Welcome to Subscribe On Youtube

2402. Meeting Rooms III

Description

You are given an integer n. There are n rooms numbered from 0 to n - 1.

You are given a 2D integer array meetings where meetings[i] = [starti, endi] means that a meeting will be held during the half-closed time interval [starti, endi). All the values of starti are unique.

Meetings are allocated to rooms in the following manner:

  1. Each meeting will take place in the unused room with the lowest number.
  2. If there are no available rooms, the meeting will be delayed until a room becomes free. The delayed meeting should have the same duration as the original meeting.
  3. When a room becomes unused, meetings that have an earlier original start time should be given the room.

Return the number of the room that held the most meetings. If there are multiple rooms, return the room with the lowest number.

A half-closed interval [a, b) is the interval between a and b including a and not including b.

 

Example 1:

Input: n = 2, meetings = [[0,10],[1,5],[2,7],[3,4]]
Output: 0
Explanation:
- At time 0, both rooms are not being used. The first meeting starts in room 0.
- At time 1, only room 1 is not being used. The second meeting starts in room 1.
- At time 2, both rooms are being used. The third meeting is delayed.
- At time 3, both rooms are being used. The fourth meeting is delayed.
- At time 5, the meeting in room 1 finishes. The third meeting starts in room 1 for the time period [5,10).
- At time 10, the meetings in both rooms finish. The fourth meeting starts in room 0 for the time period [10,11).
Both rooms 0 and 1 held 2 meetings, so we return 0.

Example 2:

Input: n = 3, meetings = [[1,20],[2,10],[3,5],[4,9],[6,8]]
Output: 1
Explanation:
- At time 1, all three rooms are not being used. The first meeting starts in room 0.
- At time 2, rooms 1 and 2 are not being used. The second meeting starts in room 1.
- At time 3, only room 2 is not being used. The third meeting starts in room 2.
- At time 4, all three rooms are being used. The fourth meeting is delayed.
- At time 5, the meeting in room 2 finishes. The fourth meeting starts in room 2 for the time period [5,10).
- At time 6, all three rooms are being used. The fifth meeting is delayed.
- At time 10, the meetings in rooms 1 and 2 finish. The fifth meeting starts in room 1 for the time period [10,12).
Room 0 held 1 meeting while rooms 1 and 2 each held 2 meetings, so we return 1.

 

Constraints:

  • 1 <= n <= 100
  • 1 <= meetings.length <= 105
  • meetings[i].length == 2
  • 0 <= starti < endi <= 5 * 105
  • All the values of starti are unique.

Solutions

Solution 1: Priority Queue (Min Heap)

We define two priority queues, representing idle meeting rooms and busy meeting rooms, respectively. Among them: the idle meeting rooms idle are sorted according to index; while the busy meeting rooms busy are sorted according to end time, index.

First, sort the meetings by start time, then traverse the meetings. For each meeting:

  • If there is a busy meeting room that is less than or equal to the start time of the current meeting, add it to the idle meeting room queue idle.
  • If there are currently idle meeting rooms, take out the meeting room with the smallest weight from the idle queue idle and add it to the busy queue busy.
  • If there are currently no idle meeting rooms, find the meeting room with the earliest end time and smallest index in the busy queue busy, and re-add it to the busy queue busy.

The time complexity is $O(m \times \log m)$, where $m$ is the number of meetings.

  • class Solution {
    public int mostBooked(int n, int[][] meetings) {
        Arrays.sort(meetings, (a, b) -> a[0] - b[0]);
        PriorityQueue<int[]> busy
            = new PriorityQueue<>((a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
        PriorityQueue<Integer> idle = new PriorityQueue<>();
        for (int i = 0; i < n; ++i) {
            idle.offer(i);
        }
        int[] cnt = new int[n];
        for (var v : meetings) {
            int s = v[0], e = v[1];
            while (!busy.isEmpty() && busy.peek()[0] <= s) {
                idle.offer(busy.poll()[1]);
            }
            int i = 0;
            if (!idle.isEmpty()) {
                i = idle.poll();
                busy.offer(new int[] {e, i});
            } else {
                var x = busy.poll();
                i = x[1];
                busy.offer(new int[] {x[0] + e - s, i});
            }
            ++cnt[i];
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (cnt[ans] < cnt[i]) {
                ans = i;
            }
        }
        return ans;
    }
}

  • using ll = long long;
using pii = pair<ll, int>;

class Solution {
public:
    int mostBooked(int n, vector<vector<int>>& meetings) {
        priority_queue<int, vector<int>, greater<int>> idle;
        priority_queue<pii, vector<pii>, greater<pii>> busy;
        for (int i = 0; i < n; ++i) idle.push(i);
        vector<int> cnt(n);
        sort(meetings.begin(), meetings.end());
        for (auto& v : meetings) {
            int s = v[0], e = v[1];
            while (!busy.empty() && busy.top().first <= s) {
                idle.push(busy.top().second);
                busy.pop();
            }
            int i = 0;
            if (!idle.empty()) {
                i = idle.top();
                idle.pop();
                busy.push({e, i});
            } else {
                auto x = busy.top();
                busy.pop();
                i = x.second;
                busy.push({x.first + e - s, i});
            }
            ++cnt[i];
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (cnt[ans] < cnt[i]) {
                ans = i;
            }
        }
        return ans;
    }
};

  • class Solution:
    def mostBooked(self, n: int, meetings: List[List[int]]) -> int:
        meetings.sort()
        busy = []
        idle = list(range(n))
        heapify(idle)
        cnt = [0] * n
        for s, e in meetings:
            while busy and busy[0][0] <= s:
                heappush(idle, heappop(busy)[1])
            if idle:
                i = heappop(idle)
                cnt[i] += 1
                heappush(busy, (e, i))
            else:
                a, i = heappop(busy)
                cnt[i] += 1
                heappush(busy, (a + e - s, i))
        ans = 0
        for i, v in enumerate(cnt):
            if cnt[ans] < v:
                ans = i
        return ans


  • func mostBooked(n int, meetings [][]int) int {
	sort.Slice(meetings, func(i, j int) bool { return meetings[i][0] < meetings[j][0] })
	idle := hp{make([]int, n)}
	for i := 0; i < n; i++ {
		idle.IntSlice[i] = i
	}
	busy := hp2{}
	cnt := make([]int, n)
	for _, v := range meetings {
		s, e := v[0], v[1]
		for len(busy) > 0 && busy[0].end <= s {
			heap.Push(&idle, heap.Pop(&busy).(pair).i)
		}
		var i int
		if idle.Len() > 0 {
			i = heap.Pop(&idle).(int)
			heap.Push(&busy, pair{e, i})
		} else {
			x := heap.Pop(&busy).(pair)
			i = x.i
			heap.Push(&busy, pair{x.end + e - s, i})
		}
		cnt[i]++
	}
	ans := 0
	for i, v := range cnt {
		if cnt[ans] < v {
			ans = i
		}
	}
	return ans
}

type hp struct{ sort.IntSlice }

func (h *hp) Push(v any) { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
	a := h.IntSlice
	v := a[len(a)-1]
	h.IntSlice = a[:len(a)-1]
	return v
}

type pair struct{ end, i int }
type hp2 []pair

func (h hp2) Len() int { return len(h) }
func (h hp2) Less(i, j int) bool {
	a, b := h[i], h[j]
	return a.end < b.end || a.end == b.end && a.i < b.i
}
func (h hp2) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
func (h *hp2) Push(v any)   { *h = append(*h, v.(pair)) }
func (h *hp2) Pop() any     { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }

All Problems

All Solutions