LeetCode 2402: Meeting Rooms III
LeetCode 2402 Solution 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)
LeetCode 2402 Solutions in Java, C++, Python
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);
}
}Interactive Code Editor for LeetCode 2402
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