1897. Meeting Room III

https://www.lintcode.com/problem/meeting-room-iii/description

这题目其实有三个问题,其实要回答的都是一个问题:至少需要几个会议室能满足彼此重叠的开会时间。

版本2的解法是:

  1. 对每个[s, e],在s点记录进入,在e+1点记录出来
  2. 需要最多的会议室的时刻,肯定是在这些时间点上
  3. 然后顺序遍历这些时间点,看每个时间点上需要的最多会议室

这题的解法非常具有启发性,值得好好学习。

class Solution:
    """
    @param intervals: an array of meeting time intervals
    @return: the minimum number of conference rooms required
    """
    def minMeetingRooms(self, intervals):
        # Write your code here
        from collections import Counter
        cnt = Counter()
        for t in intervals:
            s, e = t.start, t.end
            cnt[s] += 1
            cnt[e+1] -= 1

        ts = sorted(cnt.keys())
        ans = 0
        room = 0
        for t in ts:
            room += cnt[t]
            ans = max(ans, room)
        return ans

版本3的解法是在版本2上扩展的,我们挑选出关键的时间范围。所谓关键的时间范围是,这些时间范围占据会议室的数量已经到了临界值,只要在增加一个就会不满足条件。然后二分查找query可能重合的时间范围,判断是否真的重合。 另外一个思路是,查找query重合的时间范围,然后看这些时间范围内最大占用的会议室数量。应该可以使用区间树来做,就是稍微有点麻烦。

class Solution:
    """
    @param intervals: the intervals
    @param rooms: the sum of rooms
    @param ask: the ask
    @return: true or false of each meeting
    """

    def meetingRoomIII(self, intervals, rooms, ask):
        # Write your code here.
        from collections import Counter
        cnt = Counter()
        for s, e in intervals:
            cnt[s] += 1
            cnt[e] -= 1

        ts = sorted(cnt.keys())
        rs = []
        acc = 0
        start = 0
        for t in ts:
            rs.append((start, t, acc))
            acc += cnt[t]
            start = t

        rs = [(x, y, t) for (x, y, t) in rs if t == rooms]
        ans = []

        def overlap(r, q):
            (x, y, t) = r
            (s, e) = q
            if x >= e or y <= s:
                return False
            return True

        for q in ask:
            s, e = 0, len(rs) - 1
            while s <= e:
                m = (s + e) // 2
                if rs[m][0] >= q[0]:
                    e = m - 1
                else:
                    s = m + 1
            # check e and e + 1 overlap with q
            opts = [e, e + 1]
            ok = True
            for i in opts:
                if 0 <= i < len(rs):
                    if overlap(rs[i], q):
                        ok = False
                        break
            ans.append(ok)
        return ans