759. Employee Free Time

Question

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.

Solution

Method 1: Sort

class Solution {
      private static final Comparator<int[]> cmp = new Comparator<int[]>(){
          @Override
          public int compare(int[] a, int[] b){
              return a[0] == b[0] ? (a[1] - b[1]): (a[0] - b[0]);
          }
      };
      public int[][] employeeFreeTime(int[][][] schedule) {
          List<int[]> list = new ArrayList<>();
          for(int[][] person : schedule){	// O(N), N is the total number of intervals
              for(int[] interval : person)
                  list.add(interval);
          }
          if(list.size() == 0) return new int[0][2];
          Collections.sort(list, cmp);	//O(NlgN)
          List<int[]> res = new ArrayList<>();
          int end = list.get(0)[1];
          for(int i = 1; i < list.size(); i++){	// O(N)
              int[] cur = list.get(i);
              if(cur[0] > end){
                  res.add(new int[]{end, cur[0]});
              }
              end = Math.max(end, cur[1]);
          }
          return res.toArray(new int[0][2]);
      }
  }