Sync LeetCode submission Runtime - 779 ms (5.77%), Memory - 51.5 MB (68.69%)

Author: jimit105

2479-meeting-rooms-iii/README.md

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+<p>You are given an integer <code>n</code>. There are <code>n</code> rooms numbered from <code>0</code> to <code>n - 1</code>.</p>
+
+<p>You are given a 2D integer array <code>meetings</code> where <code>meetings[i] = [start<sub>i</sub>, end<sub>i</sub>]</code> means that a meeting will be held during the <strong>half-closed</strong> time interval <code>[start<sub>i</sub>, end<sub>i</sub>)</code>. All the values of <code>start<sub>i</sub></code> are <strong>unique</strong>.</p>
+
+<p>Meetings are allocated to rooms in the following manner:</p>
+
+<ol>
+	<li>Each meeting will take place in the unused room with the <strong>lowest</strong> number.</li>
+	<li>If there are no available rooms, the meeting will be delayed until a room becomes free. The delayed meeting should have the <strong>same</strong> duration as the original meeting.</li>
+	<li>When a room becomes unused, meetings that have an earlier original <strong>start</strong> time should be given the room.</li>
+</ol>
+
+<p>Return<em> the <strong>number</strong> of the room that held the most meetings. </em>If there are multiple rooms, return<em> the room with the <strong>lowest</strong> number.</em></p>
+
+<p>A <strong>half-closed interval</strong> <code>[a, b)</code> is the interval between <code>a</code> and <code>b</code> <strong>including</strong> <code>a</code> and <strong>not including</strong> <code>b</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 2, meetings = [[0,10],[1,5],[2,7],[3,4]]
+<strong>Output:</strong> 0
+<strong>Explanation:</strong>
+- 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. 
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 3, meetings = [[1,20],[2,10],[3,5],[4,9],[6,8]]
+<strong>Output:</strong> 1
+<strong>Explanation:</strong>
+- 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. 
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 100</code></li>
+	<li><code>1 &lt;= meetings.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>meetings[i].length == 2</code></li>
+	<li><code>0 &lt;= start<sub>i</sub> &lt; end<sub>i</sub> &lt;= 5 * 10<sup>5</sup></code></li>
+	<li>All the values of <code>start<sub>i</sub></code> are <strong>unique</strong>.</li>
+</ul>

2479-meeting-rooms-iii/solution.py

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+class Solution:
+    def mostBooked(self, n: int, meetings: List[List[int]]) -> int:
+        room_availability_time = [0] * n
+        meeting_count = [0] * n
+        for start, end in sorted(meetings):
+            min_room_availability_time = inf
+            min_available_time_room = 0
+            found_unused_room = False
+            for i in range(n):
+                if room_availability_time[i] <= start:
+                    found_unused_room = True
+                    meeting_count[i] += 1
+                    room_availability_time[i] = end
+                    break
+                if min_room_availability_time > room_availability_time[i]:
+                    min_room_availability_time = room_availability_time[i]
+                    min_available_time_room = i
+            if not found_unused_room:
+                room_availability_time[min_available_time_room] += end - start
+                meeting_count[min_available_time_room] += 1
+
+        return meeting_count.index(max(meeting_count))