Meeting Rooms II

Problem (restated)

Given meeting time intervals, find the minimum number of conference rooms required.

Intuition

Sort starts and ends; sweep: start increments rooms, end frees; track peak.

Approaches

export function minMeetingRooms(intervals: number[][]): number {
  const starts = intervals.map((x) => x[0]!).sort((a, b) => a - b);
  const ends = intervals.map((x) => x[1]!).sort((a, b) => a - b);
  let i = 0, j = 0, cur = 0, peak = 0;
  while (i < starts.length) {
    if (starts[i]! < ends[j]!) {
      cur++;
      peak = Math.max(peak, cur);
      i++;
    } else {
      cur--;
      j++;
    }
  }
  return peak;
}

export function minMeetingRooms(intervals: number[][]): number {
  const starts = intervals.map((x) => x[0]!).sort((a, b) => a - b);
  const ends = intervals.map((x) => x[1]!).sort((a, b) => a - b);
  let i = 0, j = 0, cur = 0, peak = 0;
  while (i < starts.length) {
    if (starts[i]! < ends[j]!) {
      cur++;
      peak = Math.max(peak, cur);
      i++;
    } else {
      cur--;
      j++;
    }
  }
  return peak;
}

Template connection

Intervals sweep / rooms.

Reflection

• A start is +1 and an end is −1. At the same time, processing the start before the end counts an extra room.
• The answer is the highest counter on the sweep. Three nested meetings need 3 rooms.
• An empty list is 0. Meetings that only touch share a room.