20. Meeting Rooms II

mediumAsked at Asana

Find the minimum number of meeting rooms required to host all meetings without conflicts — Asana surfaces this when its calendar-integration feature must allocate the fewest concurrent task-review slots across a team.

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

Problem

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

Constraints

1 <= intervals.length <= 10^4
0 <= starti < endi <= 10^6

Examples

Example 1

Input

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

Output

Explanation: Meetings [0,30] and [5,10] overlap — need 2 rooms. [15,20] reuses the room freed at 10.

Example 2

Input

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

Output

Explanation: No overlap — one room suffices.

Approaches

1. Separate start/end sorted arrays

Sort starts and ends independently. Use two pointers: if the next start is before the earliest end, allocate a room; otherwise reuse one.

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 of end times (optimal intuition)

Sort by start. Maintain a min-heap of meeting end times. For each new meeting, if its start >= heap minimum, pop (reuse room); otherwise push (new room). Heap size = rooms needed.

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

function minMeetingRooms(intervals) {
  if (intervals.length === 0) return 0;
  intervals.sort((a, b) => a[0] - b[0]);

  // Min-heap simulated with a sorted array (acceptable for interview)
  const heap = [];

  function heapPush(val) {
    heap.push(val);
    heap.sort((a, b) => a - b);
  }

  function heapPop() {
    return heap.shift();
  }

  heapPush(intervals[0][1]);

  for (let i = 1; i < intervals.length; i++) {
    const [start, end] = intervals[i];
    if (heap[0] <= start) {
      heapPop();
    }
    heapPush(end);
  }

  return heap.length;
}

Tradeoff:

Asana-specific tips

Asana interviewers often frame this as 'how many parallel task-review sessions do we need?' Both approaches are O(n log n) — the two-pointer solution is clever and compact, but the min-heap version maps more naturally to a real room-allocation system where rooms have identities. Mention that in production you'd use an actual priority queue library; the sorted-array stand-in is for whiteboard clarity.

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