SQL29 连续两次作答试卷的最大时间窗

1、函数

lead (str1,number) over(partition by str2 order by str3)

题目

现有试卷作答记录表exam_record（uid用户ID, exam_id试卷ID, start_time开始作答时间, submit_time交卷时间, score得分）：

请计算在2021年至少有两天作答过试卷的人中，计算该年连续两次作答试卷的最大时间窗days_window，那么根据该年的历史规律他在days_window天里平均会做多少套试卷，按最大时间窗和平均做答试卷套数倒序排序。由示例数据结果输出如下：

select uid
    ,days_window
    ,round(days_window*exam_count/total_days,2) as avg_exam_cnt
from
(select 
    uid
    ,max(datediff(next_time,start_time))+1 days_window
    ,count(start_time) as exam_count
    ,datediff(max(start_time),min(start_time))+1 total_days
from
(select uid
    ,start_time
    ,lead(start_time,1) over(partition by uid order by start_time asc) as next_time
 from exam_record
where year(start_time)=2021)a 
group by uid
having datediff(max(start_time),min(start_time))+1>1)b
order by days_window desc,avg_exam_cnt desc