leetcode - 759. 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

Solution

Flatten the free time, and sweep from left to right, if there’s an overlap, then merge the interval.

Time complexity: $o(n)$
Space complexity: $o(n)$

Code

"""
# 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]':
        new_schedule = []
        for employ_list in schedule:
            for item in employ_list:
                new_schedule.append((item.start, item.end))
        new_schedule.sort(key=lambda x: x[0])
        res = []
        prev_start, prev_end = new_schedule[0]
        for i in range(1, len(new_schedule)):
            if new_schedule[i][0] > prev_end:
                res.append([prev_end, new_schedule[i][0]])
                prev_start, prev_end = new_schedule[i]
            else:
                prev_start = min(prev_start, new_schedule[i][0])
                prev_end = max(prev_end, new_schedule[i][1])
        ret = []
        for start, end in res:
            ret.append(Interval(start, end))
        return ret