Overview

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

Implement canAttendAll(intervals).

Constraints

• Each interval is a two-element array [start, end] where start < end.
• Return true if there are no overlapping meetings, false otherwise.
• Input may be unsorted.
• Attend means the person must be present for the entire duration. Adjacent meetings ([0,5],[5,10]) do NOT conflict.
• Aim for $O(n \log n)$ time.

Examples

canAttendAll([[0,30],[5,10],[15,20]]);
// → false  — [0,30] overlaps with both others

canAttendAll([[7,10],[2,4]]);
// → true  — no overlap

canAttendAll([[1,5],[5,10]]);
// → true  — adjacent, not overlapping

Notes

• Sort by start time and compare consecutive pairs.
• If any meeting starts before the previous one ends, return false.

Solution

function canAttendAll(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;
}

Resources

• Meeting Rooms — LeetCode
• Interval Scheduling — Wikipedia