divide-and-conquer

Is Quick Sort a Divide & Conquer approach? [closed]

我们两清 提交于 2019-12-06 00:26:26
Closed . This question is opinion-based . It is not currently accepting answers. Want to improve this question? Update the question so it can be answered with facts and citations by editing this post . Closed 2 years ago . I consider Merge sort as divide and conquer because, Divide - Array is literally divided into sub arrays without any processing (compare/swap), and the problem sized is halved/Quartered/.... Conquer - merge() those sub arrays by processing (compare/swap) Code gives an impression that it is Divide&Conquer, if(hi <= lo) return; int mid = lo + (hi-lo)/2; //No (compare/swap) on

Why is the “divide and conquer” method of computing factorials so fast for large ints? [closed]

感情迁移 提交于 2019-12-04 23:39:02
问题 As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance. Closed 7 years ago . I recently decided to look at factorial algorithms for large integers, and this "divide and conquer" algorithm is faster than a simple

Divide and Conquer strategy to determine if more than 1/3 same element in list

你说的曾经没有我的故事 提交于 2019-12-04 12:58:37
I am working an a divide and conquer algorithm to determine if more than 1/3 elements in list are the same. For example: [1,2,3,4] No, all element are unique. [1,1,2,4,5] Yes, 2 of them are the same. Without sorting, is there a divide and conquer strategy ? I get stucked on how to divide... def is_valid(ids): n = len(ids) is_valid_recur(ids, n n-1) def is_valid_recur(ids, l, r): m = (l + h) // 2 return .... is_valid_recur(ids, l, m) ...is_valid_recur(ids, m+1, r): Thanks a lot! Here's a rough draft I experimented with for fun. It looks like the divide and conquer may reduce the number of

Merge skylines, divide and conquer

我的梦境 提交于 2019-12-03 15:56:49
I'm trying to solve the famous skyline problem (see gif): Input (1,11,5), (2,6,7), (3,13,9), (12,7,16), (14,3,25), (19,18,22), (23,13,29), (24,4,28) Should return, the points that are behind other buildings should be gone and the coordinates of changes in the Y-axis should be in the returning skyline: (1, 11), (3, 13), (9, 0), (12, 7), (16, 3), (19, 18), (22, 3), (23, 13), (29, 0) I'm trying to do so by using a divide and conquer approach to the algorithm as to achieve a running time of O(n lg n), but I'm stuck on the merge part. Everytime I think about it I get confused. For example, I check

Divide-And-Conquer Algorithm for Trees

北城以北 提交于 2019-12-03 13:30:31
I am trying to write a divide & conquer algorithm for trees. For the divide step I need an algorithm that partitions a given undirected Graph G=(V,E) with n nodes and m edges into sub-trees by removing a node . All subgraphs should have the property that they don't contain more than n/2 nodes (the tree should be split as equal as possible). First I tried to recursively remove all leaves from the tree to find the last remaining node, then I tried to find the longest path in G and remove the middle node of it. The given graph below shows that both approaches don't work: Is there some working

Dynamic programming and Divide and conquer

大城市里の小女人 提交于 2019-12-03 07:18:06
I was reading notes on Dynamic programming , and I encountered the following comment. If the subproblems are not independent, i.e. subproblems share subsubproblems, then a divideand-conquer algorithm repeatedly solves the common subsubproblems. Thus, it does more work than necessary What does this mean ? Can you give me examples to make the above point clear ? The author refers to the fact that many divide-and-conquer algorithms have subproblems that overlap with one another. Consider, for example, this very simple Fibonacci implementation: int Fibonacci(int n) { if (n <= 1) return n; return

algorithms: how do divide-and-conquer and time complexity O(nlogn) relate?

你说的曾经没有我的故事 提交于 2019-12-03 02:49:47
问题 In my Algorithms and Data Structures class a first divide-and-conquer algorithm namely merge sort was introduced. While implementing an algorithm for an assignment a few questions came to my mind. Does any algorithm that is implemented with the use of the divide and conquer paradigm has time complexity of O(nlogn)? Is it that the recursion part in the approach has the power to condense an algorithm that runs in like O(n^2) to O(nlogn)? What makes such an algorithm run in O(nlogn) in the first

Algorithm for Shuffling a Linked List in n log n time

爷,独闯天下 提交于 2019-12-03 02:26:50
问题 I'm trying to shuffle a linked list using a divide-and-conquer algorithm that randomly shuffles a linked list in linearithmic (n log n) time and logarithmic (log n) extra space. I'm aware that I can do a Knuth shuffle similar to that could be used in a simple array of values, but I'm not sure how I would do this with divide-and-conquer. What I mean is, what am I actually dividing? Do I just divide to each individual node in the list and then randomly assemble the list back together using some

algorithms: how do divide-and-conquer and time complexity O(nlogn) relate?

非 Y 不嫁゛ 提交于 2019-12-02 17:33:13
In my Algorithms and Data Structures class a first divide-and-conquer algorithm namely merge sort was introduced. While implementing an algorithm for an assignment a few questions came to my mind. Does any algorithm that is implemented with the use of the divide and conquer paradigm has time complexity of O(nlogn)? Is it that the recursion part in the approach has the power to condense an algorithm that runs in like O(n^2) to O(nlogn)? What makes such an algorithm run in O(nlogn) in the first place. For (3) I assume that this has something to do with recursion trees and the possible number of

Algorithm for Shuffling a Linked List in n log n time

六月ゝ 毕业季﹏ 提交于 2019-12-02 15:58:21
I'm trying to shuffle a linked list using a divide-and-conquer algorithm that randomly shuffles a linked list in linearithmic (n log n) time and logarithmic (log n) extra space. I'm aware that I can do a Knuth shuffle similar to that could be used in a simple array of values, but I'm not sure how I would do this with divide-and-conquer. What I mean is, what am I actually dividing? Do I just divide to each individual node in the list and then randomly assemble the list back together using some random value? Or do I give each node a random number and then do a mergesort on the nodes based on the