759. Employee Free Time

 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 timefor 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

/*

// Definition for an Interval.

class Interval {

public:

    int start;

    int end;

    Interval() {}

    Interval(int _start, int _end) {

        start = _start;

        end = _end;

    }

};

*/

struct comp {

    bool operator()(Interval& first, Interval& second) {

        return first.start > second.start;

    }

};

class Solution {

public:

    vector<Interval> employeeFreeTime(vector<vector<Interval>> schedule) {

        vector<Interval> ans;

        priority_queue<Interval, vector<Interval>, comp> pq;

        

        for (auto sch : schedule) {

            for (auto sc : sch) {

                pq.push(sc);

            }

        }

        

        int last_end = -1;

        while(!pq.empty()) {

            Interval it = pq.top(); pq.pop();

            if (last_end != -1 && it.start > last_end) {

                ans.push_back(Interval(last_end, it.start));

            }

            last_end = max(last_end, it.end);

        }

        return ans;

    }

};

/*   Another approach ==

class Solution {

public:

    vector<Interval> employeeFreeTime(vector<vector<Interval>> schedule) {

        vector<pair<int, int>> sched;

        for(int i=0; i< schedule.size(); i++)

        {

            for(int j=0; j< schedule[i].size(); j++)

            {

                sched.push_back({schedule[i][j].start, 1});

                sched.push_back({schedule[i][j].end,-1});

            }   

        }

        sort(sched.begin(), sched.end());

        vector<Interval> res;

        int bal=0,prev=-1;

        for(int i=0; i<sched.size(); i++)

        {

            if(prev!=-1 && bal ==0 && prev != sched[i].first)

            {

                res.push_back({prev, sched[i].first});

            }

            bal+= sched[i].second;

            prev = sched[i].first;

        }

        return res;

    }

};

*/