# $603. 连续空余座位
# https://leetcode-cn.com/problems/consecutive-available-seats/

# SQL架构
Create table If Not Exists Cinema (seat_id int primary key auto_increment, free bool);
Truncate table Cinema;
insert into Cinema (seat_id, free) values ('1', '1');
insert into Cinema (seat_id, free) values ('2', '0');
insert into Cinema (seat_id, free) values ('3', '1');
insert into Cinema (seat_id, free) values ('4', '1');
insert into Cinema (seat_id, free) values ('5', '1');

# Write your MySQL query statement below
select
    distinct c1.seat_id as seat_id
from
    Cinema c1
    join Cinema c2 on abs(c1.seat_id - c2.seat_id) = 1
    and c1.free = true
    and c2.free = true
order by
    c1.seat_id;

# clean-up
drop table Cinema;