If I have to develop an application for a data grid station of an institute. The purpose of application is to receive the data from data GRID station once in a week between 10 A.M to 10:30 A.M and then store it into a data structure and data is consist of digits only but the numbers could be very long for one entry then which data structure will be the best for given scenario from array, list, linked list, doubly linked list, queue, priority queue, stack, binary search tree, AVL trees, threaded binary tree, heap, sorted sequential array and skip list
I want to store sorted digits. The sorted data can be in ascending or descending order and the main concern is "fast and efficient searching".
From your description I gather that you don't store any other data with the digits or numbers. So basically you want to know if a number is in the set or not.
Fastest way to know this, is to have an array of flags for each number. Let's say you deal with numbers from 1 to 1000. You want to know if number 200 is in the set. Look at position 200 wether the flag is true or false. You see, this is the fastest method, because you only look up one place.
As we are talking about boolean flags here, a bit is sufficient for storage. You would decide wether to store the booleans in bits, bytes, words or whatever, depending on the number of numbers, the available memory and the machine's architecture.
Having said this, you may have to deal with so many numbers that above approach is no more feasible. It would be fastest in theory, but with limited memory, swaps to hard disk, many, many reads from it, other algorithms may prove better. You would have the choice between:
storing the numbers contiguously and perform a binary search on them
storing the numbers in a binary tree
using a hash algorithm
Which of these proves most efficient, again depends on your data and the machine.
It depends what type of searching you want to do. If you just want to know if a number is within your dataset, then a hash will be extremely fast and independent of the size of your dataset. And there is no need to sort, or even any concept of order.
If I may quote Larry Wall, author of Perl:
Doing linear scans over an associative array is like trying to club
someone to death with a loaded Uzi.
(An associative array is synonymous with a hash.)
Related
I remember learning a data structure that stored a set of integers as ranges in a tree, but it's been 10 years and I can't remember the name of the data structure, and I'm a bit fuzzy on the details. If it helps, it's a functional data structure that was taught at CMU, I believe in 15-212 (Principles of Programming) in 2002.
Basically, I want to store a set of integers, most of which are consecutive. I want to be able to query for set membership efficiently, add a range of integers efficiently, and remove a range of integers efficiently. In particular, I don't care to preserve what the original ranges are. It's better if adjacent ranges are coalesced into a single larger range.
A naive implementation would be to simply use a generic set data structure such as a HashSet or TreeSet, and add all integers in a range when adding a range, or remove all integers in a range when removing a range. But of course, that would waste a lot of memory in addition to making add and remove slow.
I'm thinking of a purely functional data structure, but for my current use I don't need it to be. IIRC, lookup, insertion, and deletion were all O(log N), where N was the number of ranges in the set.
So, can you tell me the name of the data structure I'm trying to remember, or a suitable alternative?
I found the old homework and the data structure I had in mind were Discrete Interval Encoding Trees or diets for short. They are described in detail in Diets for Fat Sets, Martin Erwig. Journal of Functional Programming, Vol. 8, No. 6, 627-632, 1998. It is basically a tree of intervals with the invariant that all of the intervals are non-overlapping and non-touching. There is a Haskell implementation in Hackage. I was hoping there would be an existing implementation for Scala, but I'm not seeing any.
The homework also included another data structure they called a Recursive Interval-Occluding Tree (RIOT), which rather than keeping only an interval at each node keeps an interval and another (possibly empty) RIOT of things removed from the interval. The assignment included benchmarks showing it did better than diets for random insertions and deletions. AFAICT it is simply something the TAs made up and never published as it no longer seems to exist anywhere on the Internets, at least not under that name.
You probably are looking for segment trees. This might be helpful: http://www.topcoder.com/tc?d1=tutorials&d2=lowestCommonAncestor&module=Static
You can also use binary search trees for the same, for which each node will have two data fields: min_val and max_val.
During insertion algorithm, you just need to call another merging operation to check if the left-child,parent,right-child create a sequence, so as to club them into a single node. This will take O(log n) time.
Other operations like deletion and look-up will take O(log n) time as usual, but special measures need to be taken while deletion.
I need a data structure that stores tuples and would allow me to do a query like: given tuple (x,y,z) of integers, find the next one (an upped bound for it). By that I mean considering the natural ordering (a,b,c)<=(d,e,f) <=> a<=d and b<=e and c<=f. I have tried MSD radix sort, which splits items into buckets and sorts them (and does this recursively for all positions in the tuples). Does anybody have any other suggestion? Ideally I would like the abouve query to happen within O(log n) where n is the number of tuples.
Two options.
Use binary search on a sorted array. If you build the keys ( assuming 32bit int)' with (a<<64)|(b<<32)|c and hold them in a simple array, packed one beside the other, you can use binary search to locate the value you are searching for ( if using C, there is even a library function to do this), and the next one is simply one position along. Worst case Performance is O(logN), and if you can do http://en.wikipedia.org/wiki/Interpolation_search then you might even approach O(log log N)
Problem with binary keys is might be tricky to add new values, might need gyrations if you will exceed available memory. But it is fast, only a few random memory accesses on average.
Alternatively, you could build a hash table by generating a key with a|b|c in some form, and then have the hash data pointing to a structure that contains the next value, whatever that might be. Possibly a little harder to create in the first place as when generating the table you need to know the next value already.
Problems with hash approach are it will likely use more memory than binary search method, performance is great if you don't get hash collisions, but then starts to drop off, although there a variations around this algorithm to help in some cases. Hash approach is possibly much easier to insert new values.
I also see you had a similar question along these lines, so I guess the guts of what I am saying is combine A,b,c to produce a single long key, and use that with binary search, hash or even b-tree. If the length of the key is your problem (what language), could you treat it as a string?
If this answer is completely off base, let me know and I will see if I can delete this answer, so you questions remains unanswered rather than a useless answer.
In cases where I have a key for each element and I don't know the index of the element into an array, hashtables perform better than arrays (O(1) vs O(n)).
Why is that? I mean: I have a key, I hash it.. I have the hash.. shouldn't the algorithm compare this hash against every element's hash? I think there's some trick behind the memory disposition, isn't it?
In cases where I have a key for each element and I don't know the
index of the element into an array, hashtables perform better than
arrays (O(1) vs O(n)).
The hash table search performs O(1) in the average case. In the worst case, the hash table search performs O(n): when you have collisions and the hash function always returns the same slot. One may think "this is a remote situation," but a good analysis should consider it. In this case you should iterate through all the elements like in an array or linked lists (O(n)).
Why is that? I mean: I have a key, I hash it.. I have the hash..
shouldn't the algorithm compare this hash against every element's
hash? I think there's some trick behind the memory disposition, isn't
it?
You have a key, You hash it.. you have the hash: the index of the hash table where the element is present (if it has been located before). At this point you can access the hash table record in O(1). If the load factor is small, it's unlikely to see more than one element there. So, the first element you see should be the element you are looking for. Otherwise, if you have more than one element you must compare the elements you will find in the position with the element you are looking for. In this case you have O(1) + O(number_of_elements).
In the average case, the hash table search complexity is O(1) + O(load_factor) = O(1 + load_factor).
Remember, load_factor = n in the worst case. So, the search complexity is O(n) in the worst case.
I don't know what you mean with "trick behind the memory disposition". Under some points of view, the hash table (with its structure and collisions resolution by chaining) can be considered a "smart trick".
Of course, the hash table analysis results can be proven by math.
With arrays: if you know the value, you have to search on average half the values (unless sorted) to find its location.
With hashes: the location is generated based on the value. So, given that value again, you can calculate the same hash you calculated when inserting. Sometimes, more than 1 value results in the same hash, so in practice each "location" is itself an array (or linked list) of all the values that hash to that location. In this case, only this much smaller (unless it's a bad hash) array needs to be searched.
Hash tables are a bit more complex. They put elements in different buckets based on their hash % some value. In an ideal situation, each bucket holds very few items and there aren't many empty buckets.
Once you know the key, you compute the hash. Based on the hash, you know which bucket to look for. And as stated above, the number of items in each bucket should be relatively small.
Hash tables are doing a lot of magic internally to make sure buckets are as small as possible while not consuming too much memory for empty buckets. Also, much depends on the quality of the key -> hash function.
Wikipedia provides very comprehensive description of hash table.
A Hash Table will not have to compare every element in the Hash. It will calculate the hashcode according to the key. For example, if the key is 4, then hashcode may be - 4*x*y. Now the pointer knows exactly which element to pick.
Whereas if it has been an array, it will have to traverse through the whole array to search for this element.
Why is [it] that [hashtables perform lookups by key better than arrays (O(1) vs O(n))]? I mean: I have a key, I hash it.. I have the hash.. shouldn't the algorithm compare this hash against every element's hash? I think there's some trick behind the memory disposition, isn't it?
Once you have the hash, it lets you calculate an "ideal" or expected location in the array of buckets: commonly:
ideal bucket = hash % num_buckets
The problem is then that another value may have already hashed to that bucket, in which case the hash table implementation has two main choice:
1) try another bucket
2) let several distinct values "belong" to one bucket, perhaps by making the bucket hold a pointer into a linked list of values
For implementation 1, known as open addressing or closed hashing, you jump around other buckets: if you find your value, great; if you find a never-used bucket, then you can store your value in there if inserting, or you know you'll never find your value when searching. There's a potential for the searching to be even worse than O(n) if the way you traverse alternative buckets ends up searching the same bucket multiple times; for example, if you use quadratic probing you try the ideal bucket index +1, then +4, then +9, then +16 and so on - but you must avoid out-of-bounds bucket access using e.g. % num_buckets, so if there are say 12 buckets then ideal+4 and ideal+16 search the same bucket. It can be expensive to track which buckets have been searched, so it can be hard to know when to give up too: the implementation can be optimistic and assume it will always find either the value or an unused bucket (risking spinning forever), it can have a counter and after a threshold of tries either give up or start a linear bucket-by-bucket search.
For implementation 2, known as closed addressing or separate chaining, you have to search inside the container/data-structure of values that all hashed to the ideal bucket. How efficient this is depends on the type of container used. It's generally expected that the number of elements colliding at one bucket will be small, which is true of a good hash function with non-adversarial inputs, and typically true enough of even a mediocre hash function especially with a prime number of buckets. So, a linked list or contiguous array is often used, despite the O(n) search properties: linked lists are simple to implement and operate on, and arrays pack the data together for better memory cache locality and access speed. The worst possible case though is that every value in your table hashed to the same bucket, and the container at that bucket now holds all the values: your entire hash table is then only as efficient as the bucket's container. Some Java hash table implementations have started using binary trees if the number of elements hashing to the same buckets passes a threshold, to make sure complexity is never worse than O(log2n).
Python hashes are an example of 1 = open addressing = closed hashing. C++ std::unordered_set is an example of closed addressing = separate chaining.
The purpose of hashing is to produce an index into the underlying array, which enables you to jump straight to the element in question. This is usually accomplished by dividing the hash by the size of the array and taking the remainder index = hash%capacity.
The type/size of the hash is typically that of the smallest integer large enough to index all of RAM. On a 32 bit system this is a 32 bit integer. On a 64 bit system this is a 64 bit integer. In C++ this corresponds to unsigned int and unsigned long long respectively. To be pedantic C++ technically specifies minimum sizes for its primitives i.e. at least 32 bits and at least 64 bits, but that's beside the point. For the sake of making code portable C++ also provides a size_t primative which corresponds to the appropriate unsigned integer. You'll see that type a lot in for loops which index into arrays, in well written code. In the case of a language like Python the integer primitive grows to whatever size it needs to be. This is typically implemented in the standard libraries of other languages under the name "Big Integer". To deal with this the Python programming language simply truncates whatever value you return from the __hash__() method down to the appropriate size.
On this score I think it's worth giving a word to the wise. The result of arithmetic is the same regardless of whether you compute the remainder at the end or at each step along the way. Truncation is equivalent to computing the remainder modulo 2^n where n is the number of bits you leave intact. Now you might think that computing the remainder at each step would be foolish due to the fact that you're incurring an extra computation at every step along the way. However this is not the case for two reasons. First, computationally speaking, truncation is extraordinarily cheap, far cheaper than generalized division. Second, and this is the real reason as the first is insufficient, and the claim would generally hold even in its absence, taking the remainder at each step keeps the number (relatively) small. So instead of something like product = 31*product + hash(array[index]), you'll want something like product = hash(31*product + hash(array[index])). The primary purpose of the inner hash() call is to take something which might not be a number and turn it into one, where as the primary purpose of the outer hash() call is to take a potentially oversized number and truncate it. Lastly I'll note that in languages like C++ where integer primitives have a fixed size this truncation step is automatically performed after every operation.
Now for the elephant in the room. You've probably realized that hash codes being generally speaking smaller than the objects they correspond to, not to mention that the indices derived from them are again generally speaking even smaller still, it's entirely possible for two objects to hash to the same index. This is called a hash collision. Data structures backed by a hash table like Python's set or dict or C++'s std::unordered_set or std::unordered_map primarily handle this in one of two ways. The first is called separate chaining, and the second is called open addressing. In separate chaining the array functioning as the hash table is itself an array of lists (or in some cases where the developer feels like getting fancy, some other data structure like a binary search tree), and every time an element hashes to a given index it gets added to the corresponding list. In open addressing if an element hashes to an index which is already occupied the data structure probes over to the next index (or in some cases where the developer feels like getting fancy, an index defined by some other function as is the case in quadratic probing) and so on until it finds an empty slot, of course wrapping around when it reaches the end of the array.
Next a word about load factor. There is of course an inherent space/time trade off when it comes to increasing or decreasing the load factor. The higher the load factor the less wasted space the table consumes; however this comes at the expense of increasing the likelihood of performance degrading collisions. Generally speaking hash tables implemented with separate chaining are less sensitive to load factor than those implemented with open addressing. This is due to the phenomenon known as clustering where by clusters in an open addressed hash table tend to become larger and larger in a positive feed back loop as a result of the fact that the larger they become the more likely they are to contain the preferred index of a newly added element. This is actually the reason why the afore mentioned quadratic probing scheme, which progressively increases the jump distance, is often preferred. In the extreme case of load factors greater than 1, open addressing can't work at all as the number of elements exceeds the available space. That being said load factors greater than 1 are exceedingly rare in general. At time of writing Python's set and dict classes employ a max load factor of 2/3 where as Java's java.util.HashSet and java.util.HashMap use 3/4 with C++'s std::unordered_set and std::unordered_map taking the cake with a max load factor of 1. Unsurprisingly Python's hash table backed data structures handle collisions with open addressing where as their Java and C++ counterparts do it with separate chaining.
Last a comment about table size. When the max load factor is exceeded, the size of the hash table must of course be grown. Due to the fact that this requires that every element there in be reindexed, it's highly inefficient to grow the table by a fixed amount. To do so would incur order size operations every time a new element is added. The standard fix for this problem is the same as that employed by most dynamic array implementations. At every point where we need to grow the table we simply increase its size by its current size. This unsurprisingly is known as table doubling.
I think you answered your own question there. "shouldn't the algorithm compare this hash against every element's hash". That's kind of what it does when it doesn't know the index location of what you're searching for. It compares each element to find the one you're looking for:
E.g. Let's say you're looking for an item called "Car" inside an array of strings. You need to go through every item and check item.Hash() == "Car".Hash() to find out that that is the item you're looking for. Obviously it doesn't use the hash when searching always, but the example stands. Then you have a hash table. What a hash table does is it creates a sparse array, or sometimes array of buckets as the guy above mentioned. Then it uses the "Car".Hash() to deduce where in the sparse array your "Car" item is actually. This means that it doesn't have to search through the entire array to find your item.
This problem is a little similar to that solved by reservoir sampling, but not the same. I think its also a rather interesting problem.
I have a large dataset (typically hundreds of millions of elements), and I want to estimate the number of unique elements in this dataset. There may be anywhere from a few, to millions of unique elements in a typical dataset.
Of course the obvious solution is to maintain a running hashset of the elements you encounter, and count them at the end, this would yield an exact result, but would require me to carry a potentially large amount of state with me as I scan through the dataset (ie. all unique elements encountered so far).
Unfortunately in my situation this would require more RAM than is available to me (nothing that the dataset may be far larger than available RAM).
I'm wondering if there would be a statistical approach to this that would allow me to do a single pass through the dataset and come up with an estimated unique element count at the end, while maintaining a relatively small amount of state while I scan the dataset.
The input to the algorithm would be the dataset (an Iterator in Java parlance), and it would return an estimated unique object count (probably a floating point number). It is assumed that these objects can be hashed (ie. you can put them in a HashSet if you want to). Typically they will be strings, or numbers.
You could use a Bloom Filter for a reasonable lower bound. You just do a pass over the data, counting and inserting items which were definitely not already in the set.
This problem is well-addressed in the literature; a good review of various approaches is http://www.edbt.org/Proceedings/2008-Nantes/papers/p618-Metwally.pdf. The simplest approach (and most compact for very high accuracy requirements) is called Linear Counting. You hash elements to positions in a bitvector just like you would a Bloom filter (except only one hash function is required), but at the end you estimate the number of distinct elements by the formula D = -total_bits * ln(unset_bits/total_bits). Details are in the paper.
If you have a hash function that you trust, then you could maintain a hashset just like you would for the exact solution, but throw out any item whose hash value is outside of some small range. E.g., use a 32-bit hash, but only keep items where the first two bits of the hash are 0. Then multiply by the appropriate factor at the end to approximate the total number of unique elements.
Nobody has mentioned approximate algorithm designed specifically for this problem, Hyperloglog.
Say I have a bunch of objects with dates and I regularly want to find all the objects that fall between two arbitrary dates. What sort of datastructure would be good for this?
A binary search tree sounds like what you're looking for.
You can use it to find all the objects in O(log(N) + K), where N is the total number of objects and K is the number of objects that are actually in that range. (provided that it's balanced). Insertion/removal is O(log(N)).
Most languages have a built-in implementation of this.
C++:
http://www.cplusplus.com/reference/stl/set/
Java:
http://java.sun.com/j2se/1.4.2/docs/api/java/util/TreeSet.html
You can find the lower bound of the range (in log(n)) and then iterate from there until you reach the upper bound.
Assuming you mean by date when you say sorted, an array will do it.
Do a binary search to find the index that's >= the start date. You can then either do another search to find the index that's <= the end date leaving you with an offset & count of items, or if you're going to process them anyway just iterate though the list until you exceed the end date.
It's hard to give a good answer without a little more detail.
What kind of performance do you need?
If linear is fine then I would just use a list of dates and iterate through the list collecting all dates that fall within the range. As Andrew Grant suggested.
Do you have duplicates in the list?
If you need to have repeated dates in your collection then most implementations of a binary tree would probably be out. Something like Java's TreeSet are set implementations and don't allow repeated elements.
What are the access characteristics? Lots of lookups with few updates, vice-versa, or fairly even?
Most datastructures have trade-offs between lookups and updates. If you're doing lots of updates then some datastructure that are optimized for lookups won't be so great.
So what are the access characteristics of the data structure, what kind of performance do you need, and what are structural characteristics that it must support (e.g. must allow repeated elements)?
If you need to make random-access modifications: a tree, as in v3's answer. Find the bottom of the range by lookup, then count upwards. Inserting or deleting a node is O(log N). stbuton makes a good point that if you want to allow duplicates (as seems plausible for datestamped events), then you don't want a tree-based set.
If you do not need to make random-access modifications: a sorted array (or vector or whatever). Find the location of the start of the range by binary chop, then count upwards. Inserting or deleting is O(N) in the middle. Duplicates are easy.
Algorithmic performance of lookups is the same in both cases, O(M + log N), where M is the size of the range. But the array uses less memory per entry, and might be faster to count through the range, because after the binary chop it's just forward sequential memory access rather than following pointers.
In both cases you can arrange for insertion at the end to be (amortised) O(1). For the tree, keep a record of the end element at the head, and you get an O(1) bound. For the array, grow it exponentially and you get amortised O(1). This is useful if the changes you make are always or almost-always "add a new event with the current time", since time is (you'd hope) a non-decreasing quantity. If you're using system time then of course you'd have to check, to avoid accidents when the clock resets backwards.
Alternative answer: an SQL table, and let the database optimise how it wants. And Google's BigTable structure is specifically designed to make queries fast, by ensuring that the result of any query is always a consecutive sequence from a pre-prepared index :-)
You want a structure that keeps your objects sorted by date, whenever you insert or remove a new one, and where finding the boundary for the segment of all objects later than or earlier than a given date is easy.
A heap seems the perfect candidate. In practical applications, heaps are simply represented by an array, where all the objects are stored in order. Seeing that sorted array as a heap is simply a way to make insertions of new objects and deletions happen in the right place, and in O(log(n)).
When you have to find all the objects between date A (excluded) and B (included), find the position of A (or the insert position, that is, the position of the earlier element later than A), and the position of B (or the insert position of B), and return all the objects between those positions (which is simply the section between those positions in the array/heap)