LeetCode 2402: Meeting Rooms III Solution

Master LeetCode problem 2402 (Meeting Rooms III), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2402. Meeting Rooms III

Hard

Array Hash Table Sorting Heap (Priority Queue)Simulation

Problem Explanation

Explanation

To solve this problem, we can use a priority queue to keep track of the end times of the meetings in the rooms. We iterate through the meetings sorted by start time, and for each meeting, we check if there are any available rooms. If there are, we assign the meeting to the room with the earliest end time. If there are no available rooms, we delay the meeting by adding it to the priority queue.

We maintain a count of meetings held in each room and return the room with the maximum count.

Time complexity: O(n log n) where n is the number of meetings Space complexity: O(n)

Solution Code

import java.util.*;

class Solution {
    public int meetingRoomIII(int n, int[][] meetings) {
        Arrays.sort(meetings, (a, b) -> a[0] - b[0]);
        PriorityQueue<int[]> rooms = new PriorityQueue<>((a, b) -> a[1] - b[1]);
        int[] count = new int[n];
        
        int maxMeetings = 0;
        for (int[] meeting : meetings) {
            while (!rooms.isEmpty() && rooms.peek()[1] <= meeting[0]) {
                int[] room = rooms.poll();
                count[room[0]]++;
                maxMeetings = Math.max(maxMeetings, count[room[0]]);
            }
            
            int room = rooms.size();
            rooms.offer(new int[]{room, meeting[1]});
        }
        
        return Arrays.binarySearch(count, maxMeetings);
    }
}

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