old_contents: # Time:  O(m * logn), m is the number of schedule, n is the number of employees, m >= n
  # Space: O(n)
  
  # Definition for an interval.
  # 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
  
new_contents: # Time:  O(m * logn), m is the number of schedule, n is the number of employees, m >= n
  # Space: O(n)
  
  # 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(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
  
subject: Update employee-free-time.py