You might have come across someplace where it is mentioned that it is faster to find elements in hashmap/dictionary/table than list/array. My question is WHY?
(inference so far I made: Why should it be faster, as far I see, in both data structure, it has to travel throughout till it reaches the required element)
Let’s reason by analogy. Suppose you want to find a specific shirt to put on in the morning. I assume that, in doing so, you don’t have to look at literally every item of clothing you have. Rather, you probably do something like checking a specific drawer in your dresser or a specific section of your closet and only look there. After all, you’re not (I hope) going to find your shirt in your sock drawer.
Hash tables are faster to search than lists because they employ a similar strategy - they organize data according to the principle that every item has a place it “should” be, then search for the item by just looking in that place. Contrast this with a list, where items are organized based on the order in which they were added and where there isn’t a a particular pattern as to why each item is where it is.
More specifically: one common way to implement a hash table is with a strategy called chained hashing. The idea goes something like this: we maintain an array of buckets. We then come up with a rule that assigns each object a bucket number. When we add something to the table, we determine which bucket number it should go to, then jump to that bucket and then put the item there. To search for an item, we determine the bucket number, then jump there and only look at the items in that bucket. Assuming that the strategy we use to distribute items ends up distributing the items more or less evenly across the buckets, this means that we won’t have to look at most of the items in the hash table when doing a search, which is why the hash table tends to be much faster to search than a list.
For more details on this, check out these lecture slides on hash tables, which fills in more of the details about how this is done.
Hope this helps!
To understand this you can think of how the elements are stored in these Data structures.
HashMap/Dictionary as you know it is a key-value data structure. To store the element, you first find the Hash value (A function which always gives a unique value to a key. For example, a simple hash function can be made by doing the modulo operation.). Then you basically put the value against this hashed key.
In List, you basically keep appending the element to the end. The order of the element insertion would matter in this data structure. The memory allocated to this data structure is not contiguous.
In Array, you can think of it as similar to List. But In this case, the memory allocated is contiguous in nature. So, if you know the value of the address for the first index, you can find the address of the nth element.
Now think of the retrieval of the element from these Data structures:
From HashMap/Dictionary: When you are searching for an element, the first thing that you would do is find the hash value for the key. Once you have that, you go to the map for the hashed value and obtain the value. In this approach, the amount of action performed is always constant. In Asymptotic notation, this can be called as O(1).
From List: You literally need to iterate through each element and check if the element is the one that you are looking for. In the worst case, your desired element might be present at the end of the list. So, the amount of action performed varies, and in the worst case, you might have to iterate the whole list. In Asymptotic notation, this can be called as O(n). where n is the number of elements in the list.
From array: To find the element in the array, what you need to know is the address value of the first element. For any other element, you can do the Math of how relative this element is present from the first index.
For example, Let's say the address value of the first element is 100. Each element takes 4 bytes of memory. The element that you are looking for is present at 3rd position. Then you know the address value for this element would be 108. Math used is
Addresses of first element + (position of element -1 )* memory used for each element.
That is 100 + (3 - 1)*4 = 108.
In this case also as you can observe the action performed is always constant to find an element. In Asymptotic notation, this can be called as O(1).
Now to compare, O(1) will always be faster than O(n). And hence retrieval of elements from HashMap/Dictionary or array would always be faster than List.
I hope this helps.
Related
Bit of a random question for you. If you have a method that has to check every single individual place inside an array, would it be okay to say that this method has notation of O(n)?
The reason i'm not sure if my answer is correct is due to the fact that as far as i'm aware O(n) relates to the number of items held in the array, while my assumption is based on the actual size of the array?
If your algorithm has to look at every item in the array, that algorithm is O(n). If doesn't really matter if the array is full or not, since you can be flexible in how you define n. It can be either the size of the array or the number of non-null elements in the array. If your algorithm has to look in empty array slots to see if they're empty or not, use the size. (If that's a real performance issue, probably a different data structure is called for.)
For a really contrived example, if it takes one hour to process each non-null array element, but one nanosecond to check for null, then you should define n to be the number of elements that actually exist, because that's what's going to dictate how the algorithm scales.
Lets say that I have a list (or a hashmap etc., whatever makes this the fastest) of objects that contain the following fields: name, time added, and time removed. The list given to me is already sorted by time removed. Now, given a time T, I want to filter (remove from the list) out all objects of the list where:
the time T is greater than an object's time removed OR T is less than an object's time added.
So after processing, the list should only contain objects where T falls in the range specified by time added and time removed.
I know I can do this easily in O(n) time by going through each individual object, but I was wondering if there was a more efficient way considering the list was already sorted by the first predicate (time removed).
*Also, I know I can easily remove all objects with time removed less than T because the list is presorted (possibly in O(log n) time since I do a binary search to find the first element that is less than and then remove the first part of the list up to that object).
(Irrelevant additional info: I will be using C++ for any code that I write)
Unfortunately you are stuck with a O(n) being your fastest option. That is unless their are hidden requirements about the difference between time added and time removed (such as a max time span) that can be exploited.
As you said you can start the search where the time removed equals (or is the first greater than) the time removed. Unfortunately you'll need to go through the rest of the list to see if time added is less than your time.
Because a comparative sort is at best O(n*log(n)) you cannot sort the objects again to improve your performance.
One thing, based on the heuristics of the application it may be beneficial to receive the data in order of date added but that is between you and wherever you get the data from.
Let's examine the data structures you offered:
A list (usually implemented as a linked list, or a dynamic array), or a hash map.
Linked List: Cannot do binary search, finding first occurance of an
element (even if list is sorted) is done in O(n), so no benefit
from the fact the data is sorted.
Dynamic Array: Removing a single element (or more) from arbitrary location requires shifting all the following elements to the left, and thus is O(n). You cannot remove elements from the list better than O(n), so no gain here from the fact the DS is sorted.
HashMap: is unsorted by definition. Also, removing k elements is O(k), no way to go around this.
So, you cannot even improve performance from O(n) to O(logn) for the same field the list was sorted by.
Some data structures such as B+ trees do allow efficient range queries, and you can pretty efficiently [O(logn)] remove a range of elements from the tree.
However, it does not help you to filter the data of the 2nd field, which the tree is unsorted by, and to filter according to it (unless there is some correlation you can exploit) - will still need O(n) time.
If all you are going to do is to later on iterate the new list, you can push the evaluation to the iteration step, but there won't be any real benefit from it - only delaying the processing to when it's needed, and avoiding it, if it is not needed.
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.
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)