Consecutive available seats ii

# 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.
# 
# Write a solution to find the length of longest consecutive sequence of available seats in the cinema.
# 
# Note:
# 
# There will always be at most one longest consecutive sequence.
# 
# If there are multiple consecutive sequences with the same length, include all of them in the output.
# 
# Return the result table ordered by first_seat_id in ascending order.
# 
# 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:
# +-----------------+----------------+-----------------------+
# | first_seat_id   | last_seat_id   | consecutive_seats_len |
# +-----------------+----------------+-----------------------+
# | 3               | 5              | 3                     |
# +-----------------+----------------+-----------------------+
# Explanation:
# Longest consecutive sequence of available seats starts from seat 3 and ends at seat 5 with a length of 3.
# Output table is ordered by first_seat_id in ascending order.

import pandas as pd

def consecutive_available_seats(cinema: pd.DataFrame) -> pd.DataFrame:
  cinema = cinema.sort_values('seat_id').reset_index(drop=True)
  free = cinema[cinema['free'] == 1]['seat_id'].tolist()
  if not free:
    return pd.DataFrame(columns=['first_seat_id', 'last_seat_id', 'consecutive_seats_len'])
  groups = []
  start = free[0]
  prev = free[0]
  for s in free[1:]:
    if s == prev + 1:
      prev = s
    else:
      groups.append((start, prev, prev - start + 1))
      start = s
      prev = s
  groups.append((start, prev, prev - start + 1))
  max_len = max(g[2] for g in groups)
  result = [(g[0], g[1], g[2]) for g in groups if g[2] == max_len]
  return pd.DataFrame(result, columns=['first_seat_id', 'last_seat_id', 'consecutive_seats_len']).sort_values('first_seat_id').reset_index(drop=True)