My Calendar I

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You are implementing a program to use as your calendar. We can add a new event if adding the event will not cause a double booking.

A double booking happens when two events have some non-empty intersection (i.e., some moment is common to both 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 MyCalendar class:

MyCalendar() 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 double booking. Otherwise, return false and do not add the event to the calendar.

Example 1:

Input ["MyCalendar", "book", "book", "book"] [[], [10, 20], [15, 25], [20, 30]] Output [null, true, false, true]

Explanation

1MyCalendar myCalendar = new MyCalendar();
2myCalendar.book(10, 20); // return True
3myCalendar.book(15, 25); // return False, It can not be booked because time 15 is already booked by another event.
4myCalendar.book(20, 30); // return True, The event can be booked, as the first event takes every time less than 20, but not including 20.

Constraints:

0 <= start < end <= 109
At most 1000 calls will be made to book.

Solution

To implement the booking behaviour, we will use binary search to find a potential insertion index, then check whether the new booking can be actually scheduled into our calendar by checking whether the new booking overlaps with calendar[idx-1] and calendar[idx].

Implementation

1class MyCalendar:
2    def __init__(self):
3        self.calendar = []
4
5    def book(self, start: int, end: int) -> bool:
6        left, right, idx = 0, len(self.calendar)-1, len(self.calendar)
7        while left <= right:
8          mid = (left + right) // 2
9          if self.calendar[mid][0] > start:
10              idx = mid
11              right = mid - 1
12          else:
13              left = mid + 1
14        # check if calendar[idx-1] or calendar[idx] overlaps with start and end
15        if (idx > 0 and self.calendar[idx-1][1] > start) or (idx < len(self.calendar) and self.calendar[idx][0] < end):
16            return False
17        self.calendar.insert(idx, (start, end))
18        return True

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Question 1 out of 10

Given a sorted array of integers and an integer called target, find the element that equals to the target and return its index. Select the correct code that fills the ___ in the given code snippet.

1def binary_search(arr, target):
2    left, right = 0, len(arr) - 1
3    while left ___ right:
4        mid = (left + right) // 2
5        if arr[mid] == target:
6            return mid
7        if arr[mid] < target:
8            ___ = mid + 1
9        else:
10            ___ = mid - 1
11    return -1
12

1public static int binarySearch(int[] arr, int target) {
2    int left = 0;
3    int right = arr.length - 1;
4
5    while (left ___ right) {
6        int mid = left + (right - left) / 2;
7        if (arr[mid] == target) return mid;
8        if (arr[mid] < target) {
9            ___ = mid + 1;
10        } else {
11            ___ = mid - 1;
12        }
13    }
14    return -1;
15}
16

1function binarySearch(arr, target) {
2    let left = 0;
3    let right = arr.length - 1;
4
5    while (left ___ right) {
6        let mid = left + Math.trunc((right - left) / 2);
7        if (arr[mid] == target) return mid;
8        if (arr[mid] < target) {
9            ___ = mid + 1;
10        } else {
11            ___ = mid - 1;
12        }
13    }
14    return -1;
15}
16

<=, left, right

<, left, right

<=, right, left

<, right, left

I don’t know

My Calendar I

Solution

Implementation

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