Hard

Employee Free TimePython

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