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
Flatten the free time, and sweep from left to right, if there’s an overlap, then merge the interval.
Time complexity: o ( n ) o(n) o(n)
Space complexity: o ( n ) o(n) 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]':
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