I'm currently working on implementing a list-type structure at work, and I need it to be crazy effective. In my search for effective data structures I stumbled across a patent for a quad liked list, and this sparked my interest enough to make me forget about my current task and start investigating the quad list instead. Unfortunately, internet was very secretive about the whole thing, and google didn't produce much in terms of usable results. The only explanation I got was the patent description that stated:
A quad linked data structure that provides bidirectional search capability for multiple related fields within a single record. The data base is searched by providing sets of pointers at intervals of N data entries to accommodate a binary search of the pointers followed by a linear search of the resultant range to locate an item of interest and its related field.
This, unfortunately, just makes me more puzzled, as I cannot wrap my head around the non-layman explanation. So therefore I turn to you all in hope that you can explain to me what this quad linked history really is, as I know not knowing will drive me up and over the walls pretty quickly.
Do you know what a quad linked list is?
I can't be sure, but it sounds a bit like a skip list.
Even if that's not what it is, you might find skip lists handy. (To the best of my knowledge they are unidirectional, however.)
I've not come across the term formally before, but from the patent description, I can make an educated guess.
A linked list is one where each node has a link to the next...
a -->-- b -->-- c -->-- d -->-- null
A doubly linked list means each node holds a link to its predecessor as well.
--<-- --<-- --<--
| | | |
a -->-- b -->-- c -->-- d -->-- null
Let's assume the list is sorted. If I want to perform binary search, I'd normally go half way down the list to find the middle node, then go into the appropriate interval and repeat. However, linked list traversal is always O(n) - I have to follow all the links. From the description, I think they're just adding additional links from a node to "skip" a fixed number of nodes ahead in the list. Something like...
--<-- --<-- --<--
| | | |
a -->-- b -->-- c -->-- d -->-- null
| |
|----------->-----------|
-----------<-----------
Now I can traverse the list more rapidly, especially if I chose the extra link targets carefully (i.e. ensure they always go back/forward half of the offset of the item they point from in the list length). I then find the rough interval I want with these links, and use the normal links to find the item.
This is a good example of why I hate software patents. It's eminently obvious stuff, wrapped in florid prose to confuse people.
I don't know if this is exactly a "quad-linked list", but it sounds like something like this:
struct Person {
// Normal doubly-linked list.
Customer *nextCustomer;
Customer *prevCustomer;
std::string firstName;
Customer *nextByFirstName;
Customer *prevByFirstName;
std::string lastName;
Customer *nextByLastName;
Customer *prevByLastName;
};
That is: you maintain several orderings through your collection. You can easily navigate in firstName order, or in lastName order. It's expensive to keep the links up to date, but it makes navigation quite quick.
Of course, this could be something completely different.
My reading of it is that a quad linked list is one which can be traversed (backwards or forwards) in O(n) in two different ways, ie sorted according to FieldX or FieldY:
(a) generating first and second sets
of link pointers, wherein the first
set of link pointers points to
successor elements of the set of
related records when the records are
ordered with respect to the fixed ID
field, and the second set of link
pointers points to predecessor
elements of the set of related records
when the records are ordered with
respect to the fixed ID field;
(b) generating third and fourth sets
of link pointers, wherein the third
set of link pointers points to
successor elements of the set of
related records when the records are
ordered with respect to the variable
ID field, and the fourth set of link
pointers points to predecessor
elements of the set of related records
when the records are ordered with
respect to the variable ID field;
So if you had a quad linked list of employees you could store it sorted by name AND sorted by age, and enumerate either in O(n).
One source of the patent is this. There are, it appears, two claims, the second of which is more nearly relevant:
A computer implemented method for organizing and searching a set of related records, wherein each record includes:
i) a fixed ID field; and
ii) a variable ID field; the method comprising the steps of:
(a) generating first and second sets of link pointers, wherein the first set of link pointers points to successor elements of the set of related records when the records are ordered with respect to the fixed ID field, and the second set of link pointers points to predecessor elements of the set of related records when the records are ordered with respect to the fixed ID field;
(b) generating third and fourth sets of link pointers, wherein the third set of link pointers points to successor elements of the set of related records when the records are ordered with respect to the variable ID field, and the fourth set of link pointers points to predecessor elements of the set of related records when the records are ordered with respect to the variable ID field;
(c) generating first and second sets of field pointers, wherein the first set of field pointers includes an ordered set of pointers that point to every Nth fixed ID field when the records are ordered with respect to the fixed ID field, and the second set of pointers includes an ordered set of pointers that point to every Nth variable ID field when the records are ordered with respect to the variable ID field;
(d) when searching for a particular record by reference to its fixed ID field, conducting a binary search of the first set of field pointers to determine an initial pointer and a final pointer defining a range within which the particular record is located;
(e) examining by linear scarch, the fixed ID fields within the range determined in step (d) to locate the particular record;
(f) when searching for a particular record by reference to its variable ID field, conducting a binary search of the second set of field pointers to determine an initial pointer and a final pointer defining a range within which the particular record is located;
(g) examining, by linear search, the variable ID fields within the range determined in step (f) to locate the particular record.
When you work through the patent gobbledegook, I think it means approximately the same as having two skip lists (one for forward search, one for backwards search) on each of two keys (hence 4 lists in total, and the name 'quad-list'). I don't think it is a very good patent - it looks to be an obvious application of skip lists to a data set where you have two keys to search on.
The description isn't particularly good, but as best I can gather, it sounds like a less-efficient skip list.
Related
I'm doing an algorithm course and I'm stuck on a question.
I need to suggest a data structure that supports the following:
build(S): build the structure S with n values in O(n*lgn)
insert(S,k): insert the value k into S in O(lgn)
delete_max(S): delete the max value from S in O(lgn)
delete_old(S,t): delete the t oldest value from S in o(lng)
add_to_new(S,d): add value d to the last value that entered to S in O(Lng)
We just learned Red-Black Trees so I think I probably need to use this structure and add something to it, or to add another structure that will help me with delete_old(S,t).
I'm struggling to understand how to update the "age" of each value after I use the "delete_old(S,t)".
Lets say that I entered to S: 11,22,44,68,79... Then 11 is the oldest, 22 is the second oldest, etc.
After I deleted the 3rd oldest (value 44), now 68 need to become the 3rd oldest and 79 will be the 4th oldest, etc- how would I update the ages of all the remaining values after the one I deleted in O(lng)?
I hope my question is clear. Any help will be much appreciated! Thanks :)
To support two orderings, you need to mix two ordered data structures.
For example, you can create a node for each value, and link it into two separate red-black trees, one ordered by value and one ordered by age. Of course the node object would need two sets of links and two isRed fields.
If you want to leverage the libraries that your language already provides, then you can make a node object that just has a value and age in it, and insert it into two separate trees.
I have contacts stored in my mobile. Lets say my contacts are
Ram
Hello
Hi
Feat
Eat
At
When I type letter 'A' I should get all the matching contacts say "Ram, Feat, Eat, At".
Now I type one more letter T. Now my total string is "AT" now my program should reuse the results of previous search for "A". Now it should return me "Feat, Eat, At"
Design and develop a program for this.
This is interview question at Samsung mobile development
I tried solving with Trie data structures. Could not get good solution for reusing already searched string results. I also tried solution with dictionary data structure, solution has same disadvantage as Trie.
question is how do I search the contacts for each letter typed reusing the search results of earlier searched string? What data structure and algorithm should be used for efficiently solving the problem.
I am not asking for program. So programming language is immaterial for me.
State machine appears to be good solution. Does anyone have suggestion?
Solution should be fast enough for million contacts.
It kind of depends on how many items you're searching. If it's a relatively small list, you can do a string.contains check on everything. So when the user types "A", you search the entire list:
for each contact in contacts
if contact.Name.Contains("A")
Add contact to results
Then the user types "T", and you sequentially search the previous returned results:
for each contact in results
if contact.Name.Contains("AT")
Add contact to new search results
Things get more interesting if the list of contacts is huge, but for the number of contacts that you'd normally have in a phone (a thousand would be a lot!), this is going to work very well.
If the interviewer said, "use the results from the previous search for the new search," then I suspect that this is the answer he was looking for. It would take longer to create a new suffix tree than to just sequentially search the previous result set.
You could optimize this a little bit by storing the position of the substring along with the contact so that all you have to do the next time around is check to see if the next character is as expected, but doing so complicates the algorithm a bit (you have to treat the first search as a special case, and you have to explicitly check string lengths, etc.), and is unlikely to provide much benefit after the first few characters because the size of the list to be searched would be pretty small. The pure sequential search with contains check is going to be plenty fast. Users wouldn't notice the few microseconds you'd save with that optimization.
Update after edit to question
If you want to do this with a million contacts, sequential search might not be the best way to go at the start. Although I'd still give it a try. "Fast enough for a million contacts" raises the question of what exactly "fast enough" means. How long does it take to search one million contacts for the existence of a single letter? How long is the user willing to wait? Remember also that you only have to show one page of contacts before the user takes another action. And you can almost certainly to that before the user presses the second key. Especially if you have a background thread doing the search while the foreground thread handles input and writing the first page of matched strings to the display.
Anyway, you could speed up the initial search by creating a bigram index. That is, for each bigram (sequence of two characters), build a list of names that contain that bigram. You'll also want to create a list of strings for each single character. So, given your list of names, you'd have:
r - ram
a - ram, feat, eat, a
m - ram
h - hello, hi
...
ra - ram
am - ram
...
at - feat, eat, at
...
etc.
I think you get the idea.
That bigram index gets stored in a dictionary or hash map. There are only 325 possible bigrams in the English language, and of course the 26 letters, so at most your dictionary is going to have 351 entries.
So you have almost instant lookup of 1- and 2-character names. How does this help you?
An analysis of Project Gutenberg text shows that the most common bigram in the English language occurs only 3.8% of the time. I realize that names won't share exactly that distribution, but that's a pretty good rough number. So after the first two characters are typed, you'll probably be working with less than 5% of the total names in your list. Five percent of a million is 50,000. With just 50,000 names, you can start using the sequential search algorithm that I described originally.
The cost of this new structure isn't too bad, although it's expensive enough that I'd certainly try the simple sequential search first, anyway. This is going to cost you an extra 2 million references to the names, in the worst case. You could reduce that to a million extra references if you build a 2-level trie rather than a dictionary. That would take slightly longer to lookup and display the one-character search results, but not enough to be noticeable by the user.
This structure is also very easy to update. To add a name, just go through the string and make entries for the appropriate characters and bigrams. To remove a name, go through the name extracting bigrams, and remove the name from the appropriate lists in the bigram index.
Look up "generalized suffix tree", e.g. https://en.wikipedia.org/wiki/Generalized_suffix_tree . For a fixed alphabet size this data structure gives asymptotically optimal solution to find all z matches of a substring of length m in a set of strings in O(z + m) time. Thus you get the same sort of benefit as if you restricted your search to the matches for the previous prefix. Also the structure has optimal O(n) space and build time where n is the total length of all your contacts. I believe you can modify the structure so that you just find the k strings that contain the substring in O(k + m) time, but in general you probably shouldn't have too many matches per contact that have a match, so this may not even be necessary.
What I'm thinking to do is, keeping track of the so far matched string. Suppose in the first step, we identify the strings those have "A" in them and we keep trace of the positions of 'A". Then in the next step we only iterate over these strings and instead of searching them full we only check for occurrence of "T" as the next character to "A" we kept trace in the previous step and so on.
Currently I am reading a book on algorithms and found this usage of sorting.
Reconstructing the original order - How can we restore the original arrangment of a set of items after we permute them for some application? Add an extra field to the data record for the item, such that i-th record sets this field to i. Carry this field along whenever you move the record, and later sort on it when you want the initial order back.
I ve been trying hard to understand what does it mean. And I failed miserably. Pls somebody help?
Suppose you have list of items in random order:
itemC, itemB, itemA, itemD
you sorted them up:
itemA, itemB, itemC, itemD
and you didn't have enough memory to store them in a separate location, so original sequence is lost. Moreover, original order is random and it will be problematic/impossible to restore it.
This article gives a solution to this problem.
Add an extra field to the data record for the item, such that i-th record sets this field to i
So, we add an extra field for each of the items:
(itemC,1), (itemB,2), (itemA,3), (itemD, 4)
And after sort we have:
(itemA,3), (itemB,2), (itemC,1), (itemD, 4)
So we can easily restore initial order sorting by additional field
Let's say you have the data in an array, because it's the simplest structure that I can use to exemplify.
So, your node (i.e., element of the array) may look like this:
(some data type) data
The algorithm is suggesting you to add an integer field, so it looks like this:
(some data type) data,
int position
And then, you fill the positions with the actual index. Something like this pseudocode:
for current: 0 to lastElement
array[current].position = current
(that's not written in any language I know of, but it should be readable)
After doing that, you shuffle it (resort it) for whatever you need to.
When you want to restore the original ordering, all you need to do is sort by the position field.
Well, basically it's saying that you need some sort of thingy to keep track of the original order (which is destroyed by the permutation). One option would be to simply reverse the permutation (check out Steve Jessop's infrmative answer here).
Another option to invert the permutation would require fewer processing steps, but more memory. More specifically, each node in your input set would have an extra ID field, and all the elements in this input set are sorted based on this field. Once you apply the permutation, it's obvious that the IDs are no longer in a sorted order. If you wish to invert the permutation, all you have to do is sort the list again based on this field.
Suppose you want to write a program that implements a simple phone book. Given a particular name, you want to be able to retrieve that person's phone number as quickly as possible. What data structure would you use to store the phone book, and why?
the text below answers your question.
In computer science, a hash table or hash map is a data structure that
uses a hash function to map identifying values, known as keys (e.g., a
person's name), to their associated values (e.g., their telephone
number). Thus, a hash table implements an associative array. The hash
function is used to transform the key into the index (the hash) of an
array element (the slot or bucket) where the corresponding value is to
be sought.
the text is from wiki:hashtable.
there are some further discussions, like collision, hash functions... check the wiki page for details.
I respect & love hashtables :) but even a balanced binary tree would be fine for your phone book application giving you in worst case a logarithmic complexity and avoiding you for having good hash functions, collisions etc. which is more suitable for huge amounts of data.
When I talk about huge data what I mean is something related to storage. Every time you fill all of the buckets in a hash-table you will need to allocate new storage and re-hash everything. This can be avoided if you know the size of the data ahead of time. Balanced trees wont let you go into these problems. Domain needs to be considered too while designing data structures, for an example for small devices storage matters a lot.
I was wondering why 'Tries' didn't come up in one of the answers,
Tries is suitable for Phone book kind of data.
Also, saving space compared to HashTable at the same cost(almost) of Retrieval efficiency, (assuming constant size alphabet & constant length Names)
Tries also facilitate the 'Prefix Matches' sometimes required while searching.
A dictionary is both dynamic and fast.
You want a dictionary, where you use the name as the key, and the number as the data stored. Check this out: http://en.wikipedia.org/wiki/Dictionary_%28data_structure%29
Why not use a singly linked list? Each node will have the name, number and link information.
One drawback is that your search might take some time since you'll have to traverse the entire list from link to link. You might order the list at the time of node insertion itself!
PS: To make the search a tad bit faster, maintain a link to the middle of the list. Search can continue to the left or right of the list based on the value of the "name" field at this node. Note that this requires a doubly linked list.
I am looking for the optimal (time and space) optimal data structure for supporting the following operations:
Add Persons (name, age) to a global data store of persons
Fetch Person with minimum and maximum age
Search for Person's age given the name
Here's what I could think of:
Keep an array of Persons, and keep adding to end of array when a new Person is to be added
Keep a hash of Person name vs. age, to assist in fetching person's age with given name
Maintain two objects minPerson and maxPerson for Person with min and max age. Update this if needed, when a new Person is added.
Now, although I keep a hash for better performance of (3), I think it may not be the best way if there are many collisions in the hash. Also, addition of a Person would mean an overhead of adding to the hash.
Is there anything that can be further optimized here?
Note: I am looking for the best (balanced) approach to support all these operations in minimum time and space.
You can get rid of the array as it doesn't provide anything that the other two structures can't do.
Otherwise, a hashtable + min/max is likely to perform well for your use case. In fact, this is precisely what I would use.
As to getting rid of the hashtable because a poor hash function might lead to collisions: well, don't use a poor hash function. I bet that the default hash function for strings that's provided by your programming language of choice is going to do pretty well out of the box.
It looks like that you need a data structure that needs fast inserts and that also supports fast queries on 2 different keys (name and age).
I would suggest keeping two data structures, one a sorted data structure (e.g. a balanced binary search tree) where the key is the age and the value is a pointer to the Person object, the other a hashtable where the key is the name and the value is a pointer to the Person object. Notice we don't keep two copies of the same object.
A balanced binary search tree would provide O(log(n)) inserts and max/min queries, while the hastable would give us O(1) (amortized) inserts and lookups.
When we add a new Person, we just add a pointer to it to both data structures. For a min/max age query, we can retrieve the Object by querying the BST. For a name query we can just query the hashtable.
Your question does not ask for updates/deletes, but those are also doable by suitably updating both data structures.
It sounds like you're expecting the name to be the unique idenitifer; otherwise your operation 3 is ambiguous (What is the correct return result if you have two entries for John Smith?)
Assuming that the uniqueness of a name is guaranteed, I would go with a plain hashtable keyed by names. Operation 1 and 3 are trivial to execute. Operation 2 could be done in O(N) time if you want to search through the data structure manually, or you can do like you suggest and keep track of the min/max and update it as you add/delete entries in the hash table.