Skip to content

Update Consecutive available seats.sql #8

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Merged
mrinal1704 merged 1 commit into from
Mar 20, 2026
+21 −1
Merged

Update Consecutive available seats.sql #8

Changes from all commits
Commits
Show all changes
1 commit
Select commit
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
22 changes: 21 additions & 1 deletion Easy/Consecutive available seats.sql
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -22,6 +22,26 @@
-- The seat_id is an auto increment int, and free is bool ('1' means free, and '0' means occupied.).
-- Consecutive available seats are more than 2(inclusive) seats consecutively available.

--My Solution
//METHOD 1
SELECT DISTINCT c1.seat_id
FROM cinema c1
JOIN cinema c2 ON ABS(c1.seat_id - c2.seat_id) = 1
Where c1.free = 1 AND c2.free = 1

//METHOD 2
SELECT c1.seat_id
FROM cinema c1
JOIN cinema c2 on c1.seat_id = c2.seat_id - 1
where c1.free = 1 AND c2.free = 1

UNION

SELECT c2.seat_id
FROM cinema c1
JOIN cinema c2 on c1.seat_id = c2.seat_id + 1
where c1.free = 1 AND c2.free = 1

-- Solution
Select seat_id
from(
Expand All @@ -30,4 +50,4 @@ lead(free,1) over() as next,
lag(free,1) over() as prev
from cinema) a
where a.free=True and (next = True or prev=True)
order by seat_id
order by seat_id