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.

 

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.

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

 

Solution:

/*
// Definition for an Interval.
class Interval {
    public int start;
    public int end;

    public Interval() {}

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

class Solution {
    public List<Interval> employeeFreeTime(List<List<Interval>> schedule) {
        List<Interval> flatten = new ArrayList<>();
        for (List<Interval> l : schedule) {
            flatten.addAll(l);
        }
        Collections.sort(flatten, (a, b) -> { return Integer.compare(a.start, b.start); });
        List<Interval> result = new ArrayList<>();
        Interval curr = flatten.get(0);
        for (int i = 1; i < flatten.size(); i ++) {
            Interval next = flatten.get(i);
            if (next.start > curr.end) {
                result.add(new Interval(curr.end, next.start));
                curr = next;
            } else {
                curr.end = Math.max(curr.end, next.end);
            }
        }
        return result;
    }
}