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mediumDSA

How do you find the minimum number of meeting rooms required?

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01

Understand the problem

Explain the meeting-rooms problem.

intervalssortingheap
02

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03

Study the solution

Sort meetings by start time and use a min-heap of end times; for each meeting, if the earliest end ≤ its start, reuse that room (pop), else allocate a new one (push). The heap's max size is the answer. O(n log n).

Solution ready — 2 min read

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04

Read the code

Sort + min-heap of end times
Run Playground
import heapq

def min_meeting_rooms(intervals):
    if not intervals:
        return 0
    intervals.sort(key=lambda x: x[0])    # by start
    heap = []                             # end times of ongoing meetings
    for start, end in intervals:
        if heap and heap[0] <= start:
            heapq.heappop(heap)           # a room freed up — reuse it
        heapq.heappush(heap, end)
    return len(heap)


# --- demo ---
print(min_meeting_rooms([[0, 30], [5, 10], [15, 20]]))   # 2
05

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