Meeting Rooms | InterviewBit | Solution Explained
I explain the solution to Meeting Rooms on InterviewBit using visuals in simple words.
0. Meeting Rooms: Problem Discussion
0.0. IO Format
- a list of intervals, where each interval is given by [s,e] which are the start and the end times of the intervals
- find the least number of meeting rooms required to conduct all the meetings
- 1 <= N <= 1e5
- 1 <= s <= e <= 1e9
Visually speaking, we can see that the maximum number of meeting rooms we require is 2, at two different places of times.
1. Meeting Rooms: Observations & Logic
The naïve approach suggests that we can go over all the times one by one, and check the number of meetings going on at each point of time.
That is, for each time = t, we go over all the N elements to find the count of meetings going on simultaneously.
The only issue is that the time complexity become
O(TN), and according to the constraints of the problem, that is a sure fire Time Limit Exceeded (TLE).
All right, so what do we do instead?
1.1. Only important times matter
The key is to realize, looking at the example shown above, is that most of the time, the number of meeting rooms is not going to change.
What I mean is that for times 0, 1, 2, 3, 4, the number of meeting rooms do not change. Then, at time = 5, the number of required meeting rooms becomes 2.
But, then again, the required rooms stays at 2 for t = 6, 7, 8, 9.
All this means is that we are doing a lot of useless computation. Thus, the realization, only the important times matter – when there is a change.
1.2. Changing times
To formalize, we want to look at what happens for each start time, s, and what happens to each end time, e.
At each start time, for any interval, the number of required meeting rooms increase by 1. Similarly, for each end time, for all intervals, the required rooms decreases by 1.
The only thing left to do is to look at this in terms of code. We do two things:
- convert these
t: changevalues to a tuple,
(t, change)– as simple as that
- next we sort the list of these tuples – to know what comes first and what comes later.
That’s it! Let’s implement it.
2. Meeting Rooms: Optimized Implementation
class Solution: # @param A : list of list of integers # @return an integer def solve(self, A): data =  # T: O(N), S: O(N) for s, e in A: data.append((s, +1)) data.append((e, -1)) data.sort() # T: O(NlogN), S: O(algo) curr = 0 ans = 0 for _, v in data: # O(N) curr += v ans = max(ans, curr) return ans
2.1. Complexity Analysis
O(NlogN), for sorting
O(NlogN), and worst case if we use merge sort
O(NlogN). There are other algorithms that different languages use that can do better.
June 13, 2022
June 13, 2022