LeetCode 253Meeting Rooms II

LeetCode 253 Meeting Rooms II Problem

``````# Definition for an interval.
# class Interval(object):
#     def __init__(self, s=0, e=0):
#         self.start = s
#         self.end = e

# class Solution(object):
#     def minMeetingRooms(self, intervals):
#         """
#         :type intervals: List[Interval]
#         :rtype: int
#         """
#         # If there is no meeting to schedule then no room needs to be allocated.
#         if not intervals:
#             return 0
#         # The heap initialization
#         free_rooms = []
#         # Sort the meetings in increasing order of their start time.
#         intervals.sort(key= lambda x: x.start)
#         # Add the first meeting. We have to give a new room to the first meeting.
#         heapq.heappush(free_rooms, intervals[0].end)
#         # For all the remaining meeting rooms
#         for i in intervals[1:]:
#             # If the room due to free up the earliest is free, assign that room to this meeting.
#             if free_rooms[0] <= i.start:
#                 heapq.heappop(free_rooms)
#             # If a new room is to be assigned, then also we add to the heap,
#             # If an old room is allocated, then also we have to add to the heap with updated end time.
#             heapq.heappush(free_rooms, i.end)
#         # The size of the heap tells us the minimum rooms required for all the meetings.
#         return len(free_rooms)

class Solution(object):
def minMeetingRooms(self, intervals):
"""
:type intervals: List[Interval]
:rtype: int
"""
timeline = []
for interval in intervals:
# meeting root + 1
timeline.append((interval.start, 1))
# meeting root - 1
timeline.append((interval.end, -1))
# sort by time
timeline.sort()
ans = curr = 0
# go through timeline
for _, v in timeline:
curr += v
# max meeting room used at this point
ans = max(ans, curr)
return ans
``````

List of all Meeting Rooms II problems

Leetcode 253 Meeting Rooms II problem solution in python3 with explanation. This is the best place to expand your knowledge and get prepared for your next interview.

self.start = s

self.end = e

class Solution(object):

def minMeetingRooms(self, intervals):

"""

:type intervals: List[Interval]

:rtype: int

"""

# If there is no meeting to schedule then no room needs to be allocated.

if not intervals:

return 0

# The heap initialization

free_rooms = []

# Sort the meetings in increasing order of their start time.

intervals.sort(key= lambda x: x.start)

# Add the first meeting. We have to give a new room to the first meeting.

heapq.heappush(free_rooms, intervals[0].end)

# For all the remaining meeting rooms

for i in intervals[1:]:

# If the room due to free up the earliest is free, assign that room to this meeting.

if free_rooms[0] <= i.start:

heapq.heappop(free_rooms)

# If a new room is to be assigned, then also we add to the heap,

# If an old room is allocated, then also we have to add to the heap with updated end time.

heapq.heappush(free_rooms, i.end)

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