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 .

Difficulty: interview

Starter code:

"""
# 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]':
        

Solutions:
class Solution:
     def intersectionSizeTwo(self, intervals):
         """
         :type intervals: List[List[int]]
         :rtype: int
         """
         intervals.sort(key=lambda x: x[1])
         if len(intervals) <= 1:
             return len(intervals[0])
         s = set()
         prev_e = None
         intersect = False
         for e in intervals:
             if not s:
                 a = e[1] - 1
                 b = e[1]
                 s.add(a)
                 s.add(b)
                 continue
             if e[0] <= a :
                 intersect = True
                 continue
                 
             if e[0] > a and  e[1] > b >= e[0] :
                 intersect = True
                 a = b
                 b = e[-1]
                 s.add(b)
                 continue
             a = e[1] - 1
             b = e[1]
             s.add(a)
             s.add(b)
         
         if not intersect:
             return 0
         return len(s)


class Solution:
     def intersectionSizeTwo(self, intervals):
         """
         :type intervals: List[List[int]]
         :rtype: int
         """
         r1=[x[0] for x in intervals]
         r2=[x[1] for x in intervals]
         intervals=sorted(intervals,key=lambda x: x[1])
         s=[intervals[0][1]-1,intervals[0][1]]
         for interval in intervals:
             if interval[0]>s[-1]:
                     s+=[interval[1]-1,interval[1]]
             elif interval[0]>s[-2]:
                 s+=[interval[1]]
         return len(s)
                 


class Solution:
     def intersectionSizeTwo(self, intervals):
         """
         :type intervals: List[List[int]]
         :rtype: int
         """
         intervals.sort(key=lambda inter: (inter[1], -inter[0]))
         res=0
         left,right=-1,-1 # current second largest, and largest
         for inter in intervals: # assume each inter[0]<inter[1]
             if inter[0]<=left: # two overlap, skip this inter, cc for intervals with the same inter[1]
                 continue # current inter[0] is not greater than previou inter[0], so at least (left,right) are overlap elements
             if inter[0]>right: # greater than current largest, no overlap, need add new [left, right]
                 res+=2
                 right=inter[1] # greedy, use the two largest num in this inter
                 left=right-1 
             elif inter[0]<=right: # one overlap with current largest
                 res+=1
                 left=right
                 right=inter[1]
         return res


class Solution:
     def intersectionSizeTwo(self, intervals):
         """
         :type intervals: List[List[int]]
         :rtype: int
         """
         intervals.sort(key=lambda x: x[1])
         if len(intervals) <= 1:
             return len(intervals[0])
         s = set()
         for e in intervals:
             if not s:
                 a = e[1] - 1
                 b = e[1]
                 s.add(a)
                 s.add(b)
                 continue
             if e[0] <= a :
                 continue
                 
             if e[0] > a and  e[1] > b >= e[0] :
                 a = b
                 b = e[-1]
                 s.add(b)
                 continue
             a = e[1] - 1
             b = e[1]
             s.add(a)
             s.add(b)
         
         return len(s)


class Solution:
     def intersectionSizeTwo(self, intervals):
         res = 0
         last2 = [-float('inf')] * 2
         intervals = sorted(intervals, key=lambda x: x[1])
         for lo, hi in intervals:
             n = sum(x < lo for x in last2)
             res += n
             if n == 2:
                 last2 = hi - 1, hi
             elif n == 1:
                 last2 = last2[1], hi
         return res