Meeting Rooms II: Min-Heap Solution for LeetCode 253

Meeting rooms ii leetcode problem #253 is the single most-asked interval question at Amazon, Bloomberg, Google, and Meta. If you are preparing for a coding interview at any of these companies, this problem should be near the top of your list.

The problem combines two fundamental concepts — interval processing and heap-based resource tracking — into one clean question. Interviewers love it because it tests whether you can move beyond brute force and apply a greedy strategy with the right data structure.

In this walkthrough, you will learn two approaches: the classic min-heap solution and the elegant sweep line alternative. Both run in O(n log n) time, but understanding both gives you flexibility when interviewers ask follow-up questions or present variations.

Given an array of meeting time intervals where each interval is a pair [start, end], find the minimum number of conference rooms required so that no two overlapping meetings share the same room.

For example, given intervals [[0,30],[5,10],[15,20]], the answer is 2. The meeting [0,30] overlaps with both [5,10] and [15,20], but [5,10] and [15,20] do not overlap with each other. So you need one room for the long meeting and one room that can be reused for the two shorter ones.

The brute force approach would check every pair of meetings for overlap, but that leads to O(n^2) time. The key insight is that sorting the meetings and using a min-heap lets you track room availability in O(n log n) time.

The min-heap approach is the most intuitive solution to the meeting rooms ii leetcode problem. The idea is simple: sort all meetings by start time, then use a min-heap to track when each active room becomes free.

Start by sorting the intervals by their start time. Initialize an empty min-heap. For each meeting, check the top of the heap — this represents the earliest time any current room becomes free.

If the earliest ending time is less than or equal to the current meeting's start time, that room is available. Pop it from the heap (the room is being reused). Regardless of whether you reused a room or not, push the current meeting's end time onto the heap.

When you have processed all meetings, the size of the heap equals the minimum number of rooms needed. Each element in the heap represents one room that is still occupied.

The sweep line approach offers an alternative perspective on the leetcode 253 solution. Instead of tracking individual rooms, you count how many meetings are active at any point in time.

Separate all start times and end times into two sorted arrays. Use two pointers to sweep through the timeline. When you encounter a start time, increment the active meeting count. When you encounter an end time, decrement it.

The maximum value the active count reaches during the sweep is the minimum number of rooms needed. This works because the peak concurrent meetings determines the resource requirement.

The sweep line approach is particularly elegant because it reduces the problem to a simple counting exercise. Some interviewers prefer to see this solution because it demonstrates a different way of thinking about interval problems.

The meeting rooms heap approach works because of a greedy insight: when a new meeting arrives, you should always try to reuse the room that becomes free the earliest. The min-heap gives you that room in O(log n) time.

Think of it this way — each element in the heap represents an occupied room, and the value is when that room becomes free. When a new meeting starts, you only need to check one room: the one that frees up soonest. If even that room is still busy, every other room is busy too, so you need a new room.

This greedy choice is always optimal. You never gain anything by holding a room in reserve. If a room is free and a meeting can use it, you should assign it. The min-heap enforces this by always surfacing the earliest available room.

This same pattern appears in many resource allocation problems. The heap tracks resource availability, and the greedy strategy assigns resources as they become free. Once you internalize this, you can apply it to car pooling, task scheduling, and server allocation problems.

Handling edge cases correctly is what separates a good interview answer from a great one. The meeting rooms explained solution must account for several boundary conditions.

The first edge case is an empty input — if there are no meetings, you need 0 rooms. The second is when all meetings overlap — for example [[1,10],[2,10],[3,10]] requires 3 rooms, one per meeting.

The third edge case is when no meetings overlap at all, like [[1,2],[3,4],[5,6]]. Here you only need 1 room because each meeting ends before the next starts. The heap never grows beyond size 1.

The trickiest edge case involves meetings that end exactly when another starts. For intervals [1,5] and [5,10], these do NOT overlap — the first meeting ends at time 5, and the second starts at time 5. They can share a room. Make sure your comparison uses less-than-or-equal, not strictly less-than.

Meeting rooms ii leetcode is more than a single problem — it teaches a reusable pattern for resource allocation under constraints. The min-heap tracks resource availability, and the greedy strategy assigns resources optimally.

The sweep line alternative teaches you to think about intervals as events on a timeline rather than as objects occupying space. This perspective unlocks solutions for problems like car pooling (LeetCode 1094), meeting scheduler (LeetCode 1229), and my calendar (LeetCode 731).

If you want to solidify these patterns, practice the related problems: Merge Intervals (#56), Non-overlapping Intervals (#435), and Task Scheduler (#621). Each one reinforces a different facet of interval and heap reasoning.

Build long-term retention by reviewing interval scheduling rooms problems with spaced repetition. YeetCode flashcards help you drill the min-heap pattern and sweep line technique so you can recall them instantly during your next interview.