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2020-10-30

253. Meeting Rooms II

Similar Question: LeetCode Question 252

Question:

Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (si < ei), find the minimum number of conference rooms required.

Example 1:

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

Example 2:

Input: [[7,10],[2,4]]
Output: 1

NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.


Solution:

Using the priority queue, check the meeting with earilest ending time. If conflicts, no deque opertation, add a second meeting to the queue. Result is the size of the queue.

class Solution {
    public int minMeetingRooms(int[][] intervals) {
        if (intervals == null || intervals.length == 0) {
            return 0;
        }
        Arrays.sort(intervals, (int[] a, int[] b) -> (a[0] - b[0]));
        PriorityQueue<int[]> pq = new PriorityQueue<>((int[] a, int[] b) -> (a[1] - b[1]));
        pq.offer(intervals[0]);
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i][0] >= pq.peek()[1]) {
                pq.poll();
            } 
            pq.offer(intervals[i]);
        }
        return pq.size();
    }
}

Use two arrays to sort the starting and ending time, if there’s a conflict increase the result.

class Solution {
    public int minMeetingRooms(int[][] intervals) {
        int[] starts = new int[intervals.length];
        int[] ends = new int[intervals.length];
        for(int i=0; i<intervals.length; i++) {
            starts[i] = intervals[i][0];
            ends[i] = intervals[i][1];
        }
        Arrays.sort(starts);
        Arrays.sort(ends);
        int result = 0;
        int endPointer = 0;
        int room = 0;
        for(int i=0; i<starts.length; ) {
            if(starts[i] < ends[endPointer]) {
                room++;
                result = Math.max(result, room);
                i++;
            } else {
                endPointer++;
                room--;
            }
        }
        return result;
    }
}

More concise way to use sort. Think as a bus problem with onboarding and offboarding process. Think the endpointer same as the peek() in prioirtyqueue. A better explanation: https://learnku.com/articles/34713

class Solution {
    public int minMeetingRooms(int[][] intervals) {
        int[] starts = new int[intervals.length];
        int[] ends = new int[intervals.length];
        for(int i=0; i<intervals.length; i++) {
            starts[i] = intervals[i][0];
            ends[i] = intervals[i][1];
        }
        Arrays.sort(starts);
        Arrays.sort(ends);
        int result = 0;
        int endPointer = 0;
        int room = 0;
        for(int i=0; i<starts.length; i++) {
            if(starts[i] < ends[endPointer]) {
                room++;
            } else {
                endPointer++;
            }
        }
        return room;
    }
}

Using hashmap to mark the time and travese the time.

class Solution {
    public int minMeetingRooms(int[][] intervals) {
        Map<Integer, Integer> map = new TreeMap<>();
        for (int[] itl : intervals) {
            map.put(itl[0], map.getOrDefault(itl[0], 0) + 1);
            map.put(itl[1], map.getOrDefault(itl[1], 0) - 1);
        }
        int room = 0, k = 0; 
        for (int v : map.values()) 
            k = Math.max(k, room += v); 
        
        return k; 
    }
}