Number of times a driver was a passenger

# 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.

import pandas as pd

def driver_passenger(rides: pd.DataFrame) -> pd.DataFrame:
  drivers = rides[['driver_id']].drop_duplicates()
  passenger_cnt = rides.groupby('passenger_id').size().reset_index(name='cnt')
  result = drivers.merge(passenger_cnt, left_on='driver_id', right_on='passenger_id', how='left')
  result['cnt'] = result['cnt'].fillna(0).astype(int)
  return result[['driver_id', 'cnt']]