Meeting Rooms II

"""
Intuition:
At any moment, the number of rooms needed equals the number of meetings currently running.
So we sort all start times and end times, and sweep through time using two pointers.
If the next start is before the earliest end, one more meeting overlaps, so rooms++.
Otherwise (start >= end), a meeting has finished and its room can be reused, so rooms--.
The maximum overlap seen during this sweep is the minimum rooms required.
"""

class Solution:
    def minMeetingRooms(self, start, end):
        start.sort()
        end.sort()
        ans=0
        i,j=0,0
        cur=0
        while i<len(start):
            if start[i]<end[j]:
                i+=1
                cur+=1
            else:
                j+=1
                cur-=1
            ans=max(ans,cur)
        return ans