[LeetCode] Meeting Rooms II

Problem Description:

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.

For example,
Given [[0, 30],[5, 10],[15, 20]],
return 2.

---

The idea is to group those non-overlapping meetings in the same room and then count how many rooms we need. You may refer to this link.

The code is as follows.

class Solution {
public:
    int minMeetingRooms(vector<Interval>& intervals) {
        sort(intervals.begin(), intervals.end(), compare);
        vector<vector<Interval>> rooms;
        int n = intervals.size();
        for (int i = 0; i < n; i++) {
            int idx = findNonOverlapping(rooms, intervals[i]);
            if (rooms.empty() || idx == -1)
                rooms.push_back({intervals[i]});
            else rooms[idx].push_back(intervals[i]);
        }
        return (int)rooms.size();
    }
private:
    static bool compare(Interval& interval1, Interval& interval2) {
        return interval1.start < interval2.start;
    }
    int findNonOverlapping(vector<vector<Interval>>& rooms, Interval& interval) {
        int n = rooms.size();
        for (int i = 0; i < n; i++)
            if (interval.start >= rooms[i].back().end)
                return i;
        return -1;
    }
};

Update: for each group of non-overlapping intervals, we just need to store the last added one instead of the full list. So we could use a vector<Interval> instead of vector<vector<Interval>> in C++. The code is now as follows.

class Solution { 
public:
    int minMeetingRooms(vector<Interval>& intervals) {
        sort(intervals.begin(), intervals.end(), compare);
        vector<Interval> rooms;
        int n = intervals.size();
        for (int i = 0; i < n; i++) {
            int idx = findNonOverlapping(rooms, intervals[i]);
            if (rooms.empty() || idx == -1)
                rooms.push_back(intervals[i]);
            else rooms[idx] = intervals[i];
        }
        return (int)rooms.size();
    } 
private:
    static bool compare(Interval& interval1, Interval& interval2) {
        return interval1.start < interval2.start;
    }
    int findNonOverlapping(vector<Interval>& rooms, Interval& interval) {
        int n = rooms.size();
        for (int i = 0; i < n; i++)
            if (interval.start >= rooms[i].end)
                return i;
        return -1;
    }
};