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;
}
}
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