Solution Explanation

This problem requires calculating the average number of sessions per user within the last 30 days leading up to and including 2019-07-27. The provided solutions use SQL queries to achieve this. Let's break down each approach:

Solution 1: Using a Common Table Expression (CTE)

This solution employs a CTE (T) to first group the data by user_id and count the distinct session_id for each user within the specified timeframe. The CTE efficiently preprocesses the data before calculating the average.

MySQL Code:

WITH
    T AS (
        SELECT
            COUNT(DISTINCT session_id) AS sessions
        FROM Activity
        WHERE activity_date <= '2019-07-27' AND DATEDIFF('2019-07-27', activity_date) < 30
        GROUP BY user_id
    )
SELECT IFNULL(ROUND(AVG(sessions), 2), 0) AS average_sessions_per_user
FROM T;

Explanation:

1. WITH T AS (...): This defines a CTE named T.
2. SELECT COUNT(DISTINCT session_id) AS sessions: This counts the unique sessions for each user. DISTINCT is crucial to avoid overcounting sessions.
3. FROM Activity: This specifies the table to query.
4. WHERE activity_date <= '2019-07-27' AND DATEDIFF('2019-07-27', activity_date) < 30: This filters the data to include only activities within the 30-day window. DATEDIFF calculates the difference in days between the activity date and 2019-07-27.
5. GROUP BY user_id: This groups the results by user ID, so COUNT(DISTINCT session_id) is calculated for each user.
6. SELECT IFNULL(ROUND(AVG(sessions), 2), 0) AS average_sessions_per_user: This calculates the average number of sessions (AVG(sessions)) from the CTE T. ROUND(..., 2) rounds the average to two decimal places. IFNULL(..., 0) handles the case where no users have activity in the specified time frame, returning 0 instead of NULL.

Solution 2: Direct Calculation

This solution is more concise, calculating the average directly without using a CTE.

MySQL Code:

SELECT
    IFNULL(
        ROUND(COUNT(DISTINCT session_id) / COUNT(DISTINCT user_id), 2),
        0
    ) AS average_sessions_per_user
FROM Activity
WHERE DATEDIFF('2019-07-27', activity_date) < 30;

Explanation:

1. SELECT IFNULL(ROUND(COUNT(DISTINCT session_id) / COUNT(DISTINCT user_id), 2), 0): This performs the calculation directly. It counts distinct sessions, divides by the count of distinct users, rounds to two decimal places, and handles the NULL case using IFNULL.
2. FROM Activity: Specifies the table.
3. WHERE DATEDIFF('2019-07-27', activity_date) < 30: Filters the data to the 30-day period.

Time Complexity Analysis

Both solutions have a similar time complexity. The database system will handle the grouping and aggregation efficiently. The dominant factor is the number of rows in the Activity table within the 30-day period. The time complexity can be considered approximately O(N), where N is the number of relevant rows in the Activity table, as the database engine likely uses optimized algorithms for grouping and aggregation. The space complexity depends on the size of the intermediate results (the CTE in Solution 1 and the distinct counts in Solution 2) but is generally proportional to the number of distinct users and sessions, making the space complexity also roughly O(N) in the worst case. In practice, the actual performance will depend on the specific database engine's optimizations and indexing.