Leetcode Problem 2402. Meeting Rooms III

2402. Meeting Rooms III

Leetcode Solutions

Heap-based Room Allocation

  1. Sort the meetings array based on the start times of the meetings.
  2. Initialize a min-heap available_rooms with all room numbers (0 to n-1).
  3. Initialize a min-heap ongoing_meetings to keep track of rooms currently in use, storing pairs of end time and room number.
  4. Initialize a frequency array room_meetings_count to count the number of meetings in each room.
  5. Iterate over the sorted meetings array: a. Release rooms from ongoing_meetings that have become available (end time <= current meeting's start time) and add them back to available_rooms. b. If available_rooms is not empty, allocate the room with the lowest number to the current meeting. c. If available_rooms is empty, take the room with the earliest end time from ongoing_meetings and update the meeting's end time. d. Add the allocated room and the updated end time to ongoing_meetings. e. Increment the count for the allocated room in room_meetings_count.
  6. After processing all meetings, find the room with the maximum count in room_meetings_count. If there is a tie, the room with the lowest number is chosen.
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Brute Force Room Allocation

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