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 .
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 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
