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:
Constraints:
1 <= schedule.length , schedule[i].length <= 500 <= schedule[i].start < schedule[i].end <= 10^8
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> res = new ArrayList<>();
PriorityQueue<Interval> pq = new PriorityQueue<>((a, b) -> (a.start - b.start));
for (List<Interval> list : schedule){
for (Interval interval : list){
pq.offer(interval);
}
}
Interval cur = pq.poll();
while(!pq.isEmpty()){
Interval next = pq.poll();
if (cur.end >= next.start){
next.end = Math.max(cur.end, next.end);
}else{
res.add(new Interval(cur.end, next.start));
}
cur = next;
}
return res;
}
}
// TC: O(nlogn)
// SC: O(n)