LeetCode #2402 — HARD

Meeting Rooms III

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an integer n. 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 = [[0,10],[1,5],[2,7],[3,4]]
Output: 0
Explanation:
- 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.

Example 2:

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

Constraints:

1 <= n <= 100
1 <= meetings.length <= 10⁵
meetings[i].length == 2
0 <= start_i < end_i <= 5 * 10⁵
All the values of start_i are unique.

Step 01

Brute Force Baseline

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

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

2
[[0,10],[1,5],[2,7],[3,4]]

Example 2

3
[[1,20],[2,10],[3,5],[4,9],[6,8]]

Core Insight

What unlocks the optimal approach

Sort meetings based on start times.
Use two min heaps, the first one keeps track of the numbers of all the rooms that are free. The second heap keeps track of the end times of all the meetings that are happening and the room that they are in.
Keep track of the number of times each room is used in an array.
With each meeting, check if there are any free rooms. If there are, then use the room with the smallest number. Otherwise, assign the meeting to the room whose meeting will end the soonest.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2402: Meeting Rooms III
class Solution {
    public int mostBooked(int n, int[][] meetings) {
        Arrays.sort(meetings, (a, b) -> a[0] - b[0]);
        PriorityQueue<int[]> busy
            = new PriorityQueue<>((a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
        PriorityQueue<Integer> idle = new PriorityQueue<>();
        for (int i = 0; i < n; ++i) {
            idle.offer(i);
        }
        int[] cnt = new int[n];
        for (var v : meetings) {
            int s = v[0], e = v[1];
            while (!busy.isEmpty() && busy.peek()[0] <= s) {
                idle.offer(busy.poll()[1]);
            }
            int i = 0;
            if (!idle.isEmpty()) {
                i = idle.poll();
                busy.offer(new int[] {e, i});
            } else {
                var x = busy.poll();
                i = x[1];
                busy.offer(new int[] {x[0] + e - s, i});
            }
            ++cnt[i];
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (cnt[ans] < cnt[i]) {
                ans = i;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2402: Meeting Rooms III
func mostBooked(n int, meetings [][]int) int {
	sort.Slice(meetings, func(i, j int) bool { return meetings[i][0] < meetings[j][0] })
	idle := hp{make([]int, n)}
	for i := 0; i < n; i++ {
		idle.IntSlice[i] = i
	}
	busy := hp2{}
	cnt := make([]int, n)
	for _, v := range meetings {
		s, e := v[0], v[1]
		for len(busy) > 0 && busy[0].end <= s {
			heap.Push(&idle, heap.Pop(&busy).(pair).i)
		}
		var i int
		if idle.Len() > 0 {
			i = heap.Pop(&idle).(int)
			heap.Push(&busy, pair{e, i})
		} else {
			x := heap.Pop(&busy).(pair)
			i = x.i
			heap.Push(&busy, pair{x.end + e - s, i})
		}
		cnt[i]++
	}
	ans := 0
	for i, v := range cnt {
		if cnt[ans] < v {
			ans = i
		}
	}
	return ans
}

type hp struct{ sort.IntSlice }

func (h *hp) Push(v any) { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
	a := h.IntSlice
	v := a[len(a)-1]
	h.IntSlice = a[:len(a)-1]
	return v
}

type pair struct{ end, i int }
type hp2 []pair

func (h hp2) Len() int { return len(h) }
func (h hp2) Less(i, j int) bool {
	a, b := h[i], h[j]
	return a.end < b.end || a.end == b.end && a.i < b.i
}
func (h hp2) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
func (h *hp2) Push(v any)   { *h = append(*h, v.(pair)) }
func (h *hp2) Pop() any     { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }

# Accepted solution for LeetCode #2402: Meeting Rooms III
class Solution:
    def mostBooked(self, n: int, meetings: List[List[int]]) -> int:
        meetings.sort(key=lambda x: x[0])
        busy = []
        idle = list(range(n))
        heapify(idle)
        cnt = [0] * n
        for s, e in meetings:
            while busy and busy[0][0] <= s:
                heappush(idle, heappop(busy)[1])
            i = 0
            if idle:
                i = heappop(idle)
                heappush(busy, (e, i))
            else:
                time_end, i = heappop(busy)
                heappush(busy, (time_end + e - s, i))
            cnt[i] += 1
        ans = 0
        for i in range(n):
            if cnt[ans] < cnt[i]:
                ans = i
        return ans

// Accepted solution for LeetCode #2402: Meeting Rooms III
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2402: Meeting Rooms III
// class Solution {
//     public int mostBooked(int n, int[][] meetings) {
//         Arrays.sort(meetings, (a, b) -> a[0] - b[0]);
//         PriorityQueue<int[]> busy
//             = new PriorityQueue<>((a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
//         PriorityQueue<Integer> idle = new PriorityQueue<>();
//         for (int i = 0; i < n; ++i) {
//             idle.offer(i);
//         }
//         int[] cnt = new int[n];
//         for (var v : meetings) {
//             int s = v[0], e = v[1];
//             while (!busy.isEmpty() && busy.peek()[0] <= s) {
//                 idle.offer(busy.poll()[1]);
//             }
//             int i = 0;
//             if (!idle.isEmpty()) {
//                 i = idle.poll();
//                 busy.offer(new int[] {e, i});
//             } else {
//                 var x = busy.poll();
//                 i = x[1];
//                 busy.offer(new int[] {x[0] + e - s, i});
//             }
//             ++cnt[i];
//         }
//         int ans = 0;
//         for (int i = 0; i < n; ++i) {
//             if (cnt[ans] < cnt[i]) {
//                 ans = i;
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2402: Meeting Rooms III
function mostBooked(n: number, meetings: number[][]): number {
    meetings.sort((a, b) => a[0] - b[0]);

    const idle = new MinPriorityQueue<number>();
    for (let i = 0; i < n; ++i) {
        idle.enqueue(i);
    }
    const busy = new PriorityQueue<[number, number]>((a, b) => {
        if (a[0] === b[0]) {
            return a[1] - b[1];
        }
        return a[0] - b[0];
    });
    const cnt: number[] = new Array(n).fill(0);
    for (const v of meetings) {
        const s = v[0],
            e = v[1];
        while (!busy.isEmpty() && busy.front()[0] <= s) {
            const i = busy.dequeue()[1];
            idle.enqueue(i);
        }
        let i = 0;
        if (!idle.isEmpty()) {
            i = idle.dequeue();
            busy.enqueue([e, i]);
        } else {
            const x = busy.dequeue();
            i = x[1];
            busy.enqueue([x[0] + e - s, i]);
        }
        ++cnt[i];
    }
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        if (cnt[ans] < cnt[i]) {
            ans = i;
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m (log m + log n)

Space

O(n + m)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.