Embedded Files
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.
# Definition for an interval.
# class Interval:
# def __init__(self, s=0, e=0):
# self.start = s
# self.end = e
class Solution:
def employeeFreeTime(self, schedule):
"""
:type schedule: List[List[Interval]]
:rtype: List[Interval]
"""
list = []
for list_sch in schedule:
for interval in list_sch:
list.append(interval)
list.sort(key = lambda x : x.start)
merged = []
previous = None if len(list) == 0 else list[0]
for point in list:
if(previous.end >= point.start):
previous.end = max(point.end,previous.end)
else:
merged.append(previous)
previous = point
if(previous != None):
merged.append(previous)
free_time = []
if(len(merged) > 1):
previous = None if len(merged) == 0 else merged[0]
for i in range(1, len(merged)):
point = merged[i]
free = Interval(previous.end, point.start)
free_time.append(free)
previous = point
return free_time
Page updated
Report abuse