# 2402. Meeting Rooms III
###### tags: `Leetcode` `Hard` `Priority Queue`
Link: https://leetcode.com/problems/meeting-rooms-iii/
## 思路

For ```[start, end]``` in the sorted ```meetings```,
we firstly release the rooms that is ready before ```start``` time.
If there is room in ```ready``` state,
we choose the room with smallest index.
Otherwise, we choose the room with smallest ```end_time``` in ```rooms```.
## Code

class Solution {
	public int mostBooked(int n, int[][] meetings) {
		Arrays.sort(meetings, (a,b)->(a[0]-b[0]));

		Queue<Integer> unused = new PriorityQueue<>();
		for (int i=0; i<n; i++) unused.add(i);

		PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> {
			if (a[0] == b[0]) return a[1] - b[1];
			return (a[0] - b[0]);
        });

		int[] count = new int[n];
        int maxFreq = 0;
        
		for (int[] meeting:meetings) {
			while (!pq.isEmpty() && meeting[0] >= pq.peek()[0]) {
				unused.add(pq.poll()[1]);
			}
            int room;
			if (!unused.isEmpty()) {
				room = unused.poll();
				count[room]++;
				pq.add(new int[]{meeting[1], room});
			} else {
				int[] last = pq.poll();
				int delayed = last[0];
				room = last[1];
				int curEnd = delayed + meeting[1] - meeting[0];
				count[room]++;
				pq.add(new int[]{curEnd, room});
			}
            maxFreq = Math.max(count[room], maxFreq);
		}
		for (int i = 0; i < n; i++) {
			if(count[i]==maxFreq) return i;
		}
        return 0;
	}
}-