# 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 forallemployees, 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`.

## Solution & Analysis

``````/**
* 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>> avails) {
List<Interval> result = new ArrayList<>();
List<Interval> timeLine = new ArrayList<>();
Collections.sort(timeLine, ((a, b) -> a.start - b.start));

Interval temp = timeLine.get(0);
for (Interval each : timeLine) {
if (temp.end < each.start) {