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 <= 500 <= schedule[i].start < schedule[i].end <= 10^8
Companies:
DoorDash, Amazon, Pinterest, Wayfair, Intuit, Microsoft, Uber, Facebook
Similar Questions:
// OJ: https://leetcode.com/problems/employee-free-time/
// Author: github.com/lzl124631x
// Time: O(NlogT + T) where N is the total number of intervals, and T is the total number of unique times.
// Space: O(T)
class Solution {
public:
vector<Interval> employeeFreeTime(vector<vector<Interval>> A) {
map<int, int> m;
for (auto &v : A) {
for (auto &it : v) {
m[it.start]++;
m[it.end]--;
}
}
vector<Interval> ans;
int cnt = 0;
for (auto it = m.begin(); it != m.end(); ++it) {
cnt += it->second;
if (cnt) continue;
int start = it->first;
++it;
if (it == m.end()) break;
cnt += it->second;
ans.emplace_back(start, it->first);
}
return ans;
}
};