Hard
Employee Free Time — Python
Full explanation · Time O(m * logn) · Space O(n)
# Time: O(m * logn), m is the number of schedule, n is the number of employees, m >= n
# Space: O(n)
import heapq
class Interval(object):
def __init__(self, s=0, e=0):
self.start = s
self.end = e
class Solution(object):
def employeeFreeTime(self, schedule):
"""
:type schedule: List[List[Interval]]
:rtype: List[Interval]
"""
result = []
min_heap = [(emp[0].start, eid, 0) for eid, emp in enumerate(schedule)]
heapq.heapify(min_heap)
last_end = -1
while min_heap:
t, eid, i = heapq.heappop(min_heap)
if 0 <= last_end < t:
result.append(Interval(last_end, t))
last_end = max(last_end, schedule[eid][i].end)
if i+1 < len(schedule[eid]):
heapq.heappush(min_heap, (schedule[eid][i+1].start, eid, i+1))
return result