LeetCode: Employee Free Time

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:

1. schedule and schedule[i] are lists with lengths in range [1, 50].
2. 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