User Activity for the Past 30 Days II

Problem

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 average number of sessions per user for a period of 30 days ending 2019-07-27 inclusively, rounded to 2 decimal places. The sessions we want to count for a user are those with at least one activity in that time period.

The result format is in the following example.

Examples

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   |
| 3       | 5          | 2019-07-21    | open_session  |
| 3       | 5          | 2019-07-21    | scroll_down   |
| 3       | 5          | 2019-07-21    | end_session   |
| 4       | 3          | 2019-06-25    | open_session  |
| 4       | 3          | 2019-06-25    | end_session   |
+---------+------------+---------------+---------------+
Output: 
+---------------------------+ 
| average_sessions_per_user |
+---------------------------+ 
| 1.33                      |
+---------------------------+
Explanation: User 1 and 2 each had 1 session in the past 30 days while user 3 had 2 sessions so the average is (1 + 1 + 2) / 3 = 1.33.

Solution

Method 1 - Using CTEs

Code

Sql

WITH myquery AS 
       (
       SELECT user_id, 
              session_id
       FROM Activity
       WHERE activity_date BETWEEN '2019-06-28' AND '2019-07-27'
       GROUP BY user_id, session_id
       )
SELECT 
    IFNULL(ROUND(COUNT(*) / COUNT(DISTINCT user_id), 2), 0.00) average_sessions_per_user
FROM myquery;

Again, not the best solution.

Method 2 - Using Mathematical functions

Code

Sql

Here's a better example:

SELECT ifnull(ROUND(COUNT(DISTINCT session_id) / COUNT(DISTINCT user_id), 2), 0.00) 
AS average_sessions_per_user
FROM Activity 
WHERE activity_date >= '2019-06-28' and activity_date <= '2019-07-27'; 

This example is better, although maybe it could have used BETWEEN. It's more simple than my solution, simply dividing all distinct Session IDs and dividing by all distinct User IDs.

I have a strange habit of over complicating things.