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1709. Biggest Window Between Visits
Description
Table: UserVisits
+-------------+------+ | Column Name | Type | +-------------+------+ | user_id | int | | visit_date | date | +-------------+------+ This table does not have a primary key, it might contain duplicate rows. This table contains logs of the dates that users visited a certain retailer.
Assume today's date is '2021-1-1'.
Write a solution that will, for each user_id, find out the largest window of days between each visit and the one right after it (or today if you are considering the last visit).
Return the result table ordered by user_id.
The query result format is in the following example.
Example 1:
Input:
UserVisits table:
+---------+------------+
| user_id | visit_date |
+---------+------------+
| 1 | 2020-11-28 |
| 1 | 2020-10-20 |
| 1 | 2020-12-3 |
| 2 | 2020-10-5 |
| 2 | 2020-12-9 |
| 3 | 2020-11-11 |
+---------+------------+
Output:
+---------+---------------+
| user_id | biggest_window|
+---------+---------------+
| 1 | 39 |
| 2 | 65 |
| 3 | 51 |
+---------+---------------+
Explanation:
For the first user, the windows in question are between dates:
- 2020-10-20 and 2020-11-28 with a total of 39 days.
- 2020-11-28 and 2020-12-3 with a total of 5 days.
- 2020-12-3 and 2021-1-1 with a total of 29 days.
Making the biggest window the one with 39 days.
For the second user, the windows in question are between dates:
- 2020-10-5 and 2020-12-9 with a total of 65 days.
- 2020-12-9 and 2021-1-1 with a total of 23 days.
Making the biggest window the one with 65 days.
For the third user, the only window in question is between dates 2020-11-11 and 2021-1-1 with a total of 51 days.
Solutions
Solution 1: Window Function
We can use the window function LEAD to obtain the date of the next visit for each user (if the date of the next visit does not exist, it is considered as 2021-1-1), and then use the DATEDIFF function to calculate the number of days between two visits. Finally, we can take the maximum value of the number of days between visits for each user.
-
# Write your MySQL query statement below WITH T AS ( SELECT user_id, DATEDIFF( LEAD(visit_date, 1, '2021-1-1') OVER ( PARTITION BY user_id ORDER BY visit_date ), visit_date ) AS diff FROM UserVisits ) SELECT user_id, MAX(diff) AS biggest_window FROM T GROUP BY 1 ORDER BY 1;