Solution Explanation

This problem requires finding the number of times each driver was a passenger in the Rides table. The solution uses a SQL query to achieve this.

Approach

The solution employs a LEFT JOIN to combine data from two derived tables. Here's a breakdown:

1. Finding Distinct Drivers: A Common Table Expression (CTE) named T is created to select all unique driver_id values from the Rides table. This ensures we consider every driver.

2. Joining with Rides Table: The CTE T is then LEFT JOINed with the Rides table using the condition t.driver_id = r.passenger_id. This joins each driver with all rows where that driver's ID appears in the passenger_id column. A LEFT JOIN ensures that all drivers are included in the result, even if they were never passengers (in which case the passenger_id will be NULL).

3. Counting Passenger Occurrences: The COUNT(passenger_id) function counts the number of times each driver_id appears as a passenger_id. COUNT() ignores NULL values, correctly giving a count of 0 for drivers who were never passengers.

4. Grouping and Result: The GROUP BY 1 clause groups the results by the first column ( driver_id), resulting in one row per driver with the corresponding passenger count. The AS cnt part assigns an alias to the count column.

MySQL Code Explained

WITH T AS (SELECT DISTINCT driver_id FROM Rides)
SELECT t.driver_id, COUNT(passenger_id) AS cnt
FROM
    T AS t
    LEFT JOIN Rides AS r ON t.driver_id = r.passenger_id
GROUP BY 1;

• WITH T AS (SELECT DISTINCT driver_id FROM Rides): This defines the CTE T, which gets the unique driver IDs.
• SELECT t.driver_id, COUNT(passenger_id) AS cnt: Selects the driver ID and the count of times they were a passenger, aliased as cnt.
• FROM T AS t LEFT JOIN Rides AS r ON t.driver_id = r.passenger_id: Performs a LEFT JOIN between T and Rides, matching drivers to their passenger records.
• GROUP BY 1: Groups the results by the first column (driver_id), aggregating the passenger counts for each driver.

Time Complexity Analysis

The time complexity of this SQL query depends on the database engine's optimization strategies. However, a reasonable estimate is:

• SELECT DISTINCT driver_id FROM Rides: O(N log N) or O(N) depending on the database's indexing and optimization, where N is the number of rows in the Rides table.
• LEFT JOIN: In the worst case, O(M*N), where M is the number of distinct drivers and N is the number of rows in Rides. However, with appropriate indexing, this can be optimized to closer to O(N).
• GROUP BY and COUNT: O(N) after the join.

Therefore, the overall time complexity is likely to be dominated by the join operation, potentially ranging from O(N) with good indexing to O(M*N) in the worst case without proper indexes. The exact complexity heavily depends on the database system's query optimizer and the presence of appropriate indexes on driver_id and passenger_id columns.

Space Complexity

The space complexity is primarily determined by the size of the intermediate result set after the LEFT JOIN, which in the worst case is proportional to the number of rows in the Rides table. So, the space complexity is O(N), again where N is the number of rows in Rides. The CTE adds minimal overhead.