2402. Meeting Rooms III - Explanation

Description

There are n rooms numbered from 0 to n - 1. You are given a 2D integer array meetings where meetings[i] = [start[i], end[i]] means that a meeting will be held during the half-closed time interval [start[i], end[i]). All the values of start[i] are unique.

Meetings are allocated to rooms in the following manner:

Each meeting will take place in the unused room with the lowest number.
If there are no available rooms, the meeting will be delayed until a room becomes free. The delayed meeting should have the same duration as the original meeting.
When a room becomes unused, meetings that have an earlier original start time should be given the room.

Return the number of the room that held the most meetings. If there are multiple rooms, return the room with the lowest number.

A half-closed interval [a, b) is the interval between a and b including a and not including b.

Example 1:

Input: n = 2, meetings = [[1,10],[2,10],[3,10],[4,10]]

Output: 0

Explanation:

At time 1, the room with number 0 is available, the first meeting is allocated to it.
At time 2, the room with number 1 is available, the second meeting is allocated to it.
At time 10, the rooms with numbers 0 and 1 becomes unused, the third meeting is allocated to room number 0 and fourth meeting is allocated to room number 1.

The room with most meetings held and have lowest number is 0.

Example 2:

Input: n = 3, meetings = [[1,20],[2,10],[3,5],[6,8],[4,9]]

Output: 1

Explanation:

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 as it is lowest.

Constraints:

1 <= n <= 100
1 <= meetings.length <= 100,000
meetings[i].length == 2
0 <= start[i] < end[i] <= 500,000
All the values of start[i] are unique.

Topics

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Prerequisites

Before attempting this problem, you should be comfortable with:

Sorting - Ordering meetings by start time to process them chronologically
Heap / Priority Queue - Efficiently finding the minimum element (earliest available room or end time)
Simulation - Tracking state changes over time as meetings are assigned and rooms become available
Greedy Selection - Choosing the optimal room based on availability and room number constraints

1. Sorting + Brute Force

Intuition

Meetings must be processed in order of their start times. For each meeting, we check rooms in order from 0 to n - 1. If a room is free by the meeting's start time, we assign the meeting there. Otherwise, we find the room that becomes free earliest and delay the meeting until that room is available. We track how many meetings each room hosts and return the room with the most meetings, breaking ties by smallest room number.

Algorithm

Sort meetings by start time.
Initialize arrays rooms (end times) and meeting_count, both of size n.
For each meeting (start, end):
- Scan rooms 0 to n - 1. If rooms[i] <= start, assign the meeting: set rooms[i] = end, increment meeting_count[i], and move to the next meeting.
- While scanning, track min_room as the room with the earliest end time.
- If no room is free, assign to min_room: set rooms[min_room] += (end - start) and increment meeting_count[min_room].
Return the index with the maximum count (smallest index on ties).

class Solution:
    def mostBooked(self, n: int, meetings: List[List[int]]) -> int:
        meetings.sort()
        rooms = [0] * n # end time of meetings in rooms
        meeting_count = [0] * n

        for s, e in meetings:
            min_room = 0
            found = False
            for i in range(n):
                if rooms[i] <= s:
                    found = True
                    meeting_count[i] += 1
                    rooms[i] = e
                    break

                if rooms[min_room] > rooms[i]:
                    min_room = i
            if found:
                continue
            meeting_count[min_room] += 1
            rooms[min_room] += e - s

        return meeting_count.index(max(meeting_count))

public class Solution {
    public int mostBooked(int n, int[][] meetings) {
        Arrays.sort(meetings, (a, b) -> Integer.compare(a[0], b[0]));
        long[] rooms = new long[n]; // end times of meetings in rooms
        int[] meetingCount = new int[n];

        for (int[] meeting : meetings) {
            int start = meeting[0], end = meeting[1];
            int minRoom = 0;
            boolean found = false;

            for (int i = 0; i < n; i++) {
                if (rooms[i] <= start) {
                    meetingCount[i]++;
                    rooms[i] = end;
                    found = true;
                    break;
                }
                if (rooms[minRoom] > rooms[i]) {
                    minRoom = i;
                }
            }
            if (found) continue;

            meetingCount[minRoom]++;
            rooms[minRoom] += end - start;
        }

        int maxIndex = 0;
        for (int i = 1; i < n; i++) {
            if (meetingCount[i] > meetingCount[maxIndex]) {
                maxIndex = i;
            }
        }
        return maxIndex;
    }
}

class Solution {
public:
    int mostBooked(int n, vector<vector<int>>& meetings) {
        sort(meetings.begin(), meetings.end());
        vector<long> rooms(n, 0); // end times of meetings in rooms
        vector<int> meetingCount(n, 0);

        for (const auto& meeting : meetings) {
            int start = meeting[0], end = meeting[1];
            int minRoom = 0;
            bool found = false;

            for (int i = 0; i < n; i++) {
                if (rooms[i] <= start) {
                    meetingCount[i]++;
                    rooms[i] = end;
                    found = true;
                    break;
                }
                if (rooms[minRoom] > rooms[i]) {
                    minRoom = i;
                }
            }

            if (found) continue;
            meetingCount[minRoom]++;
            rooms[minRoom] += end - start;
        }

        int maxIndex = 0;
        for (int i = 1; i < n; i++) {
            if (meetingCount[i] > meetingCount[maxIndex]) {
                maxIndex = i;
            }
        }
        return maxIndex;
    }
};

class Solution {
    /**
     * @param {number} n
     * @param {number[][]} meetings
     * @return {number}
     */
    mostBooked(n, meetings) {
        meetings.sort((a, b) => a[0] - b[0]);
        const rooms = new Array(n).fill(0); // end times of meetings in rooms
        const meetingCount = new Array(n).fill(0);

        for (const [start, end] of meetings) {
            let minRoom = 0;
            let found = false;

            for (let i = 0; i < n; i++) {
                if (rooms[i] <= start) {
                    meetingCount[i]++;
                    rooms[i] = end;
                    found = true;
                    break;
                }
                if (rooms[minRoom] > rooms[i]) {
                    minRoom = i;
                }
            }

            if (found) continue;
            meetingCount[minRoom]++;
            rooms[minRoom] += end - start;
        }

        return meetingCount.indexOf(Math.max(...meetingCount));
    }
}

public class Solution {
    public int MostBooked(int n, int[][] meetings) {
        Array.Sort(meetings, (a, b) => a[0].CompareTo(b[0]));
        long[] rooms = new long[n]; // end times of meetings in rooms
        int[] meetingCount = new int[n];

        foreach (var meeting in meetings) {
            int start = meeting[0], end = meeting[1];
            int minRoom = 0;
            bool found = false;

            for (int i = 0; i < n; i++) {
                if (rooms[i] <= start) {
                    meetingCount[i]++;
                    rooms[i] = end;
                    found = true;
                    break;
                }
                if (rooms[minRoom] > rooms[i]) {
                    minRoom = i;
                }
            }

            if (!found) {
                meetingCount[minRoom]++;
                rooms[minRoom] += end - start;
            }
        }

        int maxIndex = 0;
        for (int i = 1; i < n; i++) {
            if (meetingCount[i] > meetingCount[maxIndex]) {
                maxIndex = i;
            }
        }

        return maxIndex;
    }
}

func mostBooked(n int, meetings [][]int) int {
    sort.Slice(meetings, func(i, j int) bool {
        return meetings[i][0] < meetings[j][0]
    })
    rooms := make([]int64, n)
    meetingCount := make([]int, n)

    for _, meeting := range meetings {
        start, end := meeting[0], meeting[1]
        minRoom := 0
        found := false

        for i := 0; i < n; i++ {
            if rooms[i] <= int64(start) {
                meetingCount[i]++
                rooms[i] = int64(end)
                found = true
                break
            }
            if rooms[minRoom] > rooms[i] {
                minRoom = i
            }
        }

        if !found {
            meetingCount[minRoom]++
            rooms[minRoom] += int64(end - start)
        }
    }

    maxIndex := 0
    for i := 1; i < n; i++ {
        if meetingCount[i] > meetingCount[maxIndex] {
            maxIndex = i
        }
    }
    return maxIndex
}

class Solution {
    fun mostBooked(n: Int, meetings: Array<IntArray>): Int {
        meetings.sortBy { it[0] }
        val rooms = LongArray(n)
        val meetingCount = IntArray(n)

        for (meeting in meetings) {
            val start = meeting[0]
            val end = meeting[1]
            var minRoom = 0
            var found = false

            for (i in 0 until n) {
                if (rooms[i] <= start) {
                    meetingCount[i]++
                    rooms[i] = end.toLong()
                    found = true
                    break
                }
                if (rooms[minRoom] > rooms[i]) {
                    minRoom = i
                }
            }

            if (!found) {
                meetingCount[minRoom]++
                rooms[minRoom] += (end - start).toLong()
            }
        }

        var maxIndex = 0
        for (i in 1 until n) {
            if (meetingCount[i] > meetingCount[maxIndex]) {
                maxIndex = i
            }
        }
        return maxIndex
    }
}

class Solution {
    func mostBooked(_ n: Int, _ meetings: [[Int]]) -> Int {
        let meetings = meetings.sorted { $0[0] < $1[0] }
        var rooms = [Int64](repeating: 0, count: n)
        var meetingCount = [Int](repeating: 0, count: n)

        for meeting in meetings {
            let start = meeting[0], end = meeting[1]
            var minRoom = 0
            var found = false

            for i in 0..<n {
                if rooms[i] <= Int64(start) {
                    meetingCount[i] += 1
                    rooms[i] = Int64(end)
                    found = true
                    break
                }
                if rooms[minRoom] > rooms[i] {
                    minRoom = i
                }
            }

            if !found {
                meetingCount[minRoom] += 1
                rooms[minRoom] += Int64(end - start)
            }
        }

        var maxIndex = 0
        for i in 1..<n {
            if meetingCount[i] > meetingCount[maxIndex] {
                maxIndex = i
            }
        }
        return maxIndex
    }
}

Time & Space Complexity

Time complexity: $O(m\log m + n * m)$
Space complexity: $O(n)$

Where $n$ is the number of rooms and $m$ is the number of meetings.

2. Two Min-Heaps

Intuition

Scanning all rooms for each meeting is slow. Instead, we use two min-heaps: one for available rooms (ordered by room number) and one for rooms in use (ordered by end time, then room number). When a meeting starts, we first move all rooms whose meetings have ended back to the available heap. If available is empty, we pop the earliest-ending room, delay the meeting, and push that room back to available. Then we pop the smallest available room, assign the meeting, and push it to the used heap.

Algorithm

Sort meetings by start time.
Initialize a min-heap available with room numbers 0 to n - 1.
Initialize an empty min-heap used storing (end_time, room).
Initialize an array count of size n.
For each (start, end):
- While used is not empty and used[0].end_time <= start, pop from used and push the room to available.
- If available is empty, pop (end_time, room) from used, update end = end_time + (end - start), and push room to available.
- Pop room from available, push (end, room) to used, and increment count[room].
Return the index with maximum count.

class Solution:
    def mostBooked(self, n: int, meetings: List[List[int]]) -> int:
        meetings.sort()
        available = [i for i in range(n)]  # Min heap for available rooms
        used = []  # Min heap for used rooms [(end_time, room_number)]
        count = [0] * n  # Count of meetings for each room

        for start, end in meetings:
            while used and used[0][0] <= start:
                _, room = heapq.heappop(used)
                heapq.heappush(available, room)

            if not available:
                end_time, room = heapq.heappop(used)
                end = end_time + (end - start)
                heapq.heappush(available, room)

            room = heapq.heappop(available)
            heapq.heappush(used, (end, room))
            count[room] += 1

        return count.index(max(count))

public class Solution {
    public int mostBooked(int n, int[][] meetings) {
        Arrays.sort(meetings, (a, b) -> Long.compare(a[0], b[0]));
        PriorityQueue<Integer> available = new PriorityQueue<>();
        PriorityQueue<long[]> used = new PriorityQueue<>((a, b) ->
            a[0] == b[0] ? Long.compare(a[1], b[1]) : Long.compare(a[0], b[0])
        );
        for (int i = 0; i < n; i++) {
            available.offer(i);
        }
        int[] count = new int[n];

        for (int[] meeting : meetings) {
            long start = meeting[0];
            long end = meeting[1];
            while (!used.isEmpty() && used.peek()[0] <= start) {
                int room = (int) used.poll()[1];
                available.offer(room);
            }
            if (available.isEmpty()) {
                long[] current = used.poll();
                int room = (int) current[1];
                end = current[0] + (end - start);
                available.offer(room);
            }

            int room = available.poll();
            used.offer(new long[]{end, room});
            count[room]++;
        }

        int maxRoom = 0;
        for (int i = 1; i < n; i++) {
            if (count[i] > count[maxRoom]) {
                maxRoom = i;
            }
        }
        return maxRoom;
    }
}

class Solution {
public:
    int mostBooked(int n, vector<vector<int>>& meetings) {
        sort(meetings.begin(), meetings.end(), [](const vector<int>& a, const vector<int>& b) {
            return (long long)a[0] < (long long)b[0];
        });
        priority_queue<int, vector<int>, greater<int>> available;
        priority_queue<pair<long long, int>, vector<pair<long long, int>>, greater<pair<long long, int>>> used;
        for (int i = 0; i < n; i++) {
            available.push(i);
        }
        vector<int> count(n);

        for (const auto& meeting : meetings) {
            long long start = meeting[0];
            long long end = meeting[1];
            while (!used.empty() && used.top().first <= start) {
                int room = used.top().second;
                used.pop();
                available.push(room);
            }
            if (available.empty()) {
                auto current = used.top();
                used.pop();
                end = current.first + (end - start);
                available.push(current.second);
            }

            int room = available.top();
            available.pop();
            used.push({end, room});
            count[room]++;
        }

        int maxRoom = 0;
        for (int i = 1; i < n; i++) {
            if (count[i] > count[maxRoom]) {
                maxRoom = i;
            }
        }
        return maxRoom;
    }
};

class Solution {
    /**
     * @param {number} n
     * @param {number[][]} meetings
     * @return {number}
     */
    mostBooked(n, meetings) {
        meetings.sort((a, b) => a[0] - b[0]);
        const available = new PriorityQueue((a, b) => a - b);
        const used = new PriorityQueue((a, b) =>
            a[0] === b[0] ? a[1] - b[1] : a[0] - b[0],
        );
        for (let i = 0; i < n; i++) {
            available.enqueue(i);
        }
        const count = new Array(n).fill(0);

        for (const [start, end] of meetings) {
            while (!used.isEmpty() && used.front()[0] <= start) {
                const room = used.dequeue()[1];
                available.enqueue(room);
            }

            let room;
            let newEnd = end;
            if (available.isEmpty()) {
                const [endTime, usedRoom] = used.dequeue();
                newEnd = endTime + (end - start);
                available.enqueue(usedRoom);
            }
            room = available.dequeue();
            used.enqueue([newEnd, room]);
            count[room]++;
        }

        return count.indexOf(Math.max(...count));
    }
}

public class Solution {
    public int MostBooked(int n, int[][] meetings) {
        Array.Sort(meetings, (a, b) => a[0].CompareTo(b[0]));

        var available = new SortedSet<int>();
        for (int i = 0; i < n; i++) {
            available.Add(i);
        }

        var used = new PriorityQueue<(long end, int room), (long end, int room)>(
            Comparer<(long end, int room)>.Create((a, b) =>
                a.end != b.end ? a.end.CompareTo(b.end) : a.room.CompareTo(b.room))
        );

        int[] count = new int[n];

        foreach (var meeting in meetings) {
            long start = meeting[0], end = meeting[1];

            while (used.Count > 0 && used.Peek().end <= start) {
                var (_, room) = used.Dequeue();
                available.Add(room);
            }

            if (available.Count == 0) {
                var (prevEnd, room) = used.Dequeue();
                end = prevEnd + (end - start);
                available.Add(room);
            }

            int assignedRoom = available.Min;
            available.Remove(assignedRoom);
            used.Enqueue((end, assignedRoom), (end, assignedRoom));
            count[assignedRoom]++;
        }

        int maxRoom = 0;
        for (int i = 1; i < n; i++) {
            if (count[i] > count[maxRoom]) {
                maxRoom = i;
            }
        }

        return maxRoom;
    }
}

func mostBooked(n int, meetings [][]int) int {
    sort.Slice(meetings, func(i, j int) bool {
        return meetings[i][0] < meetings[j][0]
    })

    available := &IntHeap{}
    heap.Init(available)
    for i := 0; i < n; i++ {
        heap.Push(available, i)
    }

    used := &RoomHeap{}
    heap.Init(used)
    count := make([]int, n)

    for _, meeting := range meetings {
        start, end := int64(meeting[0]), int64(meeting[1])

        for used.Len() > 0 && (*used)[0][0] <= start {
            room := heap.Pop(used).([2]int64)[1]
            heap.Push(available, int(room))
        }

        if available.Len() == 0 {
            top := heap.Pop(used).([2]int64)
            end = top[0] + (end - start)
            heap.Push(available, int(top[1]))
        }

        room := heap.Pop(available).(int)
        heap.Push(used, [2]int64{end, int64(room)})
        count[room]++
    }

    maxRoom := 0
    for i := 1; i < n; i++ {
        if count[i] > count[maxRoom] {
            maxRoom = i
        }
    }
    return maxRoom
}

type IntHeap []int
func (h IntHeap) Len() int           { return len(h) }
func (h IntHeap) Less(i, j int) bool { return h[i] < h[j] }
func (h IntHeap) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
func (h *IntHeap) Push(x any)        { *h = append(*h, x.(int)) }
func (h *IntHeap) Pop() any {
    old := *h
    n := len(old)
    x := old[n-1]
    *h = old[0 : n-1]
    return x
}

type RoomHeap [][2]int64
func (h RoomHeap) Len() int { return len(h) }
func (h RoomHeap) Less(i, j int) bool {
    if h[i][0] == h[j][0] {
        return h[i][1] < h[j][1]
    }
    return h[i][0] < h[j][0]
}
func (h RoomHeap) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
func (h *RoomHeap) Push(x any)   { *h = append(*h, x.([2]int64)) }
func (h *RoomHeap) Pop() any {
    old := *h
    n := len(old)
    x := old[n-1]
    *h = old[0 : n-1]
    return x
}

class Solution {
    fun mostBooked(n: Int, meetings: Array<IntArray>): Int {
        meetings.sortBy { it[0] }
        val available = PriorityQueue<Int>()
        val used = PriorityQueue<LongArray>(compareBy({ it[0] }, { it[1] }))

        for (i in 0 until n) {
            available.offer(i)
        }
        val count = IntArray(n)

        for (meeting in meetings) {
            var start = meeting[0].toLong()
            var end = meeting[1].toLong()

            while (used.isNotEmpty() && used.peek()[0] <= start) {
                val room = used.poll()[1].toInt()
                available.offer(room)
            }

            if (available.isEmpty()) {
                val current = used.poll()
                end = current[0] + (end - start)
                available.offer(current[1].toInt())
            }

            val room = available.poll()
            used.offer(longArrayOf(end, room.toLong()))
            count[room]++
        }

        var maxRoom = 0
        for (i in 1 until n) {
            if (count[i] > count[maxRoom]) {
                maxRoom = i
            }
        }
        return maxRoom
    }
}

class Solution {
    func mostBooked(_ n: Int, _ meetings: [[Int]]) -> Int {
        let meetings = meetings.sorted { $0[0] < $1[0] }
        var available = Array(0..<n)
        var used: [(end: Int64, room: Int)] = []
        var count = [Int](repeating: 0, count: n)

        func heapifyAvailable() {
            available.sort()
        }

        func heapifyUsed() {
            used.sort { ($0.end, $0.room) < ($1.end, $1.room) }
        }

        heapifyAvailable()

        for meeting in meetings {
            let start = Int64(meeting[0])
            var end = Int64(meeting[1])

            heapifyUsed()
            while !used.isEmpty && used[0].end <= start {
                let room = used.removeFirst().room
                available.append(room)
                heapifyAvailable()
            }

            if available.isEmpty {
                heapifyUsed()
                let current = used.removeFirst()
                end = current.end + (end - start)
                available.append(current.room)
                heapifyAvailable()
            }

            let room = available.removeFirst()
            used.append((end: end, room: room))
            count[room] += 1
        }

        var maxRoom = 0
        for i in 1..<n {
            if count[i] > count[maxRoom] {
                maxRoom = i
            }
        }
        return maxRoom
    }
}

Time & Space Complexity

Time complexity: $O(m\log m + m \log n)$
Space complexity: $O(n)$

Where $n$ is the number of rooms and $m$ is the number of meetings.

3. One Min-Heap

Intuition

We can simplify to a single heap that tracks (end_time, room). Initially all rooms have end time 0. Before assigning a meeting, we update any rooms whose end time is before the meeting's start to have end time equal to the start (they become available). Then we pop the room with the smallest end time (and smallest room number on ties), assign the meeting, and push the room back with the new end time.

Algorithm

Sort meetings by start time.
Initialize a min-heap available with (0, room) for each room 0 to n - 1.
Initialize an array count of size n.
For each (start, end):
- While available[0].end_time < start, pop (_, room) and push (start, room).
- Pop (end_time, room) from the heap.
- Push (end_time + (end - start), room) back to the heap.
- Increment count[room].
Return the index with maximum count.

class Solution:
    def mostBooked(self, n: int, meetings: List[List[int]]) -> int:
        meetings.sort()
        available = []
        count = [0] * n

        for i in range(n):
            heapq.heappush(available, (0, i))

        for start, end in meetings:
            while available and available[0][0] < start:
                end_time, room = heapq.heappop(available)
                heapq.heappush(available, (start, room))

            end_time, room = heapq.heappop(available)
            heapq.heappush(available, (end_time + (end - start), room))
            count[room] += 1

        return count.index(max(count))

public class Solution {
    public int mostBooked(int n, int[][] meetings) {
        Arrays.sort(meetings, (a, b) -> Integer.compare(a[0], b[0]));
        PriorityQueue<long[]> available = new PriorityQueue<>((a, b) ->
            a[0] == b[0] ? Long.compare(a[1], b[1]) : Long.compare(a[0], b[0])
        );
        for (int i = 0; i < n; i++) {
            available.offer(new long[]{0, i});
        }
        int[] count = new int[n];

        for (int[] meeting : meetings) {
            int start = meeting[0], end = meeting[1];
            while (!available.isEmpty() && available.peek()[0] < start) {
                long[] earliest = available.poll();
                available.offer(new long[]{start, earliest[1]});
            }

            long[] room = available.poll();
            long endTime = room[0] + (end - start);
            available.offer(new long[]{endTime, room[1]});
            count[(int) room[1]]++;
        }

        int maxRoom = 0;
        for (int i = 1; i < n; i++) {
            if (count[i] > count[maxRoom]) {
                maxRoom = i;
            }
        }
        return maxRoom;
    }
}

class Solution {
public:
    int mostBooked(int n, vector<vector<int>>& meetings) {
        sort(meetings.begin(), meetings.end());
        priority_queue<pair<long long, int>, vector<pair<long long, int>>, greater<pair<long long, int>>> available;
        for (int i = 0; i < n; i++) {
            available.push({0, i});
        }
        vector<int> count(n);

        for (const auto& meeting : meetings) {
            int start = meeting[0], end = meeting[1];
            while (!available.empty() && available.top().first < start) {
                auto [end_time, room] = available.top();
                available.pop();
                available.push({start, room});
            }

            auto [end_time, room] = available.top();
            available.pop();
            available.push({end_time + (end - start), room});
            count[room]++;
        }

        return max_element(count.begin(), count.end()) - count.begin();
    }
};

class Solution {
    /**
     * @param {number} n
     * @param {number[][]} meetings
     * @return {number}
     */
    mostBooked(n, meetings) {
        meetings.sort((a, b) => a[0] - b[0]);
        const available = new PriorityQueue((a, b) =>
            a[0] === b[0] ? a[1] - b[1] : a[0] - b[0],
        );
        for (let i = 0; i < n; i++) {
            available.enqueue([0, i]);
        }
        const count = new Array(n).fill(0);

        for (const [start, end] of meetings) {
            while (!available.isEmpty() && available.front()[0] < start) {
                const [endTime, room] = available.dequeue();
                available.enqueue([start, room]);
            }

            const [endTime, room] = available.dequeue();
            available.enqueue([endTime + (end - start), room]);
            count[room]++;
        }

        return count.indexOf(Math.max(...count));
    }
}

public class Solution {
    public int MostBooked(int n, int[][] meetings) {
        Array.Sort(meetings, (a, b) => a[0].CompareTo(b[0]));

        var pq = new PriorityQueue<(long end, int room), (long end, int room)>(
            Comparer<(long end, int room)>.Create((a, b) =>
                a.end != b.end ? a.end.CompareTo(b.end) : a.room.CompareTo(b.room))
        );

        for (int i = 0; i < n; i++) {
            pq.Enqueue((0, i), (0, i));
        }

        int[] count = new int[n];

        foreach (var meeting in meetings) {
            int start = meeting[0], end = meeting[1];

            List<(long end, int room)> temp = new List<(long end, int room)>();

            // Make sure rooms are not idle before meeting start
            while (pq.Count > 0 && pq.Peek().end < start) {
                var room = pq.Dequeue();
                temp.Add((start, room.room));
            }

            foreach (var item in temp) {
                pq.Enqueue(item, item);
            }

            var current = pq.Dequeue();
            long newEnd = current.end + (end - start);
            pq.Enqueue((newEnd, current.room), (newEnd, current.room));
            count[current.room]++;
        }

        int maxRoom = 0;
        for (int i = 1; i < n; i++) {
            if (count[i] > count[maxRoom]) {
                maxRoom = i;
            }
        }

        return maxRoom;
    }
}

func mostBooked(n int, meetings [][]int) int {
    sort.Slice(meetings, func(i, j int) bool {
        return meetings[i][0] < meetings[j][0]
    })

    available := &RoomHeap{}
    heap.Init(available)
    for i := 0; i < n; i++ {
        heap.Push(available, [2]int64{0, int64(i)})
    }
    count := make([]int, n)

    for _, meeting := range meetings {
        start, end := int64(meeting[0]), int64(meeting[1])

        for available.Len() > 0 && (*available)[0][0] < start {
            top := heap.Pop(available).([2]int64)
            heap.Push(available, [2]int64{start, top[1]})
        }

        room := heap.Pop(available).([2]int64)
        newEnd := room[0] + (end - start)
        heap.Push(available, [2]int64{newEnd, room[1]})
        count[room[1]]++
    }

    maxRoom := 0
    for i := 1; i < n; i++ {
        if count[i] > count[maxRoom] {
            maxRoom = i
        }
    }
    return maxRoom
}

type RoomHeap [][2]int64
func (h RoomHeap) Len() int { return len(h) }
func (h RoomHeap) Less(i, j int) bool {
    if h[i][0] == h[j][0] {
        return h[i][1] < h[j][1]
    }
    return h[i][0] < h[j][0]
}
func (h RoomHeap) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
func (h *RoomHeap) Push(x any)   { *h = append(*h, x.([2]int64)) }
func (h *RoomHeap) Pop() any {
    old := *h
    n := len(old)
    x := old[n-1]
    *h = old[0 : n-1]
    return x
}

class Solution {
    fun mostBooked(n: Int, meetings: Array<IntArray>): Int {
        meetings.sortBy { it[0] }
        val available = PriorityQueue<LongArray>(compareBy({ it[0] }, { it[1] }))
        for (i in 0 until n) {
            available.offer(longArrayOf(0, i.toLong()))
        }
        val count = IntArray(n)

        for (meeting in meetings) {
            val start = meeting[0].toLong()
            val end = meeting[1].toLong()

            while (available.isNotEmpty() && available.peek()[0] < start) {
                val earliest = available.poll()
                available.offer(longArrayOf(start, earliest[1]))
            }

            val room = available.poll()
            val endTime = room[0] + (end - start)
            available.offer(longArrayOf(endTime, room[1]))
            count[room[1].toInt()]++
        }

        var maxRoom = 0
        for (i in 1 until n) {
            if (count[i] > count[maxRoom]) {
                maxRoom = i
            }
        }
        return maxRoom
    }
}

Time & Space Complexity

Time complexity:
- $O(m \log m + m \log n)$ time in average case.
- $O(m \log m + m * n)$ time in worst case.
Space complexity: $O(n)$

Where $n$ is the number of rooms and $m$ is the number of meetings.

Common Pitfalls

Forgetting to Sort Meetings by Start Time

Meetings must be processed in chronological order. Without sorting, you cannot correctly determine which room becomes available first or when meetings need to be delayed.

Integer Overflow When Calculating Delayed End Times

When meetings are delayed, end times can grow very large. Using int instead of long can cause overflow. Always use long or equivalent 64-bit integer types for end times.

Incorrect Tie-Breaking for Room Selection

When multiple rooms are available, you must choose the room with the smallest index. When multiple rooms become free at the same time, you must also prefer the smaller index. Failing to implement this tie-breaking correctly leads to wrong answers.

Not Preserving Meeting Duration When Delaying

When a meeting is delayed because no room is available, the meeting duration must remain the same. The new end time should be earliest_available_time + (original_end - original_start), not simply the original end time.

Confusing Available vs Used Room Logic with Two Heaps

When using two heaps, it is easy to forget to move rooms from the used heap back to the available heap when their meetings end. Always check and transfer rooms whose end time is at or before the current meeting's start time before attempting to assign a room.