17. Meeting Rooms

easyAsked at Booking

Determine if a person can attend all meetings — Booking applies the same interval-overlap check when validating that a property's booking windows never collide on the same room.

By Alex Chen, Founder, InterviewChamp.AI · Last verified 2026-05-26

Problem

Given an array of meeting time intervals where intervals[i] = [start_i, end_i], determine if a person could attend all meetings (no two intervals overlap).

Constraints

0 <= intervals.length <= 10^4
intervals[i].length == 2
0 <= start_i < end_i <= 10^6

Examples

Example 1

Input

intervals = [[0,30],[5,10],[15,20]]

Output

false

Explanation: [0,30] overlaps with both [5,10] and [15,20].

Example 2

Input

intervals = [[7,10],[2,4]]

Output

true

Approaches

1. Brute force

Check every pair of intervals for overlap.

Time: O(n^2)
Space: O(1)

function canAttendMeetings(intervals) {
  for (let i = 0; i < intervals.length; i++) {
    for (let j = i + 1; j < intervals.length; j++) {
      const [s1, e1] = intervals[i];
      const [s2, e2] = intervals[j];
      if (s1 < e2 && s2 < e1) return false;
    }
  }
  return true;
}

Tradeoff:

2. Sort then linear scan

Sort by start time; overlap exists only if a start falls before the previous end. Single pass after sort.

Time: O(n log n)
Space: O(1)

function canAttendMeetings(intervals) {
  intervals.sort((a, b) => a[0] - b[0]);
  for (let i = 1; i < intervals.length; i++) {
    if (intervals[i][0] < intervals[i - 1][1]) return false;
  }
  return true;
}

Tradeoff:

Booking-specific tips

Booking cares that you immediately reach for sorting by start time — it mirrors how their calendar engine validates reservation windows. State the overlap condition (s2 < e1) precisely; off-by-one on boundary equality trips many candidates.

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Output

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