Heap / Priority Queue - Meeting Rooms II

Given an array of meeting time intervals [start, end], find the minimum number of conference rooms required.

Input: [[0,30],[5,10],[15,20]] → Output: 2

Input: [[7,10],[2,4]] → Output: 1

Sort intervals by start time. Use a min-heap of end times. For each meeting, if earliest-ending room is free, reuse it. Otherwise open a new room.

Sort intervals by start time.
Min-heap stores end times of occupied rooms.
For each interval: if heap.top() <= interval.start, poll (room freed).
Push interval.end to heap.
Heap size at end = number of rooms needed.

import java.util.*; class Solution { public int minMeetingRooms(int[][] intervals) { Arrays.sort(intervals, (a, b) -> a[0] - b[0]); PriorityQueue<Integer> minHeap = new PriorityQueue<>(); for (int[] interval : intervals) { if (!minHeap.isEmpty() && minHeap.peek() <= interval[0]) minHeap.poll(); minHeap.offer(interval[1]); } return minHeap.size(); } }

Time Complexity: O(N log N)
Space Complexity: O(N)