Number of Times a Driver Was a Passenger - Practice Coding Problems

Number of Times a Driver Was a Passenger - Problem

Database Medium

In a ride-sharing service, we have a database table Rides that tracks all completed trips. Each row represents a single ride with three key pieces of information:

ride_id - unique identifier for each ride
driver_id - ID of the person who drove
passenger_id - ID of the person who was the passenger

Your task: Some people in the system act as both drivers and passengers at different times. We need to find out for each person who has been a driver, how many times they have also been a passenger in other rides.

Output format: Return a result table with two columns:

driver_id - ID of each driver
cnt - number of times that driver was also a passenger

Note: If a driver has never been a passenger, they should still appear in the result with a count of 0.

Input & Output

example_1.sql — Basic Example

$ Input: Rides table: +----------+----------+--------------+ | ride_id | driver_id | passenger_id | +----------+----------+--------------+ | 1 | 10 | 20 | | 2 | 20 | 10 | | 3 | 30 | 40 | +----------+----------+--------------+

› Output: +----------+-----+ | driver_id | cnt | +----------+-----+ | 10 | 1 | | 20 | 1 | | 30 | 0 | +----------+-----+

💡 Note: Driver 10 was a passenger 1 time (in ride 2), driver 20 was a passenger 1 time (in ride 1), and driver 30 was never a passenger (cnt = 0).

example_2.sql — Multiple Passenger Rides

$ Input: Rides table: +----------+----------+--------------+ | ride_id | driver_id | passenger_id | +----------+----------+--------------+ | 1 | 10 | 20 | | 2 | 30 | 10 | | 3 | 30 | 20 | | 4 | 40 | 10 | +----------+----------+--------------+

› Output: +----------+-----+ | driver_id | cnt | +----------+-----+ | 10 | 0 | | 30 | 0 | | 40 | 0 | +----------+-----+

💡 Note: None of the drivers (10, 30, 40) appear as passengers in any ride, so all have cnt = 0.

example_3.sql — Same Person Multiple Roles

$ Input: Rides table: +----------+----------+--------------+ | ride_id | driver_id | passenger_id | +----------+----------+--------------+ | 1 | 10 | 20 | | 2 | 20 | 10 | | 3 | 10 | 30 | | 4 | 40 | 10 | +----------+----------+--------------+

› Output: +----------+-----+ | driver_id | cnt | +----------+-----+ | 10 | 2 | | 20 | 1 | | 40 | 0 | +----------+-----+

💡 Note: Driver 10 was a passenger 2 times (rides 2 and 4), driver 20 was a passenger 1 time (ride 1), and driver 40 was never a passenger.

Visualization

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Understanding the Visualization

Identify All Drivers

Extract unique driver IDs from all rides - these are our people of interest

Match with Passenger Records

For each driver, find all rides where they appear as a passenger

Count and Preserve Zeros

Count passenger appearances, ensuring drivers with zero passenger rides still appear in results

Key Takeaway

🎯 Key Insight: Use LEFT JOIN to preserve all drivers in the result, ensuring those who were never passengers still appear with count = 0

Time & Space Complexity

Time Complexity

⏱️

O(n log n)

Single pass through data with efficient join operations, typically O(n log n) due to internal sorting for GROUP BY

⚡ Linearithmic

Space Complexity

O(n)

Space for result set plus temporary space for join operations

⚡ Linearithmic Space

Constraints

1 ≤ rides.length ≤ 10⁴
1 ≤ ride_id, driver_id, passenger_id ≤ 10⁶
driver_id ≠ passenger_id for each ride
ride_id values are unique

Asked in

U Uber 45 L Lyft 30 a Amazon 25 D DoorDash 20

The optimal solution uses a LEFT JOIN to match all unique drivers with their appearances as passengers. The key insight is using LEFT JOIN to preserve all drivers (even those who were never passengers) and GROUP BY with COUNT to efficiently count passenger occurrences in a single query.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Subquery for Each Driver)	O(n²)	O(n)	For each driver, run a separate subquery to count their passenger appearances
Left Join (Optimal SQL Solution)	O(n log n)	O(n)	Use LEFT JOIN to match drivers with passenger records in a single efficient query

Brute Force (Subquery for Each Driver) — Algorithm Steps

Get all distinct driver IDs from the rides table
For each driver, run a subquery to count their appearances as passenger_id
Combine the results to show driver_id and their passenger count

Visualization

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Step-by-Step Walkthrough

Get Unique Drivers

Extract all distinct driver_id values

For Each Driver

Run subquery to count passenger appearances

Combine Results

Merge driver_id with their passenger counts

Code -

solution.c — C

// SQL solution using subquery approach
char* query = "\
SELECT \
    driver_id, \
    ( \
        SELECT COUNT(*) \
        FROM Rides r2 \
        WHERE r2.passenger_id = r1.driver_id \
    ) AS cnt \
FROM ( \
    SELECT DISTINCT driver_id \
    FROM Rides \
) r1 \
ORDER BY driver_id;"

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of the n unique drivers, we scan through n rides to count passenger appearances

⚠ Quadratic Growth

Space Complexity

O(n)

Space for storing the result set with one row per unique driver

⚡ Linearithmic Space

Constraints

1 ≤ rides.length ≤ 10⁴
1 ≤ ride_id, driver_id, passenger_id ≤ 10⁶
driver_id ≠ passenger_id for each ride
ride_id values are unique

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Subquery for Each Driver) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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