Please edit the question to make it a lot clearer what you are asking. Stack Exchange network consists of 177 Q&A communities including Well, $O(n\log n + n) = O(n\log n)$. Should the answer not be 4?This problem considers that two intervals overlap with 1 starts at time t1 and the other end at the same time t1.this can be solved in O(nlogn) with greedy algo. Anybody can answer [S] Fit a simple fitted rectangle i.e., parallel to the axes X and Y you may use minmax function for X and Y of the given points (e.g., polygon's vertices) The question asks for "the most efficient" algorithm.
Write a function which produces the set of merged intervals for the given set of intervals. First (Incorrect) Attempt The topic here is greedy algorithms, so an attractive algorithm might be to simply choose intervals that Observe that there is an optimal placement of two vertical lines each of which go through some segment endpoints 2. I need to find the most efficient greedy algorithm for this problem with $O(n\log n)$ time complexity.My thought is to sort the elements in ascending order (using merge sort for minimum time complexity) and then, starting from the first element, use the current element as the lowest limit, and (element+1) as the highest limit.This has a running time of $O(n\log n + n)$. And I'm not sure what better means. If a point is overlapped by just two intervals, it is fine. The best answers are voted up and rise to the top Thus, we print on a new line 4. and after several days looking at this problem off and on, using idiotic loops to compare subsets and getting correct/incorrect answers, I realized I was just looking at combinations. And there are 2 sets of intervals ($[1.5,2.5]$ and $[2.8,3.8]$). Learn more about Stack Overflow the company site design / logo © 2020 Stack Exchange Inc; user contributions licensed under The first problem can be solved in [math]O(N\cdot\log(N))[/math] time per test case. For any given set of real numbers, find the minimum set of intervals with length 1 that include all elements. How would we know that you don't care about efficiency, and only care about a better solution? (3,4) (6,6) (10,10) (11,11) (14,14) (15,15) ----------- (1,5) (6,9) (10,14)We use cookies to ensure you have the best browsing experience on our website. Anybody can ask a question

For set s2, we can choose all intervals without having more than two of them overlap at any given point. Please read our

It is very similar to the "activity selection problem" that can be found in Wiki. Start here for a quick overview of the site And "find the most efficient algorithm" is undecidable problem in general.Thanks for the answer, but my question is whether there is a greedy algorithm that gives a better solution while having O(nlogn) at the same time, not how to improve the time complexity of the algorithm I thought..@Pantelis, I'm confused. For example, for the set: $${\{1.5,2.3,2.4,2.5,2.8,3.3,3.6,3.8}\}$$ It only takes a minute to sign up.For any given set of real numbers, find the minimum set of intervals with length 1 that include all elements. A simple approach is to start from the first interval and compare it with all other intervals for overlapping, if it overlaps with any other interval, then remove the other interval from list and merge the other into the first interval. Discuss the workings and policies of this site My proposal for solution: Find the centroid of the polygon (see Finding center of geometry of object?

This is incorrect, right? Compress segments' That is, there is an optimal solution with end-point at the smallest number.

I am not able to get it .A set of intervals is given.

1. This is a collection of my HackerRank, Codeforces, Geeks for Geeks, and LeetCode problem solutions in C++, Java, and Python 3. determines the smallest set of unit-length closed intervals that contains all of the given points.

In this case, it is a typical method for greedy algorithm: showing your first decision is correct. Contribute to tsyogesh40/HackerRank-solutions development by creating an account on GitHub. Formally, given a set S of n planar points, find the circle C of smallest radius such that all points in S are contained in either C or its boundary. By clicking “Post Your Answer”, you agree to our To subscribe to this RSS feed, copy and paste this URL into your RSS reader. can somebody please explain the problem more clearly. I find the following 9 are the least overlapping.

By using our site, you acknowledge that you have read and understand our Computer Science Stack Exchange is a question and answer site for students, researchers and practitioners of computer science. These are not meant to be competition style implementations but clean and easily understandable To show that the problem has a lower bound of $\Omega (n\log n)$, you can reduce the Thanks for contributing an answer to Computer Science Stack Exchange!



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