Minimum number of Meeting Rooms required to Accomodate all Meetings in MySQL - mysql

Question

I have the following columns in a table called meetings: meeting_id - int, start_time - time, end_time - time. Assuming that this table has data for one calendar day only, how many minimum number of rooms do I need to accomodate all the meetings. Room size/number of people attending the meetings don't matter.
Here's the solution:
select * from
(select t.start_time,
t.end_time,
count(*) - 1 overlapping_meetings,
count(*) minimum_rooms_required,
group_concat(distinct concat(y.start_time,' to ',t.end_time)
separator ' // ') meeting_details from
(select 1 meeting_id, '08:00' start_time, '09:15' end_time union all
select 2, '13:20', '15:20' union all
select 3, '10:00', '14:00' union all
select 4, '13:55', '16:25' union all
select 5, '14:00', '17:45' union all
select 6, '14:05', '17:45') t left join
(select 1 meeting_id, '08:00' start_time, '09:15' end_time union all
select 2, '13:20', '15:20' union all
select 3, '10:00', '14:00' union all
select 4, '13:55', '16:25' union all
select 5, '14:00', '17:45' union all
select 6, '14:05', '17:45') y
on t.start_time between y.start_time and y.end_time
group by start_time, end_time) z;
My question - is there anything wrong with this answer? Even if there's nothing wrong with this, can someone share a better answer?

Answer 1

Let's say you have a table called 'meeting' like this -
Then You can use this query to get the minimum number of meeting Rooms required to accommodate all Meetings.
select max(minimum_rooms_required)
from (select count(*) minimum_rooms_required
from meetings t
left join meetings y on t.start_time >= y.start_time and t.start_time < y.end_time group by t.id
) z;
This looks clearer and simple and works fine.

Answer 2

Meetings can "overlap". So, GROUP BY start_time, end_time can't figure this out.
Not every algorithm can be done in SQL. Or, at least, it may be grossly inefficient.
I would use a real programming language for the computation, leaving the database for what it is good at -- being a data repository.
Build a array of 1440 (minutes in a day) entries; initialize to 0.
Foreach meeting:
Foreach minute in the meeting (excluding last minute):
increment element in array.
Find the largest element in the array -- the number of rooms needed.

Answer 3

CREATE TABLE [dbo].[Meetings](
[id] [int] NOT NULL,
[Starttime] [time](7) NOT NULL,
[EndTime] [time](7) NOT NULL) ON [PRIMARY] )GO
sample data set:
INSERT INTO Meetings VALUES (1,'8:00','09:00')
INSERT INTO Meetings VALUES (2,'8:00','10:00')
INSERT INTO Meetings VALUES (3,'10:00','11:00')
INSERT INTO Meetings VALUES (4,'11:00','12:00')
INSERT INTO Meetings VALUES (5,'11:00','13:00')
INSERT INTO Meetings VALUES (6,'13:00','14:00')
INSERT INTO Meetings VALUES (7,'13:00','15:00')
To Find Minimum number of rooms required run the below query:
create table #TempMeeting
(
id int,Starttime time,EndTime time,MeetingRoomNo int,Rownumber int
)
insert into #TempMeeting select id, Starttime,EndTime,0 as MeetingRoomNo,ROW_NUMBER()
over (order by starttime asc) as Rownumber from Meetings
declare #RowCounter int
select top 1 #RowCounter=Rownumber from #TempMeeting order by Rownumber
WHILE #RowCounter<=(Select count(*) from #TempMeeting)
BEGIN
update #TempMeeting set MeetingRoomNo=1
where Rownumber=(select top 1 Rownumber from #TempMeeting where
Rownumber>#RowCounter and Starttime>=(select top 1 EndTime from #TempMeeting
where Rownumber=#RowCounter)and MeetingRoomNo=0)set #RowCounter=#RowCounter+1
END
select count(*) from #TempMeeting where MeetingRoomNo=0

Answer 4

Consider a table meetings with columns id, start_time and end_time. Then the following query should give correct answer.
with mod_meetings as (select id, to_timestamp(start_time, 'HH24:MI')::TIME as start_time,
to_timestamp(end_time, 'HH24:MI')::TIME as end_time from meetings)
select CASE when max(a_cnt)>1 then max(a_cnt)+1
when max(a_cnt)=1 and max(b_cnt)=1 then 2 else 1 end as rooms
from
(select count(*) as a_cnt, a.id, count(b.id) as b_cnt from mod_meetings a left join mod_meetings b
on a.start_time>b.start_time and a.start_time<b.end_time group by a.id) join_table;

Answer 5

Sample DATA:
DROP TABLE IF EXISTS meeting;
CREATE TABLE "meeting" (
"meeting_id" INTEGER NOT NULL UNIQUE,
"start_time" TEXT NOT NULL,
"end_time" TEXT NOT NULL,
PRIMARY KEY("meeting_id")
);
INSERT INTO meeting values (1,'08:00','14:00');
INSERT INTO meeting values (2,'09:00','10:30');
INSERT INTO meeting values (3,'11:00','12:00');
INSERT INTO meeting values (4,'12:00','13:00');
INSERT INTO meeting values (5,'10:15','11:00');
INSERT INTO meeting values (6,'12:00','13:00');
INSERT INTO meeting values (7,'10:00','10:30');
INSERT INTO meeting values (8,'11:00','13:00');
INSERT INTO meeting values (9,'11:00','14:00');
INSERT INTO meeting values (10,'12:00','14:00');
INSERT INTO meeting values (11,'10:00','14:00');
INSERT INTO meeting values (12,'12:00','14:00');
INSERT INTO meeting values (13,'10:00','14:00');
INSERT INTO meeting values (14,'13:00','14:00');
Solution:
DROP VIEW IF EXISTS Final;
CREATE VIEW Final AS SELECT time, group_concat(event), sum(num) num from (
select start_time time, 's' event, 1 num from meeting
union all
select end_time time, 'e' event, -1 num from meeting)
group by 1
order by 1;
select max(room) AS Min_Rooms_Required FROM (
select
a.time,
sum(b.num) as room
from
Final a
, Final b
where a.time >= b.time
group by a.time
order by a.time
);

Answer 6

Here's the explanation to gashu's nicely working code (or otherwise a non-code explanation of how to solve it with any language).
Firstly, if the variable 'minimum_rooms_required' would be renamed to 'overlap' it would make the whole thing much easier to understand. Because for each of the start or end times we want to know the numbers of overlapping ongoing meetings. When we found the maximum, this means there's no way of getting around with less than the overlapping amount, because well they overlap.
By the way, I think there might be a mistake in the code. It should check for t.start_time or t.end_time between y.start_time and y.end_time. Counterexample: meeting 1 starts at 8:00, ends at 11:00 and meeting 2 starts at 10:00, ends at 12:00.
(I'd post it as a comment to the gashu's answerbut I don't have enough reputation)

Answer 7

I'd go for Lead() analytic function
select
sum(needs_room_ind) as min_rooms
from (
select
id,
start_time,
end_time,
case when lead(start_time,1) over (order by start_time asc) between start_time
and end_time then 1 else 0 end as needs_room_ind
from
meetings
) a

Answer 8

IMO, I wanna to take the difference between how many meeting are started and ended at the same time when each meeting_id is started (assuming meeting starts and ends on time)
my code was just like this :
with alpha as
(
select a.meeting_id,a.start_time,
count(distinct b.meeting_id) ttl_meeting_start_before,
count(distinct c.meeting_id) ttl_meeting_end_before
from meeting a
left join
(
select meeting_id,start_time from meeting
) b
on a.start_time > b.start_time
left join
(
select meeting_id,end_time from meeting
) c
on a.start_time > c.end_time
group by a.meeting_id,a.start_time
)
select max(ttl_meeting_start_before-ttl_meeting_end_before) max_meeting_room
from alpha