Calender 2

 

Approach 2: Line Sweep

Intuition

The previous approach works well for the given problem, where we need to avoid triple bookings. However, if the requirements change such as checking for four overlapping bookings, the method becomes less flexible. We'd need to introduce additional lists, for example, to track triple bookings, making the solution harder to maintain and extend.

To address this, we can use a more flexible and standard solution: the Line Sweep algorithm. This approach is common for interval-related problems and can easily handle changes, such as checking for four or more overlapping bookings.

The Line Sweep algorithm works by marking when bookings start and end. For each booking (start, end), we mark the start point by increasing its count by 1 (indicating a booking begins), and we mark the end point by decreasing its count by 1 (indicating a booking ends). These marks are stored in a map, which keeps track of the number of bookings starting or ending at each point.

Once all bookings are processed, we compute the prefix sum over the map. The prefix sum at any point tells us how many active bookings overlap at that moment. If the sum at any point exceeds 2, it means we have a triple booking. At this point, the function should return false to prevent adding a new booking. If no triple booking is found, the function returns true, and the booking is allowed.

This approach is easily extendible. If we wanted to check for four or more bookings instead of three, we would simply adjust the threshold from 2 to 3 when calculating the prefix sum. This flexibility makes the Line Sweep method a more robust solution for variations of the problem.

fig

Algorithm

  1. Class MyCalendarTwo will have two data members, maxOverlappedBooking which is the maximum number of concurrent bookings possible at a time, and bookingCount which is a map from integer to integer with the time point as the key and number of bookings as the value.

  2. Initialize maxOverlappedBooking as 2, as we need to check for triple booking.

  3. Define the function book(start, end) as:

    • Increase the number of bookings for the time start and decrease the number of bookings for end by 1 in the map bookingCount.
    • Iterate over each key-value pair in the map in ascending order of keys to find the prefix sum. Add the value in the map to the count overlappedBooking.
    • If overlappedBooking is more than two, it implies that this is triple booking. Hence, we should return false. Also, we need to revert the changes in the map as this booking shouldn't be added.
    • If we reach here, it implies no triple booking and hence returns true.

Comments

Post a Comment

Popular posts from this blog

30. Substring with Concatenation of All Words

Problem Statement You are given a string s and an array of strings words. All the strings of words are of the same length. A concatenated substring in s is a substring that contains all the strings of any permutation of words concatenated. For example, if words = ["ab","cd","ef"], then "abcdef", "abefcd", "cdabef", "cdefab", "efabcd", and "efcdab" are all concatenated strings. "acdbef" is not a concatenated substring because it is not the concatenation of any permutation of words. Return the starting indices of all the concatenated substrings in s. You can return the answer in any order. Example 1: Input: s = "barfoothefoobarman", words = ["foo","bar"] Output: [0,9] Explanation: Since words.length == 2 and words[i].length == 3, the concatenated substring has to be of length 6. The substring starting at 0 is "barfoo". It is the concaten...
Read more

Lowest common ancestor in a binary Tree

Approach Define a recursive function rec that takes three parameters: root (the current node), p (first node to find the lowest common ancestor for), and q (second node to find the lowest common ancestor for). Check if the current node root is null or if it is either p or q . If any of these conditions is true, return the current node root as the lowest common ancestor. Recursively call the rec function for the left subtree of the current node and assign the result to a variable l . Recursively call the rec function for the right subtree of the current node and assign the result to a variable r . Check if both l and r are not null. If so, it means that p and q are found on different subtrees of the current node, and the current node root is their lowest common ancestor. Return the current node root . If l is null, it means that both p and q are on the right subtree (or not present in the tree). Return r . If none of the above conditions are met, it means ...
Read more

New start day 1 : 30/09/2020 (codeforces 71A) Way to long strings

 Question:  source (codeforces: https://codeforces.com/problemset/problem/71/A) Way too long strings   Sometimes some words like " localization " or " internationalization " are so long that writing them many times in one text is quite tiresome. Let's consider a word  too long if its length is  strictly more  than  10  characters. All too long words should be replaced with a special abbreviation. This abbreviation is made like this: we write down the first and the last letter of a word and between them, we write the number of letters between the first and the last letters. That number is in the decimal system and doesn't contain any leading zeroes. Thus, " localization " will be spelled as " l10n ", and " internationalization » will be spelled as " i18n ". You are suggested to automatize the process of changing the words with abbreviations. At that, all too long words should be replaced by the abbreviation and the words th...
Read more