leetcode - 253. Meeting Rooms II

Description

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

Example 1:

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

Example 2:

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

Constraints:

复制代码
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
相关推荐
追随者永远是胜利者5 小时前
(LeetCode-Hot100)253. 会议室 II
java·算法·leetcode·go
Jason_Honey25 小时前
【平安Agent算法岗面试-二面】
人工智能·算法·面试
程序员酥皮蛋5 小时前
hot 100 第三十五题 35.二叉树的中序遍历
数据结构·算法·leetcode
追随者永远是胜利者5 小时前
(LeetCode-Hot100)207. 课程表
java·算法·leetcode·go
香芋Yu6 小时前
【大模型面试突击】08_推理范式与思维链
面试·职场和发展
仰泳的熊猫6 小时前
题目1535:蓝桥杯算法提高VIP-最小乘积(提高型)
数据结构·c++·算法·蓝桥杯
那起舞的日子7 小时前
动态规划-Dynamic Programing-DP
算法·动态规划
闻缺陷则喜何志丹7 小时前
【前后缀分解】P9255 [PA 2022] Podwyżki|普及+
数据结构·c++·算法·前后缀分解
每天吃饭的羊7 小时前
时间复杂度
数据结构·算法·排序算法
小李独爱秋8 小时前
模拟面试:用自己的话解释一下lvs的工作原理
linux·运维·面试·职场和发展·操作系统·lvs