https://neetcode.io/problems/meeting-schedule-ii?list=neetcode150
initial
as long as u get the pattern its solveable. i think sort via start time and storing each end times in a heap and if current iterating start time is greater or equal to the minimum end time in heap, we pop that in heap and add the new end date? else we just add the current end date to heap cuz we need extra room? That is exactly correct
sol
class Solution {
public int minMeetingRooms(List<Interval> intervals) {
PriorityQueue<Integer> pq = new PriorityQueue<>();
intervals.sort(Comparator.comparingInt(interval->interval.start));
for(Interval interval:intervals){
if(pq.isEmpty()){
pq.offer(interval.end);
continue;
}
if(interval.start>=pq.peek()){
pq.poll();
}
pq.offer(interval.end);
}
return pq.size();
}
}complexity
n log n cuz of sort
worst case n space
