LeetCode [759] Employee Free Time

759. Employee Free Time

Hard

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:

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]]

 

Constraints:

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

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bool comp1(const vector<int> &lhs, const vector<int> &rhs) { return lhs[0] < rhs[0]; } class Solution { public: vector<vector<int>> employeeFreeTime(vector<vector<vector<int>>> &schedule) { vector<vector<int>> ss; for (auto s : schedule) for (auto v : s) ss.push_back(v); sort(ss.begin(), ss.end(), comp1); int n = ss.size(), right = ss[0][1]; vector<vector<int>> ret; for(int i=1; i<n; ++i) { if(ss[i][0]>right) ret.push_back(vector<int>{right, ss[i][0]}); right = max(right, ss[i][1]); } return ret; } };

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/* // Definition for an Interval. class Interval { public int start; public int end; public Interval() {} public Interval(int _start, int _end) { start = _start; end = _end; } }; */ class Solution { public List<Interval> employeeFreeTime(List<List<Interval>> avails) { List<Interval> result = new ArrayList<>(); List<Interval> timeLine = new ArrayList<>(); avails.forEach(e -> timeLine.addAll(e)); Collections.sort(timeLine, ((a, b) -> a.start - b.start)); Interval temp = timeLine.get(0); for(Interval each : timeLine) { if(temp.end < each.start) { result.add(new Interval(temp.end, each.start)); temp = each; }else{ temp = temp.end < each.end ? each : temp; } } return result; } }