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NeetCode #836LC-252EasyIntervalsBlind 75NC 150NC 250NC Algo100
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#836 · #252 · Meeting Rooms(会议室)

📌 Problem Statement & Constraints

You are given an array of meeting time intervals intervals. Determine whether a person could attend all meetings, i.e. no two intervals overlap. Constraints: 0 <= intervals.length <= 10^4, 0 <= intervals[i][0] < intervals[i][1] <= 10^6.

💡 Core Algorithmic Approaches

  1. Sort the intervals by start time.
  2. Two meetings conflict exactly when the later one starts before the earlier one ends.
  3. So a single adjacent scan suffices: check intervals[i][0] < intervals[i-1][1] for each pair.
  4. Because the array is sorted by start, only adjacent pairs can possibly overlap.

💻 Benchmark Python3 Implementation

class Solution:
    def canAttendMeetings(self, intervals: List[List[int]]) -> bool:
        intervals.sort(key=lambda x: x[0])
        for i in range(1, len(intervals)):
            if intervals[i][0] < intervals[i - 1][1]:
                return False                # conflict
        return True

⚡ Complexity Deep Dive

⏱️ Time Complexity
O(n log n): the sort dominates; the scan is O(n).
💾 Space Complexity
O(1) beyond the sort.

⚠️ Interview Pitfalls & Follow-ups

  • Using <= for the conflict test: a meeting ending at 5 and another starting at 5 do not conflict, so the test is strict <.
  • Comparing non-adjacent pairs: after sorting by start, adjacency is sufficient.
  • Assuming the input is sorted: it is not, so the sort is required.
  • Forgetting the empty input: with no intervals the loop is skipped and true is returned, which is correct.