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19. Meeting Rooms II

mediumAsked at Dropbox

Find the minimum number of conference rooms needed for a set of meetings — Dropbox maps this directly to how many concurrent file-version locks the sync layer must hold at peak conflict time.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Problem

Given an array of meeting time intervals where intervals[i] = [start_i, end_i], return the minimum number of conference rooms required to hold all meetings without overlap.

Constraints

  • 1 <= intervals.length <= 10^4
  • 0 <= start_i < end_i <= 10^6

Examples

Example 1

Input
intervals = [[0,30],[5,10],[15,20]]
Output
2

Explanation: Meetings [5,10] and [15,20] can share a room, but [0,30] overlaps both and needs its own room.

Example 2

Input
intervals = [[7,10],[2,4]]
Output
1

Explanation: The two meetings do not overlap, so one room suffices.

Approaches

1. Sorted events (split starts/ends)

Separate start and end times, sort each independently. Use two pointers; when the next meeting starts before the earliest ending one, allocate a new room. Otherwise, reclaim the freed room.

Time
O(n log n)
Space
O(n)
function minMeetingRooms(intervals) {
  const starts = intervals.map(i => i[0]).sort((a, b) => a - b);
  const ends = intervals.map(i => i[1]).sort((a, b) => a - b);
  let rooms = 0;
  let endPtr = 0;
  for (let i = 0; i < starts.length; i++) {
    if (starts[i] < ends[endPtr]) {
      rooms++;
    } else {
      endPtr++;
    }
  }
  return rooms;
}

Tradeoff:

2. Min-heap (priority queue)

Sort meetings by start time. Use a min-heap of end times. For each meeting, if it starts after the heap's minimum end time, reuse that room (pop). Otherwise add a new room. Heap size is the answer.

Time
O(n log n)
Space
O(n)
class MinHeap {
  constructor() { this.h = []; }
  push(v) {
    this.h.push(v);
    let i = this.h.length - 1;
    while (i > 0) {
      const p = Math.floor((i - 1) / 2);
      if (this.h[p] <= this.h[i]) break;
      [this.h[p], this.h[i]] = [this.h[i], this.h[p]];
      i = p;
    }
  }
  pop() {
    const top = this.h[0];
    const last = this.h.pop();
    if (this.h.length > 0) {
      this.h[0] = last;
      let i = 0;
      while (true) {
        let s = i;
        const l = 2*i+1, r = 2*i+2;
        if (l < this.h.length && this.h[l] < this.h[s]) s = l;
        if (r < this.h.length && this.h[r] < this.h[s]) s = r;
        if (s === i) break;
        [this.h[s], this.h[i]] = [this.h[i], this.h[s]];
        i = s;
      }
    }
    return top;
  }
  peek() { return this.h[0]; }
  get size() { return this.h.length; }
}

function minMeetingRooms(intervals) {
  intervals.sort((a, b) => a[0] - b[0]);
  const heap = new MinHeap();
  for (const [start, end] of intervals) {
    if (heap.size > 0 && heap.peek() <= start) {
      heap.pop();
    }
    heap.push(end);
  }
  return heap.size;
}

Tradeoff:

Dropbox-specific tips

Dropbox often extends this to ask how you'd handle room preferences or priorities. The two-pointer approach is the cleanest to explain under time pressure; but if you have time, mentioning the heap variant shows you can generalize to weighted constraints — which matters in Dropbox's resource-scheduling context.

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