Update 0253-meeting-rooms-ii.py

16_Intervals/05_Meeting_Rooms_II/0253-meeting-rooms-ii.py

@@ -5,10 +5,10 @@
 We can solve this problem using a priority queue (min-heap). First, we sort the intervals based on their start times. We then iterate through the sorted intervals and maintain a min-heap of end times of ongoing meetings. For each interval, if the start time is greater than or equal to the smallest end time in the min-heap, it means the current meeting can reuse an existing room, so we pop the smallest end time from the min-heap. If not, we need to allocate a new room. After processing all intervals, the size of the min-heap gives us the minimum number of meeting rooms required.
 
 Time Complexity:
-- The time complexity of this approach is O(n log n) due to the sorting step and the heap operations, where n is the number of intervals.
+- The time complexity of this approach is O(n log n) due to the sorting step 
 
 Space Complexity:
-- The space complexity is O(n), as we store the end times in the min-heap.
+- The space complexity is O(n), additional space used for start, end
 """
 class Solution:
     def minMeetingRooms(self, intervals: List[Interval]) -> int:
@@ -20,12 +20,13 @@ def minMeetingRooms(self, intervals: List[Interval]) -> int:
 
         res, rooms = 0,0
         s, e = 0,0
-        while s < len(intervals):
-            if start[s] < end [s]:
-                s += 1
-                rooms += 1
-            else:
-                e += 1
+        while s < len(intervals): 
+            if start[s] < end [s]: # new meeting starting while another going
+                s += 1 #move to next starting time
+                rooms += 1 
+            else: #meeting has ended
+                e += 1 #move to next end time
+                rooms -= 1 #free up room
             res = max(res, rooms)
         return res