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.

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]]

(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.

Note:

1. schedule and schedule[i] are lists with lengths in range [1, 50].

2. 0 <= schedule[i].start < schedule[i].end <= 10^8.

// Can also be done in NlogK with PQ, where K = Number of employees
/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
class Solution {
    public List<Interval> employeeFreeTime(List<List<Interval>> schedule) {
        if (schedule == null || schedule.size() == 0)
            return null;
        // Putting all intervals in a single list
        List<Interval> list = new ArrayList<>();
        for (List<Interval> l : schedule) {
            list.addAll(l);
        }
        // Merging overlapping intervals
        List<Interval> merged = merge(list);
        if (merged.size() == 1)
            return null;
        List<Interval> result = new ArrayList<>();
        for (int i = 1; i < merged.size(); i++) {
            Interval prev = merged.get(i - 1);
            Interval cur = merged.get(i);
            if (prev.end < cur.start) {
                result.add(new Interval(prev.end, cur.start));
            }
        }
        return result;
    }
    // Helper function to Sort & Merge overlapping intervals
    private static List<Interval> merge(List<Interval> intervals) {
        Collections.sort(intervals, (a, b) -> a.start - b.start);
        if (intervals.size() <= 1)
            return intervals;
        List<Interval> result = new ArrayList<>();
        for (Interval itvl : intervals) {
            if (result.size() == 0)
                result.add(itvl);
            else {
                Interval it = result.get(result.size() - 1);
                if (it.end >= itvl.start) {
                    Interval interval = new Interval(Math.min(it.start, itvl.start), Math.max(it.end, itvl.end));
                    result.add(interval);
                    result.remove(result.size() - 2);
                } else
                    result.add(itvl);
            }
        }
        return result;
    }
}