I am faced with an application where I have to design a container that has random access (or at least better than O(n)) has inexpensive (O(1)) insert and removal, and stores the data according to the order (rank) specified at insertion.
For example if I have the following array:
[2, 9, 10, 3, 4, 6]
I can call the remove on index 2 to remove 10 and I can also call the insert on index 1 by inserting 13.
After those two operations I would have:
[2, 13, 9, 3, 4, 6]
The numbers are stored in a sequence and insert/remove operations require an index parameter to specify where the number should be inserted or which number should be removed.
My question is, what kind of data structures, besides a Linked List and a vector, could maintain something like this? I am leaning towards a Heap that prioritizes on the next available index. But I have been seeing something about a Fusion Tree being useful (but more in a theoretical sense).
What kind of Data structures would give me the most optimal running time while still keeping memory consumption down? I have been playing around with an insertion order preserving hash table, but it has been unsuccessful so far.
The reason I am tossing out using a std:: vector straight up is because I must construct something that out preforms a vector in terms of these basic operations. The size of the container has the potential to grow to hundreds of thousands of elements, so committing to shifts in a std::vector is out of the question. The same problem lines with a Linked List (even if doubly Linked), traversing it to a given index would take in the worst case O (n/2), which is rounded to O (n).
I was thinking of a doubling linked list that contained a Head, Tail, and Middle pointer, but I felt that it wouldn't be much better.
In a basic usage, to be able to insert and delete at arbitrary position, you can use linked lists. They allow for O(1) insert/remove, but only provided that you have already located the position in the list where to insert. You can insert "after a given element" (that is, given a pointer to an element), but you can not as efficiently insert "at given index".
To be able to insert and remove an element given its index, you will need a more advanced data structure. There exist at least two such structures that I am aware of.
One is a rope structure, which is available in some C++ extensions (SGI STL, or in GCC via #include <ext/rope>). It allows for O(log N) insert/remove at arbitrary position.
Another structure allowing for O(log N) insert/remove is a implicit treap (aka implicit cartesian tree), you can find some information at http://codeforces.com/blog/entry/3767, Treap with implicit keys or https://codereview.stackexchange.com/questions/70456/treap-with-implicit-keys.
Implicit treap can also be modified to allow to find minimal value in it (and also to support much more operations). Not sure whether rope can handle this.
UPD: In fact, I guess that you can adapt any O(log N) binary search tree (such as AVL or red-black tree) for your request by converting it to "implicit key" scheme. A general outline is as follows.
Imagine a binary search tree which, at each given moment, stores the consequitive numbers 1, 2, ..., N as its keys (N being the number of nodes in the tree). Every time we change the tree (insert or remove the node) we recalculate all the stored keys so that they are still from 1 to the new value of N. This will allow insert/remove at arbitrary position, as the key is now the position, but it will require too much time for all keys update.
To avoid this, we will not store keys in the tree explicitly. Instead, for each node, we will store the number of nodes in its subtree. As a result, any time we go from the tree root down, we can keep track of the index (position) of current node — we just need to sum the sizes of subtrees that we have to our left. This allows us, given k, locate the node that has index k (that is, which is the k-th in the standard order of binary search tree), on O(log N) time. After this, we can perform insert or delete at this position using standard binary tree procedure; we will just need to update the subtree sizes of all the nodes changed during the update, but this is easily done in O(1) time per each node changed, so the total insert or remove time will be O(log N) as in original binary search tree.
So this approach allows to insert/remove/access nodes at given position in O(log N) time using any O(log N) binary search tree as a basis. You can of course store the additional information ("values") you need in the nodes, and you can even be able to calculate the minimum of these values in the tree just by keeping the minimum value of each node's subtree.
However, the aforementioned treap and rope are more advanced as they allow also for split and merge operations (taking a substring/subarray and concatenating two strings/arrays).
Consider a skip list, which can implement linear time rank operations in its "indexable" variation.
For algorithms (pseudocode), see A Skip List Cookbook, by Pugh.
It may be that the "implicit key" binary search tree method outlined by #Petr above is easier to get to, and may even perform better.
Related
Given an input stream of numbers ranging from 1 to 10^5 (non-repeating) we need to be able to tell at each point how many numbers smaller than this have been previously encountered.
I tried to use the set in C++ to maintain the elements already encountered and then taking upper_bound on the set for the current number. But upper_bound gives me the iterator of the element and then again I have to iterate through the set or use std::distance which is again linear in time.
Can I maintain some other data structure or follow some other algorithm in order to achieve this task more efficiently?
EDIT : Found an older question related to fenwick trees that is helpful here. Btw I have solved this problem now using segment trees taking hints from #doynax comment.
How to use Binary Indexed tree to count the number of elements that is smaller than the value at index?
Regardless of the container you are using, it is very good idea to enter them as sorted set so at any point we can just get the element index or iterator to know how many elements are before it.
You need to implement your own binary search tree algorithm. Each node should store two counters with total number of its child nodes.
Insertion to binary tree takes O(log n). During the insertion counters of all parents of that new element should be incremented O(log n).
Number of elements that are smaller than the new element can be derived from stored counters O(log n).
So, total running time O(n log n).
Keep your table sorted at each step. Use binary search. At each point, when you are searching for the number that was just given to you by the input stream, binary search is going to find either the next greatest number, or the next smallest one. Using the comparison, you can find the current input's index, and its index will be the numbers that are less than the current one. This algorithm takes O(n^2) time.
What if you used insertion sort to store each number into a linked list? Then you can count the number of elements less than the new one when finding where to put it in the list.
It depends on whether you want to use std or not. In certain situations, some parts of std are inefficient. (For example, std::vector can be considered inefficient in some cases due to the amount of dynamic allocation that occurs.) It's a case-by-case type of thing.
One possible solution here might be to use a skip list (relative of linked lists), as it is easier and more efficient to insert an element into a skip list than into an array.
You have to use the skip list approach, so you can use a binary search to insert each new element. (One cannot use binary search on a normal linked list.) If you're tracking the length with an accumulator, returning the number of larger elements would be as simple as length-index.
One more possible bonus to using this approach is that std::set.insert() is log(n) efficient already without a hint, so efficiency is already in question.
I have an immutable base list of items that I want to perform a number of operations on: edit, add, delete, read. The actual operations will be queued up and performed elsewhere (sent up to the server and a new list will be sent down), but I want a representation of what the list would look like with the current set of operations applied to the base list.
My current implementation keeps a vector of ranges and where they map to. So an unedited list has one range from 0 to length that maps directly to the base list. If an add is performed in index 5, then we have 3 ranges: 0-4 maps to base list 0-4. 5 maps to the new item, and 6-(length+1) maps to 5-length. This works, however with a lot of adds and deletes, reads degrades to O(n).
I've thought of using hashmaps but the shifts in ranges that can occur with inserts and deletes presents a challenge. Is there some way to achieve this so that reads are around O(1) still?
If you had a roughly balanced tree of ranges, where each node kept a record of the total number of elements below it in the tree, you could answer reads in worst case time the depth of the tree, which should be about log(n). Perhaps a https://en.wikipedia.org/wiki/Treap would be one of the easier balanced trees to implement for this.
If you had a lot of repetitious reads and few modifications, you might gain by keeping a hashmap of the results of reads since the last modification, clearing it on modification.
A skip list is a data structure in which the elements are stored in sorted order and each node of the list may contain more than 1 pointer, and is used to reduce the time required for a search operation from O(n) in a singly linked list to O(lg n) for the average case. It looks like this:
Reference: "Skip list" by Wojciech Muła - Own work. Licensed under Public domain via Wikimedia Commons - http://commons.wikimedia.org/wiki/File:Skip_list.svg#mediaviewer/File:Skip_list.svg
It can be seen as an analogy to a ruler:
In a skip list, searching an element and deleting one is fine, but when it comes to insertion, it becomes difficult, because according to Data Structures and Algorithms in C++: 2nd edition by Adam Drozdek:
To insert a new element, all nodes following the node just inserted have to be restructured; the number of pointers and the value of pointers have to be changed.
I can construe from this that although choosing a node with random number of pointers based on the likelihood of nodes to insert a new element, doesn't create a perfect skip list, it gets really close to it when large number of elements (~9 million for example) are considered.
My question is: Why can't we insert the new element in a new node, determine its number of pointers based on the previous node, attach it to the end of the list, and then use efficient sorting algorithms to sort just the data present in the nodes, thereby maintaining the perfect structure of the skip list and also achieving the O(lg n) insert complexity?
Edit: I haven't tried any code yet, I'm just presenting a view. Simply because implementing a skip list is somewhat difficult. Sorry.
There is no need to modify any following nodes when you insert a node. See the original paper, Skip Lists: A Probabilistic Alternative to Balanced Trees, for details.
I've implemented a skip list from that reference, and I can assure you that my insertion and deletion routines do not modify any nodes forward of the insertion point.
I haven't read the book you're referring to, but out of context the passage you highlight is just flat wrong.
You have a problem on this point and then use efficient sorting algorithms to sort just the data present in the nodes. Sorting the data will have complexity O(n*lg(n)) and thus it will increase the complexity of insertion. In theory you can choose "perfect" number of links for each node being inserted, but even if you do that, when you perform remove operations, the perfectness will be "broken". Using the randomized approach is close enough to perfect structure to perform well.
You need to have function / method that search for location.
It need to do following:
if you insert unique keys, it need to locate the node. then you keep everything, just change the data (baggage). e.g. node->data = data.
if you allow duplicates, or if key is not found, then this function / method need to give you previous node on each height (lane). Then you determine height of new node and insert it after the found nodes.
Here is my C realisation:
https://github.com/nmmmnu/HM2/blob/master/hm_skiplist.c
You need to check following function:
static const hm_skiplist_node_t *_hm_skiplist_locate(const hm_skiplist_t *l, const char *key, int complete_evaluation);
it stores the position inside hm_skiplist_t struct.
complete_evaluation is used to save time in case you need the data and you are not intend to insert / delete.
I need a data structure in c++ STL for performing insertion, searching and retrieval of kth element in log(n)
(Note: k is a variable and not a constant)
I have a class like
class myClass{
int id;
//other variables
};
and my comparator is just based on this id and no two elements will have the same id.
Is there a way to do this using STL or I've to write log(n) functions manually to maintain the array in sorted order at any point of time?
Afaik, there is no such datastructure. Of course, std::set is close to this, but not quite. It is a red black tree. If each node of this red black tree was annotated with the tree weight (the number of nodes in the subtree rooted at this node), then a retrieve(k) query would be possible. As there is no such weight annotation (as it takes valuable memory and makes insert/delete more complex as weights have to be updated), it is impossible to answer such a query efficently with any search tree.
If you want to build such a datastructure, use a conventional search tree implementation (red-black,AVL,B-Tree,...) and add a weight field to each node that counts the number of entries in its subtree. Then searching for the k-th entry is quite simple:
Sketch:
Check the weight of the child nodes, and find the child c which has the largest weight (accumulated from left) that is not greater than k
Subtract from k all weights of children that are left of c.
Descend down to c and call this procedure recursively.
In case of a binary search tree, the algorithm is quite simple since each node only has two children. For a B-tree (which is likely more efficient), you have to account as many children as the node contains.
Of course, you must update the weight on insert/delete: Go up the tree from the insert/delete position and increment/decrement the weight of each node up to the root. Also, you must exchange the weights of nodes when you do rotations (or splits/merges in the B-tree case).
Another idea would be a skip-list where the skips are annotated with the number of elements they skip. But this implementation is not trivial, since you have to update the skip length of each skip above an element that is inserted or deleted, so adjusting a binary search tree is less hassle IMHO.
Edit: I found a C implementation of a 2-3-4 tree (B-tree), check out the links at the bottom of this page: http://www.chiark.greenend.org.uk/~sgtatham/algorithms/cbtree.html
You can not achieve what you want with simple array or any other of the built-in containers. You can use a more advanced data structure for instance a skip list or a modified red-black tree(the backing datastructure of std::set).
You can get the k-th element of an arbitrary array in linear time and if the array is sorted you can do that in constant time, but still the insert will require shifting all the subsequent elements which is linear in the worst case.
As for std::set you will need additional data to be stored at each node to be able to get the k-th element efficiently and unfortunately you can not modify the node structure.
I have to implement a data structure which supports the following three functions. The data is a pair(a,b) of two double values and the data is concentrated in a particular region. Let's say with values of 'a' in the range 500-600.
Insert(double a, double b) - Insert the data, a pair(double,double) in the data structure. If the first element of the pair already exists, update its second element to the new value.
Delete(double a) - Delete the data containing the first element = a.
PrintData(int count) - Print the value of the data which has the count-th largest value. Value is compared according to data.first.
The input file contains a series of Insert, Delete and PrintData operations. Currently, I have implemented the data structure as a height balanced binary search tree using STL-Map but it is not fast enough.
Is there any other implementation which is faster than a Map.
We can use caching to store the most common PrintData queries.
I'd recommend 2 binary search trees (BSTs) - one being the map from a to b (sorted by a), the other should be sorted by b.
The second will need to be a custom BST - you'll need to let each node store a count of the number of nodes in the subtree with it as root - these counts can be updated in O(log n), and will allow for O(log n) queries to get the k-th largest element.
When doing an insert, you'll simply look up b's value in the first BST first, then remove that value from the second, then update the first and insert the new value into the second.
For a delete, you'll simply look up b's value in the first BST (and remove that pair), then remove that value from the second.
All mentioned operations should take O(log n).
Caching
If you are, for instance, only going to query the top 10 elements, you could maintain another BST containing only those 10 elements (or even just an optionally-sorted array, since there's only 10 elements), which we'll then query instead of the second BST above.
When inserting, also insert into this structure if the value is greater than the smallest one, and remove the smallest.
When removing, we need to look up the next largest value and insert it into the small BST. Although this could also be done lazily - when removing, just remove it from this BST - don't fill it up to 10 again. When querying, if there are enough elements in this BST, we can just look up using this one, otherwise we find all the values in the big BST required to fill this BST up, then we query.
This would result in best-case O(1) query (worst-case O(log n)), while the other operations will still be O(log n).
Although the added complexity is probably not worth it - O(log n) is pretty fast, even for a large n.
Building on this idea, we could only have this small BST along with the BST mapping a to b - this would require that we check all values to find the required ones during a query after a removal, so it would only really be beneficial if there aren't a whole lot of removals.
I would recommend an indexed skip list. That will give you O(log n) insert and delete, and O(log n) access to the nth largest value (assuming that you maintain the list in descending order).
Skip list isn't any more difficult to implement than a self-balancing binary tree, and gives much better performance in some situations. Well worth considering.
The original skip list paper.