华为大数据面试SQL-合并日期重叠的活动

一、题目

已知有表记录了每个大厅的活动开始日期和结束日期，每个大厅可以有多个活动。请编写一个SQL查询合并在同一个大厅举行的所有重叠的活动，如果两个活动至少有一天相同，那他们就是重叠的

样例数据

+----------+-------------+-------------+
| hall_id  | start_date  |  end_date   |
+----------+-------------+-------------+
| 1        | 2023-01-13  | 2023-01-20  |
| 1        | 2023-01-14  | 2023-01-17  |
| 1        | 2023-01-14  | 2023-01-16  |
| 1        | 2023-01-18  | 2023-01-25  |
| 1        | 2023-01-20  | 2023-01-26  |
| 2        | 2022-12-09  | 2022-12-23  |
| 2        | 2022-12-13  | 2022-12-17  |
| 2        | 2022-12-20  | 2022-12-24  |
| 2        | 2022-12-25  | 2022-12-30  |
| 3        | 2022-12-01  | 2023-01-30  |
+----------+-------------+-------------+

结果

+----------+-------------+-------------+
| hall_id  | start_date  |  end_date   |
+----------+-------------+-------------+
| 1        | 2023-01-13  | 2023-01-26  |
| 2        | 2022-12-09  | 2022-12-24  |
| 2        | 2022-12-25  | 2022-12-30  |
| 3        | 2022-12-01  | 2023-01-30  |
+----------+-------------+-------------+

解释:有三个大厅。 大厅 1:

两个活动["2823-01-13"，"2023-01-20"]和[“2023-01-14"，"2023-01-17"]重叠，我们将它们合并到一个活动中[“2023-01-13"，"2023-01-20"]。

大厅 2：

["2022-12-25"，"2022-12-30"]不与任何其他活动重叠，所以我们保持原样。

二、分析

我们首先按照hall_id分组，根据start_date、end_date 升序排列，按照start_date 进行了升序排列，所以当前行的start_date一定晚于前一行的start_date,对当前行的start_date 和截止到上一行的最大end_date进行比较，如果当前行的start_date 小于等于截止到前一行最大end_date 代表有交叉，可以合并，否则代表不可合并。判断出是否可以合并之后，具体操作合并就转化成类似连续问题了。

维度	评分
题目难度	⭐️⭐️⭐️⭐️⭐️
题目清晰度	⭐️⭐️⭐️⭐️⭐️
业务常见度	⭐️⭐️⭐️⭐️

三、SQL

1.使用max()函数开窗，获得截止到当前行之前的活动最后日期

这里我们使用max()，按照hall_id进行分组，然后根据start_date和end_date进行排序，利用聚合函数开窗，开窗函数内有排序则聚合到当前行的特性，进行处理。其中我们这边需要聚合到当前行的上一行。

执行SQL

select
    hall_id,
    start_date,
    end_date,
    max(end_date) over (partition by hall_id order by start_date asc,end_date asc rows between unbounded preceding and 1 preceding) as max_end_date
from t1_hall_event

执行结果

+----------+-------------+-------------+---------------+
| hall_id  | start_date  |  end_date   | max_end_date  |
+----------+-------------+-------------+---------------+
| 1        | 2023-01-13  | 2023-01-20  | NULL          |
| 1        | 2023-01-14  | 2023-01-16  | 2023-01-20    |
| 1        | 2023-01-14  | 2023-01-17  | 2023-01-20    |
| 1        | 2023-01-18  | 2023-01-25  | 2023-01-20    |
| 1        | 2023-01-20  | 2023-01-26  | 2023-01-25    |
| 2        | 2022-12-09  | 2022-12-23  | NULL          |
| 2        | 2022-12-13  | 2022-12-17  | 2022-12-23    |
| 2        | 2022-12-20  | 2022-12-24  | 2022-12-23    |
| 2        | 2022-12-25  | 2022-12-30  | 2022-12-24    |
| 3        | 2022-12-01  | 2023-01-30  | NULL          |
+----------+-------------+-------------+---------------+

2.根据当前行的start_day与max_end_date进行比较，得出是否可以合并标记；

执行SQL

select hall_id,
       start_date,
       end_date,
       max_end_date,
       if(start_date <= max_end_date, 0, 1) as is_merge --0:合并，1:不合并
from (select hall_id,
             start_date,
             end_date,
             max(end_date)
                 over (partition by hall_id order by start_date asc,end_date asc rows between unbounded preceding and 1 preceding) as max_end_date
      from t1_hall_event) t

执行结果

+----------+-------------+-------------+---------------+-----------+
| hall_id  | start_date  |  end_date   | max_end_date  | is_merge  |
+----------+-------------+-------------+---------------+-----------+
| 1        | 2023-01-13  | 2023-01-20  | NULL          | 1         |
| 1        | 2023-01-14  | 2023-01-16  | 2023-01-20    | 0         |
| 1        | 2023-01-14  | 2023-01-17  | 2023-01-20    | 0         |
| 1        | 2023-01-18  | 2023-01-25  | 2023-01-20    | 0         |
| 1        | 2023-01-20  | 2023-01-26  | 2023-01-25    | 0         |
| 2        | 2022-12-09  | 2022-12-23  | NULL          | 1         |
| 2        | 2022-12-13  | 2022-12-17  | 2022-12-23    | 0         |
| 2        | 2022-12-20  | 2022-12-24  | 2022-12-23    | 0         |
| 2        | 2022-12-25  | 2022-12-30  | 2022-12-24    | 1         |
| 3        | 2022-12-01  | 2023-01-30  | NULL          | 1         |
+----------+-------------+-------------+---------------+-----------+

3.连续问题，使用sum()over()进行分组；

执行SQL

select hall_id,
       start_date,
       end_date,
       max_end_date,
       is_merge,
       sum(is_merge) over (partition by hall_id order by start_date asc,end_date asc) as group_id
from (select hall_id,
             start_date,
             end_date,
             max_end_date,
             if(start_date <= max_end_date, 0, 1) as is_merge --0:合并，1:不合并
      from (select hall_id,
                   start_date,
                   end_date,
                   max(end_date)
                       over (partition by hall_id order by start_date asc,end_date asc rows between unbounded preceding and 1 preceding) as max_end_date
            from t1_hall_event) t) tt

执行结果

+----------+-------------+-------------+---------------+-----------+-----------+
| hall_id  | start_date  |  end_date   | max_end_date  | is_merge  | group_id  |
+----------+-------------+-------------+---------------+-----------+-----------+
| 1        | 2023-01-13  | 2023-01-20  | NULL          | 1         | 1         |
| 1        | 2023-01-14  | 2023-01-16  | 2023-01-20    | 0         | 1         |
| 1        | 2023-01-14  | 2023-01-17  | 2023-01-20    | 0         | 1         |
| 1        | 2023-01-18  | 2023-01-25  | 2023-01-20    | 0         | 1         |
| 1        | 2023-01-20  | 2023-01-26  | 2023-01-25    | 0         | 1         |
| 2        | 2022-12-09  | 2022-12-23  | NULL          | 1         | 1         |
| 2        | 2022-12-13  | 2022-12-17  | 2022-12-23    | 0         | 1         |
| 2        | 2022-12-20  | 2022-12-24  | 2022-12-23    | 0         | 1         |
| 2        | 2022-12-25  | 2022-12-30  | 2022-12-24    | 1         | 2         |
| 3        | 2022-12-01  | 2023-01-30  | NULL          | 1         | 1         |
+----------+-------------+-------------+---------------+-----------+-----------+

4.完成合并，得到最终结果

取每个组内的start_day 的最小值作为活动开始日期，end_day的最大值作为活动结束日期，得到最终结果。注意分组条件为hall_id+group_id

执行SQL

select hall_id,
       min(start_date) as start_date,
       max(end_date)   as end_date
from (select hall_id,
             start_date,
             end_date,
             max_end_date,
             is_merge,
             sum(is_merge) over (partition by hall_id order by start_date asc,end_date asc) as group_id
      from (select hall_id,
                   start_date,
                   end_date,
                   max_end_date,
                   if(start_date <= max_end_date, 0, 1) as is_merge --0:合并，1:不合并
            from (select hall_id,
                         start_date,
                         end_date,
                         max(end_date)
                             over (partition by hall_id order by start_date asc,end_date asc rows between unbounded preceding and 1 preceding) as max_end_date
                  from t1_hall_event) t) tt) ttt
group by hall_id, group_id --注意这里的分组，有group_id

执行结果

+----------+-------------+-------------+
| hall_id  | start_date  |  end_date   |
+----------+-------------+-------------+
| 1        | 2023-01-13  | 2023-01-26  |
| 2        | 2022-12-09  | 2022-12-24  |
| 2        | 2022-12-25  | 2022-12-30  |
| 3        | 2022-12-01  | 2023-01-30  |
+----------+-------------+-------------+

四、建表语句和数据插入

--建表语句
CREATE TABLE IF NOT EXISTS t1_hall_event (
    hall_id STRING, --大厅ID
    start_date STRING, -- 营销活动开始日期
    end_date STRING -- 营销活动结束日期
);
--数据插入
insert into  t1_hall_event(hall_id, start_date, end_date) values
('1','2023-01-13','2023-01-20'),
('1','2023-01-14','2023-01-17'),
('1','2023-01-14','2023-01-16'),
('1','2023-01-18','2023-01-25'),
('1','2023-01-20','2023-01-26'),
('2','2022-12-09','2022-12-23'),
('2','2022-12-13','2022-12-17'),
('2','2022-12-20','2022-12-24'),
('2','2022-12-25','2022-12-30'),
('3','2022-12-01','2023-01-30');

本文同步在微信公众号”数据仓库技术“和个人博客”数据仓库技术 (opens in a new tab)“发表；

1.部门人员数据分析 1.计算每个用户的受欢迎程度