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Commit 725e8f9

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

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@@ -54,7 +54,7 @@ Result 表:
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:GROUP BY**
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**方法一:分组求最小值**
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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。
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# Write your MySQL query statement below
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SELECT player_id, min(event_date) AS first_login
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FROM Activity
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GROUP BY player_id;
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GROUP BY 1;
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```
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<!-- tabs:end -->

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

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## Solutions
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**Solution 1: Group By + Min Function**
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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.
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<!-- tabs:start -->
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### **SQL**
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# Write your MySQL query statement below
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SELECT player_id, min(event_date) AS first_login
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FROM Activity
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GROUP BY player_id;
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GROUP BY 1;
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```
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<!-- tabs:end -->
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# Write your MySQL query statement below
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SELECT player_id, min(event_date) AS first_login
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FROM Activity
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GROUP BY player_id;
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GROUP BY 1;

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

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@@ -61,6 +61,10 @@ customer_number 为 '3' 的顾客有两个订单,比顾客 '1' 或者 '2' 都
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:分组 + 排序**
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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。
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<!-- tabs:start -->
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### **SQL**

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

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## Solutions
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**Solution 1: Group By + Sorting**
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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.
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<!-- tabs:start -->
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### **SQL**

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

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:左连接 + 分组 + CASE 表达式**
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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。
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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:
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- 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ドル$ 分;
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- 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ドル$ 分;
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- 如果主队和客队进球数相同,则积分加 1ドル$ 分;
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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。
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<!-- tabs:start -->
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### **SQL**
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```sql
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# Write your MySQL query statement below
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SELECT
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t.team_id,
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t.team_name,
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SUM(
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team_id,
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team_name,
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sum(
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CASE
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WHEN t.team_id = m.host_team
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AND m.host_goals > m.guest_goals THEN 3
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WHEN m.host_goals = m.guest_goals THEN 1
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WHEN t.team_id=m.guest_team
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ANDm.guest_goals>m.host_goals THEN 3
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WHEN team_id = host_team
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AND host_goals > guest_goals THEN 3
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WHEN team_id = guest_team
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AND guest_goals > host_goals THEN 3
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WHEN host_goals = guest_goals THEN 1
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ELSE 0
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END
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) AS num_points
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FROM
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TeamsAS t
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LEFT JOIN Matches AS m ON t.team_id = m.host_team OR t.team_id = m.guest_team
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GROUP BY t.team_id
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ORDER BY num_points DESC, team_id;
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Teams
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LEFT JOIN Matches ON team_id = host_team OR team_id = guest_team
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GROUP BY 1
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ORDER BY 3 DESC, 1;
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```
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<!-- tabs:end -->

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

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## Solutions
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**Solution 1: Left Join + Group By + Case Expression**
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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.
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Next, we group by `team_id` and use a `CASE` expression to calculate the points for each team according to the following rules:
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- If the team is the host team and has more goals than the guest team, add 3ドル$ points to the team's score.
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- If the team is the guest team and has more goals than the host team, add 3ドル$ points to the team's score.
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- If the host team and the guest team have the same number of goals, add 1ドル$ point to the team's score.
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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.
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### **SQL**
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```sql
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# Write your MySQL query statement below
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SELECT
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t.team_id,
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t.team_name,
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SUM(
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team_id,
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team_name,
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sum(
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CASE
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WHEN t.team_id = m.host_team
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AND m.host_goals > m.guest_goals THEN 3
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WHEN m.host_goals = m.guest_goals THEN 1
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WHEN t.team_id=m.guest_team
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ANDm.guest_goals>m.host_goals THEN 3
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WHEN team_id = host_team
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AND host_goals > guest_goals THEN 3
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WHEN team_id = guest_team
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AND guest_goals > host_goals THEN 3
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WHEN host_goals = guest_goals THEN 1
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ELSE 0
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END
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) AS num_points
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FROM
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TeamsAS t
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LEFT JOIN Matches AS m ON t.team_id = m.host_team OR t.team_id = m.guest_team
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GROUP BY t.team_id
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ORDER BY num_points DESC, team_id;
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Teams
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LEFT JOIN Matches ON team_id = host_team OR team_id = guest_team
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GROUP BY 1
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ORDER BY 3 DESC, 1;
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```
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<!-- tabs:end -->
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# Write your MySQL query statement below
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SELECT
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t.team_id,
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t.team_name,
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SUM(
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team_id,
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team_name,
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sum(
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CASE
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WHEN t.team_id = m.host_team
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AND m.host_goals > m.guest_goals THEN 3
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WHEN m.host_goals = m.guest_goals THEN 1
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WHEN t.team_id=m.guest_team
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ANDm.guest_goals>m.host_goals THEN 3
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WHEN team_id = host_team
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AND host_goals > guest_goals THEN 3
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WHEN team_id = guest_team
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AND guest_goals > host_goals THEN 3
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WHEN host_goals = guest_goals THEN 1
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ELSE 0
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END
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) AS num_points
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FROM
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TeamsAS t
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LEFT JOIN Matches AS m ON t.team_id = m.host_team OR t.team_id = m.guest_team
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GROUP BY t.team_id
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ORDER BY num_points DESC, team_id;
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Teams
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LEFT JOIN Matches ON team_id = host_team OR team_id = guest_team
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GROUP BY 1
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ORDER BY 3 DESC, 1;

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

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:等值连接 + CASE 表达式**
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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`。
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### **SQL**
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right_operand,
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CASE
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WHEN (
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(e.operator = '=' AND v1.value = v2.value)
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OR (e.operator = '>' AND v1.value > v2.value)
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OR (e.operator = '<' AND v1.value < v2.value)
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(operator = '=' AND v1.value = v2.value)
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OR (operator = '>' AND v1.value > v2.value)
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OR (operator = '<' AND v1.value < v2.value)
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) THEN 'true'
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ELSE 'false'
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END AS value
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FROM
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Expressions AS e
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LEFT JOIN Variables AS v1 ON e.left_operand = v1.name
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LEFT JOIN Variables AS v2 ON e.right_operand = v2.name;
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JOIN Variables AS v1 ON e.left_operand = v1.name
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JOIN Variables AS v2 ON e.right_operand = v2.name;
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```
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<!-- tabs:end -->

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

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## Solutions
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**Solution 1: Equi-Join + CASE Expression**
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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`.
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### **SQL**
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right_operand,
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CASE
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WHEN (
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(e.operator = '=' AND v1.value = v2.value)
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OR (e.operator = '>' AND v1.value > v2.value)
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OR (e.operator = '<' AND v1.value < v2.value)
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(operator = '=' AND v1.value = v2.value)
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OR (operator = '>' AND v1.value > v2.value)
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OR (operator = '<' AND v1.value < v2.value)
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) THEN 'true'
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ELSE 'false'
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END AS value
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FROM
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Expressions AS e
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LEFT JOIN Variables AS v1 ON e.left_operand = v1.name
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LEFT JOIN Variables AS v2 ON e.right_operand = v2.name;
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JOIN Variables AS v1 ON e.left_operand = v1.name
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JOIN Variables AS v2 ON e.right_operand = v2.name;
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```
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<!-- tabs:end -->

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