# 759 Employee Free Time

## 759. [Employee Free Time](https://leetcode.com/problems/employee-free-time/description/)

## 1. Question

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`.

## 2. Implementation

**(1) Heap**

```java
/**
 * 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) {
        List<Interval> res = new ArrayList<>();

        if (schedule == null || schedule.size() == 0) {
            return res;
        }

        List<Interval> list = new ArrayList<>();

        for (List<Interval> time : schedule) {
            for (Interval interval : time) {
                list.add(interval);
            }
        }

        Collections.sort(list, new Comparator<Interval>() {
            @Override
            public int compare(Interval i1, Interval i2) {
                return i1.start == i2.start ? i1.end - i2.end : i1.start - i2.start;
            }
        });

        PriorityQueue<Integer> endTime = new PriorityQueue<>(Collections.reverseOrder());
        endTime.add(list.get(0).end);

        for (int i = 1; i < list.size(); i++) {
            if (endTime.size() > 0 && endTime.peek() < list.get(i).start) {
                res.add(new Interval(endTime.peek(), list.get(i).start));
            }
            endTime.add(list.get(i).end);
        }
        return res;
    }
}
```

## 3. Time & Space Complexity

**Heap:** 时间复杂度O(nlogn), n为所有employee的工作时间段，空间复杂度O(n)


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