Problem description:

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.

(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.

Example 1:

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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:

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Input: schedule = [[[1,3],[6,7]],[[2,4]],[[2,5],[9,12]]]
Output: [[5,6],[7,9]]

Constraints:

  • 1 <= schedule.length , schedule[i].length <= 50
  • 0 <= schedule[i].start < schedule[i].end <= 10^8

Solution:

takes into account that all employees lists are already sorted. heap helps to iterate schedule in a sorted manner for O(n*log(k)).

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"""
# 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]':
        heap = []
        for i in range(len(schedule)):
            heapq.heappush(heap, (schedule[i][0].start, i, 0))
        
        res = []
        prev_end = 0
        while heap:
            start, i, j = heapq.heappop(heap)
            if not prev_end:
                prev_end = schedule[i][j].end
            
            if j+1 < len(schedule[i]):
                heapq.heappush(heap, (schedule[i][j+1].start, i, j+1))
            
            if start > prev_end:
                res.append(Interval(prev_end, start))
            prev_end = max(schedule[i][j].end, prev_end)

        return res

time complexity: $O(nlogk)$, n is number of intervals across all employees, k is number of employees
space complexity: $O(k)$
reference:
related problem: