0759. Employee Free Time

https://leetcode.com/problems/employee-free-time

Description

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

ac

class Solution {
    public List<Interval> employeeFreeTime(List<List<Interval>> schedule) {
        List<Interval> res = new ArrayList<Interval>();
        // edge cases
        if (schedule == null || schedule.size() == 0) return res;
        
        PriorityQueue<Interval> pq = new PriorityQueue<>((a, b) -> {return a.start - b.start;});
        schedule.forEach(pq::offer);
        
        Interval prev = pq.poll();
        while (!pq.isEmpty()) {
            Interval curr = pq.poll();
            if (prev.end < curr.start) { // a gap found
                Interval free = new Interval(prev.end, curr.start);
                res.add(free);
                prev = curr;
            } else {
                prev.end = Math.max(prev.end, curr.end);
            }
        }
        
        return res;
    }
}