_operand = v1.name
+    JOIN Variables AS v2 ON e.right_operand = v2.name;

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

@@ -96,6 +96,10 @@ Id为4的商品(LC-T-Shirt)的存货量为 4x10x20 = 800
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：等值连接 + 分组求和**
+
+我们可以使用等值连接将 `Warehouse` 表和 `Products` 表按照 `product_id` 进行连接，并按照仓库名称进行分组，然后使用 `SUM` 函数计算每个仓库的存货量。
+
 <!-- tabs:start -->
 
 ### **SQL**
@@ -104,11 +108,11 @@ Id为4的商品(LC-T-Shirt)的存货量为 4x10x20 = 800
 # Write your MySQL query statement below
 SELECT
     name AS warehouse_name,
-    SUM(units * Width * Length * Height) AS volume
+    sum(width * length * height * units) AS volume
 FROM
-    Warehouse AS w
-    JOIN Products AS p ON w.product_id = p.product_id
-GROUP BY name;
+    Warehouse
+    JOIN Products USING (product_id)
+GROUP BY 1;
 ```
 
 <!-- tabs:end -->

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

@@ -92,6 +92,10 @@ LCHouse3: 1 unit of LC-T-Shirt.
 
 ## Solutions
 
+**Solution 1: Inner Join + Group By + Sum Function**
+
+We can use an inner join to join the `Warehouse` table and the `Products` table on the condition of `product_id`, and then group by warehouse name to calculate the inventory of each warehouse using the `SUM` function.
+
 <!-- tabs:start -->
 
 ### **SQL**
@@ -100,11 +104,11 @@ LCHouse3: 1 unit of LC-T-Shirt.
 # Write your MySQL query statement below
 SELECT
     name AS warehouse_name,
-    SUM(units * Width * Length * Height) AS volume
+    sum(width * length * height * units) AS volume
 FROM
-    Warehouse AS w
-    JOIN Products AS p ON w.product_id = p.product_id
-GROUP BY name;
+    Warehouse
+    JOIN Products USING (product_id)
+GROUP BY 1;
 ```
 
 <!-- tabs:end -->

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

@@ -1,8 +1,8 @@
 # Write your MySQL query statement below
 SELECT
     name AS warehouse_name,
-    SUM(units * Width * Length * Height) AS volume
+    sum(width * length * height * units) AS volume
 FROM
-    Warehouse AS w
-    JOIN Products AS p ON w.product_id = p.product_id
-GROUP BY name;
+    Warehouse
+    JOIN Products USING (product_id)
+GROUP BY 1;

solution/1700-1799/1741.Find Total Time Spent by Each Employee/README.md

@@ -65,17 +65,19 @@ Employees 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 `emp_id` 和 `event_day` 进行分组，然后计算每个分组的总时间。总时间等于每个分组的 `out_time` 减去 `in_time` 的和。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
-SELECT
-    event_day AS day,
-    emp_id,
-    SUM(out_time - in_time) AS total_time
+# Write your MySQL query statement below
+SELECT event_day AS day, emp_id, sum(out_time - in_time) AS total_time
 FROM Employees
-GROUP BY emp_id, event_day;
+GROUP BY 1, 2;
 ```
 
 <!-- tabs:end -->