Is Number of Times a Driver Was a Passenger easy or hard?

The problem is rated Medium but the SQL logic is straightforward if you understand LEFT JOIN and GROUP BY. The main challenge is ensuring drivers with zero passenger appearances are still included in the result.

Number of Times a Driver Was a Passenger Python/Java solution

This problem is categorized under SQL/Database on LeetCode, so the expected solution is written in SQL (commonly MySQL). Python or Java are not typically used unless you manually process the table rows, which would be less efficient than a database aggregation query.

How to solve Number of Times a Driver Was a Passenger in O(n)?

Select distinct driver_id values first, then LEFT JOIN the Rides table where passenger_id equals that driver_id. Group the result by driver_id and count matching ride_id values. This ensures each ride is processed once while still returning drivers with zero passenger rides.

What is the best approach for Number of Times a Driver Was a Passenger?

The most reliable solution builds a list of distinct drivers and LEFT JOINs it with the Rides table on passenger_id. Aggregating with COUNT(ride_id) returns how many times each driver appears as a passenger. This approach runs in O(n) time with a single scan and correctly includes drivers who were never passengers.

Is Number of Times a Driver Was a Passenger asked at Google/Amazon/Meta?

Problems like this commonly appear in SQL interview rounds at companies such as Amazon, Meta, and data-focused teams at Google. They test understanding of joins, aggregation, and handling missing matches using LEFT JOIN.

What data structure is used in Number of Times a Driver Was a Passenger?

The solution relies on relational database operations rather than traditional data structures. Key concepts include SQL tables, joins, grouping, and aggregation functions like COUNT(). Internally the database engine may use hash joins or indexes for efficiency.

What is the time complexity of Number of Times a Driver Was a Passenger?

The optimal SQL solution runs in O(n) time where n is the number of rows in the Rides table. The database scans the table once for the join and aggregation. Memory usage is O(d) for storing the distinct set of drivers.

Description Solution Video 8:15

2238. Number of Times a Driver Was a Passenger

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Table: Rides

+--------------+------+
| Column Name  | Type |
+--------------+------+
| ride_id      | int  |
| driver_id    | int  |
| passenger_id | int  |
+--------------+------+
ride_id is the column with unique values for this table.
Each row of this table contains the ID of the driver and the ID of the passenger that rode in ride_id.
Note that driver_id != passenger_id.

Write a solution to report the ID of each driver and the number of times they were a passenger.

Return the result table in any order.

The result format is in the following example.

Example 1:

Input: 
Rides table:
+---------+-----------+--------------+
| ride_id | driver_id | passenger_id |
+---------+-----------+--------------+
| 1       | 7         | 1            |
| 2       | 7         | 2            |
| 3       | 11        | 1            |
| 4       | 11        | 7            |
| 5       | 11        | 7            |
| 6       | 11        | 3            |
+---------+-----------+--------------+
Output: 
+-----------+-----+
| driver_id | cnt |
+-----------+-----+
| 7         | 2   |
| 11        | 0   |
+-----------+-----+
Explanation: 
There are two drivers in all the given rides: 7 and 11.
The driver with ID = 7 was a passenger two times.
The driver with ID = 11 was never a passenger.

Asked at 1 company

Amazon

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Database

8:15

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2238. Number of Times a Driver Was a Passenger

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