731 My Calendar Ii
731 My Calendar Ii
My Calendar II 
You are implementing a program to use as your calendar. We can add a new event if adding the event will not cause a triple booking.
A triple booking happens when three events have some non-empty intersection (i.e., some moment is common to all the three events.).
The event can be represented as a pair of integers start and end that represents a booking on the half-open interval [start, end), the range of real numbers x such that start <= x < end.
Implement the MyCalendarTwo class:
1
2
MyCalendarTwo() Initializes the calendar object.
boolean book(int start, int end) Returns true if the event can be added to the calendar successfully without causing a **triple booking**. Otherwise, return false and do not add the event to the calendar.
Example 1:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
**Input**
["MyCalendarTwo", "book", "book", "book", "book", "book", "book"]
[[], [10, 20], [50, 60], [10, 40], [5, 15], [5, 10], [25, 55]]
**Output**
[null, true, true, true, false, true, true]
**Explanation**
MyCalendarTwo myCalendarTwo = new MyCalendarTwo();
myCalendarTwo.book(10, 20); // return True, The event can be booked.
myCalendarTwo.book(50, 60); // return True, The event can be booked.
myCalendarTwo.book(10, 40); // return True, The event can be double booked.
myCalendarTwo.book(5, 15); // return False, The event cannot be booked, because it would result in a triple booking.
myCalendarTwo.book(5, 10); // return True, The event can be booked, as it does not use time 10 which is already double booked.
myCalendarTwo.book(25, 55); // return True, The event can be booked, as the time in [25, 40) will be double booked with the third event, the time [40, 50) will be single booked, and the time [50, 55) will be double booked with the second event.
Constraints:
1
2
0 <= start < end <= 109
At most 1000 calls will be made to book.
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
class MyCalendarTwo:
def __init__(self):
self.cal = []
def book(self, start: int, end: int) -> bool:
overlap = []
for idx1, k in enumerate(self.cal):
if k[0] <= start and start < k[1]:
s = start
e = min(end, k[1])
for idx2,l in enumerate(self.cal):
if l[0] <= s and s < l[1] and idx1 != idx2:
#print(start, end, l[0], l[1], s , e)
return False
if l[0] < e and e <= l[1] and idx1 != idx2:
#print(start, end, l[0], l[1], s , e)
return False
if l[0] < e and l[0] >= s and idx1 != idx2:
return False
elif k[0] < end and end <= k[1]:
s = max(k[0],start)
e = end
for idx2,l in enumerate(self.cal):
if l[0] <= s and s < l[1] and idx1 != idx2:
#print(start, end, l[0], l[1], s , e)
return False
if l[0] < e and e <= l[1] and idx1 != idx2:
#print(start, end, l[0], l[1], s , e)
return False
if l[0] < e and l[0] >= s and idx1 != idx2:
return False
elif k[0] < end and k[0] >= start:
s = k[0]
e = k[1]
for idx2,l in enumerate(self.cal):
if l[0] <= s and s < l[1] and idx1 != idx2:
#print(start, end, l[0], l[1], s , e)
return False
if l[0] < e and e <= l[1] and idx1 != idx2:
#print(start, end, l[0], l[1], s , e)
return False
if l[0] < e and l[0] >= s and idx1 != idx2:
return False
self.cal.append((start, end))
return True
# Your MyCalendarTwo object will be instantiated and called as such:
# obj = MyCalendarTwo()
# param_1 = obj.book(start,end)
This post is licensed under CC BY 4.0 by the author.