[SQL문제풀기] User Activity for the Past 30 Days I

Can you solve this real interview question? User Activity for the Past 30 Days I - Table: Activity +---------------+---------+ | Column Name | Type | +---------------+---------+ | user_id | int | | session_id | int | | activity_date | date | | activity_type | enum | +---------------+---------+ This table may have duplicate rows. The activity_type column is an ENUM (category) of type ('open_session', 'end_session', 'scroll_down', 'send_message'). The table shows the user activities for a social media website. Note that each session belongs to exactly one user.   Write a solution to find the daily active user count for a period of 30 days ending 2019-07-27 inclusively. A user was active on someday if they made at least one activity on that day. Return the result table in any order. The result format is in the following example. Note: Any activity from ('open_session', 'end_session', 'scroll_down', 'send_message') will be considered valid activity for a user to be considered active on a day.   Example 1: Input: Activity table: +---------+------------+---------------+---------------+ | user_id | session_id | activity_date | activity_type | +---------+------------+---------------+---------------+ | 1 | 1 | 2019-07-20 | open_session | | 1 | 1 | 2019-07-20 | scroll_down | | 1 | 1 | 2019-07-20 | end_session | | 2 | 4 | 2019-07-20 | open_session | | 2 | 4 | 2019-07-21 | send_message | | 2 | 4 | 2019-07-21 | end_session | | 3 | 2 | 2019-07-21 | open_session | | 3 | 2 | 2019-07-21 | send_message | | 3 | 2 | 2019-07-21 | end_session | | 4 | 3 | 2019-06-25 | open_session | | 4 | 3 | 2019-06-25 | end_session | +---------+------------+---------------+---------------+ Output: +------------+--------------+ | day | active_users | +------------+--------------+ | 2019-07-20 | 2 | | 2019-07-21 | 2 | +------------+--------------+ Explanation: Note that we do not care about days with zero active users.