Solution Explanation:

This problem requires generating a report summarizing Hopper company statistics for each month of 2020. The report needs two key metrics: the number of active drivers at the end of each month and the number of accepted rides during that month. We'll achieve this using SQL queries.

Approach:

The solution uses a combination of Common Table Expressions (CTEs) to break down the problem into manageable parts.

1. Months CTE: This CTE generates a series of numbers from 1 to 12, representing the months of the year. This simplifies iterating through each month.

2. Ride CTE: This CTE calculates the number of accepted rides for each month in 2020 by joining Rides and AcceptedRides tables, filtering for the year 2020, and grouping by the month.

3. Main Query: The main query joins the Months CTE with the Drivers table to determine the number of active drivers at the end of each month. It uses a LEFT JOIN with the Ride CTE to incorporate the accepted rides count for each month. The IFNULL function handles cases where a month has no accepted rides, returning 0 instead of NULL. The final result is grouped by month and ordered accordingly.

Time Complexity Analysis:

The time complexity is dominated by the joins and aggregations in the query. The Months CTE takes O(1) time as it's a fixed-size sequence. The Ride CTE's complexity is determined by the joins and GROUP BY operations, which are typically O(N log N) or O(N) depending on the database engine's optimization strategies (N being the number of rows in the joined tables). The main query's join and GROUP BY operations will have a similar complexity. Overall, the dominant complexity is likely O(N log N) or O(N), where N represents the number of rows in the largest of the three tables (Drivers, Rides, and AcceptedRides).

Space Complexity Analysis:

The space complexity is determined by the intermediate CTEs and the final result set. The Months CTE uses constant space (O(1)). The Ride CTE's space depends on the size of the intermediate result after the join and grouping, potentially O(N) in the worst case (N being the number of rows after joining Rides and AcceptedRides). The final result set is of size O(12), which is constant as it always has 12 rows (one for each month). Therefore, the overall space complexity is dominated by the Ride CTE, making it O(N) in the worst case.

Code in MySQL:

The MySQL code provided in the solution section is already well-structured and efficient. It directly implements the approach described above.

This detailed explanation provides a clear understanding of the solution's logic, efficiency, and performance characteristics. The code itself is already well-commented and optimized for MySQL.