-
Notifications
You must be signed in to change notification settings - Fork 2
/
EmployeeFreeTime.java
119 lines (100 loc) · 3.7 KB
/
EmployeeFreeTime.java
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
package Leetcode;
import java.util.ArrayList;
import java.util.List;
import java.util.PriorityQueue;
/**
* @author kalpak
*
* 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
*/
public class EmployeeFreeTime {
static class Interval {
public int start;
public int end;
public Interval() {}
public Interval(int _start, int _end) {
start = _start;
end = _end;
}
};
public static List<Interval> employeeFreeTime(List<List<Interval>> schedule) {
List<Interval> result = new ArrayList<>();
PriorityQueue<Interval> startTimeMinHeap = new PriorityQueue<>((a, b) -> a.start - b.start);
schedule.forEach(i -> startTimeMinHeap.addAll(i));
Interval temp = startTimeMinHeap.poll();
while(!startTimeMinHeap.isEmpty()) {
if(temp.end < startTimeMinHeap.peek().start) { // disjoint interval
result.add(new Interval(temp.end, startTimeMinHeap.peek().start));
temp = startTimeMinHeap.poll();
} else {
// overlapping or intersecting intervals
temp = temp.end < startTimeMinHeap.peek().end ? startTimeMinHeap.peek() : temp;
startTimeMinHeap.poll();
}
}
return result;
}
public static void printResult(List<Interval> result) {
for(Interval r :result)
System.out.println(r.start + ", " + r.end);
System.out.println();
}
public static void main(String[] args) {
List<List<Interval>> interval = new ArrayList<>();
Interval i1 = new Interval(1, 2);
Interval i2 = new Interval(5, 6);
List<Interval> employee1 = new ArrayList<>();
employee1.add(i1);
employee1.add(i2);
interval.add(employee1);
Interval i3 = new Interval(1, 3);
Interval i4 = new Interval(4, 10);
List<Interval> employee2 = new ArrayList<>();
employee2.add(i3);
employee2.add(i4);
interval.add(employee2);
printResult(employeeFreeTime(interval));;
interval = new ArrayList<>();
i1 = new Interval(1, 3);
i2 = new Interval(6, 7);
employee1 = new ArrayList<>();
employee1.add(i1);
employee1.add(i2);
interval.add(employee1);
i3 = new Interval(2, 4);
i4 = new Interval(2, 5);
Interval i5 = new Interval(9, 12);
employee2 = new ArrayList<>();
employee2.add(i3);
employee2.add(i4);
employee2.add(i5);
interval.add(employee2);
printResult(employeeFreeTime(interval));
}
}