Consecutive available seats

# Table: Cinema
# 
# +-------------+------+
# | Column Name | Type |
# +-------------+------+
# | seat_id     | int  |
# | free        | bool |
# +-------------+------+
# seat_id is an auto-increment column for this table.
# Each row of this table indicates whether the ith seat is free or not. 1 means free while 0 means occupied.
# 
#  
# 
# Find all the consecutive available seats in the cinema.
# 
# Return the result table ordered by seat_id in ascending order.
# 
# The test cases are generated so that more than two seats are consecutively available.
# 
# The result format is in the following example.
#
# Example 1:
# Input:
# Cinema table:
# +---------+------+
# | seat_id | free |
# +---------+------+
# | 1       | 1    |
# | 2       | 0    |
# | 3       | 1    |
# | 4       | 1    |
# | 5       | 1    |
# +---------+------+
# Output:
# +---------+
# | seat_id |
# +---------+
# | 3       |
# | 4       |
# | 5       |
# +---------+

import pandas as pd

def consecutive_available_seats(cinema: pd.DataFrame) -> pd.DataFrame:
  free = cinema[cinema['free'] == 1]
  prev_free = free['seat_id'] - 1
  next_free = free['seat_id'] + 1
  free_set = set(free['seat_id'])
  mask = free['seat_id'].apply(lambda x: (x - 1) in free_set or (x + 1) in free_set)
  result = free.loc[mask, ['seat_id']].sort_values('seat_id')
  return result