2022-10-07 Daily-Challenge
Today I have done leetcode's October LeetCoding Challenge with cpp.
October LeetCoding Challenge 7
Description
My Calendar III
A k-booking happens when k events have some non-empty intersection (i.e., there is some time that is common to all k events.)
You are given some events [start, end), after each given event, return an integer k representing the maximum k-booking between all the previous events.
Implement the MyCalendarThree class:
MyCalendarThree()Initializes the object.int book(int start, int end)Returns an integerkrepresenting the largest integer such that there exists ak-booking in the calendar.
Example 1:
Input
["MyCalendarThree", "book", "book", "book", "book", "book", "book"]
[[], [10, 20], [50, 60], [10, 40], [5, 15], [5, 10], [25, 55]]
Output
[null, 1, 1, 2, 3, 3, 3]
Explanation
MyCalendarThree myCalendarThree = new MyCalendarThree();
myCalendarThree.book(10, 20); // return 1, The first event can be booked and is disjoint, so the maximum k-booking is a 1-booking.
myCalendarThree.book(50, 60); // return 1, The second event can be booked and is disjoint, so the maximum k-booking is a 1-booking.
myCalendarThree.book(10, 40); // return 2, The third event [10, 40) intersects the first event, and the maximum k-booking is a 2-booking.
myCalendarThree.book(5, 15); // return 3, The remaining events cause the maximum K-booking to be only a 3-booking.
myCalendarThree.book(5, 10); // return 3
myCalendarThree.book(25, 55); // return 3
Constraints:
0 <= start < end <= 109- At most
400calls will be made tobook.
Solution
class MyCalendarThree {
map<int, int> count;
public:
MyCalendarThree() { }
int book(int start, int end) {
count[start] += 1;
count[end] -= 1;
int answer = 0;
int current = 0;
for(const auto &[pos, cnt] : count) {
current += cnt;
answer = max(answer, current);
}
return answer;
}
};
// Accepted
// 98/98 cases passed (292 ms)
// Your runtime beats 22.72 % of cpp submissions
// Your memory usage beats 94.41 % of cpp submissions (262.3 MB)