bucket-sort

american flag sort optimization

由 假装没事ソ 提交于 2019-12-20 03:46:15
问题 I am trying to implement American Bucket Sort. Wiki says "first to count the number of objects that will fall in each bin, and second to place each object in its bucket." In second phase, when placing objects in proper buckets, Do I need to use auxiliary array? Is there a way to do this by swapping array elements in linear time? 回答1: Assuming you mean http://en.wikipedia.org/wiki/American_flag_sort, then as the article points out at the top, you can run this in-place (although then this is

Can the Duplicate Characters in a string be Identified and Quantified in O(n)?

由 非 Y 不嫁゛ 提交于 2019-12-08 15:45:27
问题 This comment suggests that there is a O(n) alternative to my O(n log n) solution to this problem: Given string str("helloWorld") the expected output is: l = 3 o = 2 My solution was to do this: sort(begin(str), end(str)); for(auto start = adjacent_find(cbegin(str), cend(str)), finish = upper_bound(start, cend(str), *start); start != cend(str); start = adjacent_find(finish, cend(str)), finish = upper_bound(start, cend(str), *start)) { cout << *start << " = " << distance(start, finish) << endl;

Mapping fields of weakly related tables in SQL

由 南楼画角 提交于 2019-12-07 08:32:46
问题 I'm searching for an SQL-Query that can map a set of items of an individual size to a set off buckets of individual size. I would like to satisfy the following conditions: The size of a bucket has to be bigger or equal the size of an item. Every bucket can contain only one item or it is left empty. Every item can only be placed in one bucket. No item can be split to multiple buckets. I want to fill the buckets in a way, that the smallest unused buckets are filled first. Then initial item and

Mapping fields of weakly related tables in SQL

由 此生再无相见时 提交于 2019-12-05 13:13:24
I'm searching for an SQL-Query that can map a set of items of an individual size to a set off buckets of individual size. I would like to satisfy the following conditions: The size of a bucket has to be bigger or equal the size of an item. Every bucket can contain only one item or it is left empty. Every item can only be placed in one bucket. No item can be split to multiple buckets. I want to fill the buckets in a way, that the smallest unused buckets are filled first. Then initial item and bucket sets can be ordered by size or id, but are not incremental Sizes and ids of initial bucket and

Radix sort vs Counting sort vs Bucket sort. What's the difference?

由 为君一笑 提交于 2019-12-03 01:50:49
问题 I am reading the definitions of radix, counting and bucket sorts and it seems that all of them are just the code below: public static void sort(int[] a, int maxVal){ int [] bucket=new int[maxVal+1]; for (int i=0; i<bucket.length; i++){ bucket[i]=0; } for (int i=0; i<a.length; i++){ bucket[a[i]]++; } int outPos=0; for (int i=0; i<bucket.length; i++){ for (int j=0; j<bucket[i]; j++){ a[outPos++]=i; } } } I know I can't be right, so what am I missing? Show the code if you think that can help

Radix sort vs Counting sort vs Bucket sort. What's the difference?

由 北城余情 提交于 2019-12-02 14:02:58
I am reading the definitions of radix, counting and bucket sorts and it seems that all of them are just the code below: public static void sort(int[] a, int maxVal){ int [] bucket=new int[maxVal+1]; for (int i=0; i<bucket.length; i++){ bucket[i]=0; } for (int i=0; i<a.length; i++){ bucket[a[i]]++; } int outPos=0; for (int i=0; i<bucket.length; i++){ for (int j=0; j<bucket[i]; j++){ a[outPos++]=i; } } } I know I can't be right, so what am I missing? Show the code if you think that can help explaining in Java or C. Konstantin Vladimirov Let's start with some rewriting your code in C, because C

What is the worst case complexity for bucket sort?

由 天涯浪子 提交于 2019-11-30 15:07:56
问题 I just read the Wikipedia page about Bucket sort. In this article they say that the worst case complexity is O(n²). But I thought the worst case complexity was O(n + k) where k are the number of buckets. This is how I calculate this complexity: Add the element to the bucket. Using a linked list this is O(1) Going through the list and put the elements in the correct bucket = O(n) Merging the buckets = O(k) O(1) * O(n) + O(k) = O(n + k) Am I missing something? 回答1: What if the algorithm decides

What is the worst case complexity for bucket sort?

由 ε祈祈猫儿з 提交于 2019-11-30 13:28:48
I just read the Wikipedia page about Bucket sort . In this article they say that the worst case complexity is O(n²). But I thought the worst case complexity was O(n + k) where k are the number of buckets. This is how I calculate this complexity: Add the element to the bucket. Using a linked list this is O(1) Going through the list and put the elements in the correct bucket = O(n) Merging the buckets = O(k) O(1) * O(n) + O(k) = O(n + k) Am I missing something? What if the algorithm decides that every element belongs in the same bucket? In that case, the linked list in that bucket needs to be
订阅 bucket-sort