Identity Employee Free Time

Similar Problems:

- CheatSheet: Leetcode For Code Interview
- CheatSheet: Common Code Problems & Follow-ups
- Tag: #interval, #linesweep, #heap

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:

- schedule and schedule[i] are lists with lengths in range [1, 50].
- 0 <= schedule[i].start < schedule[i].end <= 10^8.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

## https://code.dennyzhang.com/employee-free-time ## Basic Ideas: heap + greedy ## ## skyline problem: line sweep ## ## For each interval, (s, e) ## s: block the following interval ## e: release the following interval ## ## Per requirement, list of list can be used by list ## ## Complexity: Time O(n*log(n)), Space O(n) """ # Definition for an Interval. class Interval: def __init__(self, start: int = None, end: int = None): self.start = start self.end = end """ class Solution: def employeeFreeTime(self, schedule: '[[Interval]]') -> '[Interval]': res = [] intervals = [] for l in schedule: for v in l: intervals.append([v.start, 0]) intervals.append([v.end, 1]) # first sort by time, then type (start, then close) intervals.sort() cover = 0 # how many cover not closed prev = -1 for cur in intervals: if prev != -1 and cover == 0: res.append(Interval(prev, cur[0])) prev = cur[0] cover += 1 if (cur[1]==0) else -1 return res