761. Employee Free Time

• Difficulty: Hard

• Topics: Heap, Greedy

• Similar Questions:

  • Merge Intervals

  • Interval List Intersections

Problem:

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:

1. schedule and schedule[i] are lists with lengths in range [1, 50].
2. 0 <= schedule[i].start < schedule[i].end <= 10^8.

NOTE: input types have been changed on June 17, 2019. Please reset to default code definition to get new method signature.

Solutions:

/*
// Definition for an Interval.
class Interval {
public:
    int start;
    int end;

    Interval() {}

    Interval(int _start, int _end) {
        start = _start;
        end = _end;
    }
};
*/
class Solution {
public:
    vector<Interval*> employeeFreeTime(vector<vector<Interval*>> schedule) {
        priority_queue<Node, vector<Node>> pq;
        for (int i = 0; i < schedule.size(); ++i) {
            if (schedule[i].size() > 0) {
                pq.push({schedule[i][0]->start, true, i, 0});
                pq.push({schedule[i][0]->end, false, i, 0});
            }
        }

        vector<Interval*> ret;
        int count = 0;
        int start = -1;
        while (!pq.empty()) {
            Node node = pq.top(); pq.pop();
            if (!node.isStart && node.index + 1 < schedule[node.employee].size()) {
                pq.push({schedule[node.employee][node.index + 1]->start, true, node.employee, node.index + 1});
                pq.push({schedule[node.employee][node.index + 1]->end, false, node.employee, node.index + 1});
            }

            if (node.isStart) {
                ++count;
            } else {
                --count;
            }
            if (node.isStart && count == 1 && start != -1) { // node.isStart check is important!
                ret.push_back(new Interval(start, node.timestamp));
                start = -1;
            } else if (!node.isStart && count == 0) {
                start = node.timestamp;
            }
        }

        return ret;
    }

private:
    struct Node {
        int timestamp;
        bool isStart;
        int employee;
        int index;

        Node(int timestamp, bool isStart, int employee, int index) {
            this->timestamp = timestamp;
            this->isStart = isStart;
            this->employee = employee;
            this->index = index;
        }

        bool operator< (const Node& rhs) const {
            if (timestamp != rhs.timestamp) {
                return timestamp > rhs.timestamp;
            }

            return !isStart;
        }
    };

};