759. Employee Free Time

We are given a list schedule of employees, which represents the working time for each employee.

Each employee has a list of non-overlapping Intervals, and these intervals are in sorted order.

Return the list of finite intervals representing common, positive-length free time for all employees, also in sorted order.

(Even though we are representing Intervals in the form [x, y], the objects inside are Intervals, not lists or arrays. For example, schedule[0][0].start = 1, schedule[0][0].end = 2, and schedule[0][0][0] is not defined). Also, we wouldn't include intervals like [5, 5] in our answer, as they have zero length.

Example 1:

Input: schedule = [[[1,2],[5,6]],[[1,3]],[[4,10]]]
Output: [[3,4]]
Explanation: There are a total of three employees, and all common
free time intervals would be [-inf, 1], [3, 4], [10, inf].
We discard any intervals that contain inf as they aren't finite.

Example 2:

Input: schedule = [[[1,3],[6,7]],[[2,4]],[[2,5],[9,12]]]
Output: [[5,6],[7,9]]

Constraints:

• 1 <= schedule.length , schedule[i].length <= 50

• 0 <= schedule[i].start < schedule[i].end <= 10^8

class Interval {
public:
    int start;
    int end;

    Interval() {}

    Interval(int _start, int _end) {
        start = _start;
        end = _end;
    }
};

// Greedy
vector<Interval> employeeFreeTime(vector<vector<Interval>> schedule) { // time: O(mn * log(mn)); space: O(mn)
    vector<Interval> res, timeline;
    for (const vector<Interval>& s : schedule) {
        for (const Interval& i : s) {
            timeline.push_back(i);    
        }
    }
    sort(timeline.begin(), timeline.end(), [](const Interval& a, const Interval& b) {
        return a.start < b.start;
    });
    Interval tmp = timeline[0];
    for (const Interval& i : timeline) {
        if (tmp.end < i.start) {
            res.push_back(Interval(tmp.end, i.start));
            tmp = i;
        } else {
            tmp = tmp.end < i.end ? i : tmp;
        }
    }
    return res;
}