leetcode - 253. Meeting Rooms II

KpLn_HJL2023-10-12 9:03

Description

Given an array of meeting time intervals intervals where intervals $i$ = $starti, endi$ , return the minimum number of conference rooms required.

Example 1:

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Input: intervals = [[0,30],[5,10],[15,20]]
Output: 2

Example 2:

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Input: intervals = [[7,10],[2,4]]
Output: 1

Constraints:

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1 <= intervals.length <= 10^4
0 <= starti < endi <= 10^6

Solution

Heap

Solved after hints...

Think about how we would approach this problem in a very simplistic way. We will allocate rooms to meetings that occur earlier in the day v/s the ones that occur later on, right?

If you've figured out that we have to sort the meetings by their start time, the next thing to think about is how do we do the allocation?

There are two scenarios possible here for any meeting. Either there is no meeting room available and a new one has to be allocated, or a meeting room has freed up and this meeting can take place there.

So use a min-heap to store the ending time, every time we visit a new interval, compare the start time with the earliest ending time. If the start time begins later than the earliest ending time, then we could free up the room and allocate the room to the new interval. Otherwise we need to assign a new room for the new interval.

Time complexity: o ( n log ⁡ n ) o(n\log n ) o(nlogn)

Space complexity: o ( n ) o(n) o(n)

Sort + sweep

For all start, +1 at the point, and -1 for all ending points. Then sweep through all the points.

Time complexity: o ( n log ⁡ n ) o(n\log n) o(nlogn)

Space complexity: o ( 1 ) o(1) o(1)

Code

Heap

python3 复制代码

class Solution:
    def minMeetingRooms(self, intervals: List[List[int]]) -> int:
        heap = []
        intervals.sort(key=lambda x: x[0])
        res = 0
        for i in range(len(intervals)):
            if heap and heap[0] <= intervals[i][0]:
                heapq.heappop(heap)
            heapq.heappush(heap, intervals[i][1])
            res = max(res, len(heap))
        return res

Sort + sweep

python3 复制代码

class Solution:
    def minMeetingRooms(self, intervals: List[List[int]]) -> int:
        meetings = {}
        for start, end in intervals:
            meetings[start] = meetings.get(start, 0) + 1
            meetings[end] = meetings.get(end, 0) - 1
        points = sorted(meetings)
        res = 0
        room = 0
        for each_point in points:
            room += meetings[each_point]
            res = max(res, room)
        return res