solve: meetingRoomsIi

Author: yolophg

meeting-rooms-ii/yolophg.py

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+ # Time Complexity: O(n log n) -> sorting takes O(n log n), and heap operations take O(log n) per interval.
+ # Space Complexity: O(n) -> in the worst case, store all meetings in the heap.
+
+class Solution:
+    def minMeetingRooms(self, intervals: List[List[int]]) -> int:        
+        if not intervals:
+            return 0
+        
+        # sort meetings by start time.
+        intervals.sort() 
+        
+        # heap to keep track of meeting end times.
+        min_heap = [] 
+        
+        for start, end in intervals:
+            if min_heap and min_heap[0] <= start:
+                # remove the meeting that has ended.
+                heapq.heappop(min_heap) 
+            # add the current meeting's end time.
+            heapq.heappush(min_heap, end) 
+        
+        return len(min_heap)