I have to maintain a list of unordered integers , where number of integers are unknown. It may increase or decrease over the time. I need to update this list of integers frequently. I have tried using vector . But it is really slow . Array appears to be faster , but since the length of list is not fixed, it takes significant amount of time to resize it . Please suggest any other option .
Use a hash table, if order of the values in unimportant. Time is O(1). I'm pretty sure you'll find an implementation in the standard template libraries.
Failing that, a splay tree is extremely fast, especially if you want to keep the list ordered: amortized cost of O(ln n) per operation, with a very low constant factor. I think C++ stdlib map is something like this.
Know thy data structures.
If you are interested in Dynamic increments of Arrays size you can do this .
current =0;
x = (int**)malloc(temp * sizeof(int*));
x[current]=(int*)malloc(RequiredLength * sizeof(int));
So add elements to array and when elements are filled in x[current]
You can add more space for elements by doing
x[++current]=(int*)malloc(RequiredLength * sizeof(int));
Doing this you can accommodate for RequiredLength more elements .
You can repeat this upto 1024 times which means 1024*RequiredLength elements can be
accommodated , here it gives you chance to increase size of array whenever you want it .
You can always access the n th element by X[ n / 1024 ][ n % 1024] ;
Considering your comments, it looks like it is std::set or std::unordered_set fits your needs better than std::vector.
If sequential data structures fails to meet requirements, you could try looking at trees (binary, AVL, m-way, red-black ect ...). I would suggest you try to implement AVL tree since it yields a balanced or near balanced binary search tree which would optimize your operation. For more on AVL tree: http://en.wikipedia.org/wiki/AVL_tree
well,deque has no resize cost,but if it's unordered,it's search time is linear ，and its delete and insert operation time in the middle of its self is even worth than vector.
if you don't need search by the value of the number,hashmap or map may be your choice .No resize cost.,then you set the key of the map to number's index,and the value to the number's value.the search and insert operation is better than linear.
std::list is definitely created for such problems, adding and deleting elements in list do not necessitate memory re-allocations like in vector. However, due to the noncontagious memory allocation of the list, searching elements may prove to be a painful experience ofcourse but if you do not search its entries frequently, it can be used.
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 need to use a data structure, implementable in C++, that can do basic operations, such as lookup, insertion and deletion, in constant time. I, however, also need to be able to find the maximum value in constant time.
This data structure should probably be sorted to find the maximum values and I have looked into red-black trees, however they have logarithmic-time operations.
I would propose
You could use a hash table which gives O(1) expected time
Regarding the maximum, you could store it in attribute and be aware at each insertion if the maximum changes. With the deletion is some more complicated because if the maximum is deleted, then you must perform a linear search, but this only would happen if the maximum is deleted. Any other element could be deleted in O(1) expected time
Yes I agree with Irleon.You can use a hash table to perform these operations.Let us analyze this step by step:
1.If we take arrays,the time complexity of insertion will be O(1) at the end.
2.Take linked lists and it will be O(n) due to the traversal that you need to do.
3.Take binary search trees and it will be O(logn) where logn is the height of the tree.
4.Now we can use hash tables.We know that it works on keys and values.So,here the key will be 'number_to_be_inserted % n' where 'n' is the number of elements we have.
But as the list grows on the same index,you will be needing to traverse the list.So it will O(numbers_at_that_index).
Same will be the case in deletion operation.
Ofcourse there are other cases to consider in case of collisions ,but we can ignore that for now and we will get our basic hash table.
If you could do such a thing, then you could sort in linear time: simply insert all of your items, then, do the following n times:
Find maximum
Print maximum
Delete maximum
Therefore, in a model of computation in which you can't sort in linear time, you also can't solve your problem with all operations in O(1) time.
I'm currently working on an embedded device project where I'm running into performance problems. Profiling has located an O(N) operation that I'd like to eliminate.
I basically have two arrays int A[N] and short B[N]. Entries in A are unique and ordered by external constraints. The most common operation is to check if a particular value a appears in A[]. Less frequently, but still common is a change to an element of A[]. The new value is unrelated to the previous value.
Since the most common operation is the find, that's where B[] comes in. It's a sorted array of indices in A[], such that A[B[i]] < A[B[j]] if and only if i<j. That means that I can find values in A using a binary search.
Of course, when I update A[k], I have to find k in B and move it to a new position, to maintain the search order. Since I know the old and new values of A[k], that's just a memmove() of a subset of B[] between the old and new position of k. This is the O(N) operation that I need to fix; since the old and new values of A[k] are essentially random I'm moving on average about N/2 N/3 elements.
I looked into std::make_heap using [](int i, int j) { return A[i] < A[j]; } as the predicate. In that case I can easily make B[0] point to the smallest element of A, and updating B is now a cheap O(log N) rebalancing operation. However, I generally don't need the smallest value of A, I need to find if any given value is present. And that's now a O(N log N) search in B. (Half of my N elements are at heap depth log N, a quarter at (log N)-1, etc), which is no improvement over a dumb O(N) search directly in A.
Considering that std::set has O(log N) insert and find, I'd say that it should be possible to get the same performance here for update and find. But how do I do that? Do I need another order for B? A different type?
B is currently a short [N] because A and B together are about the size of my CPU cache, and my main memory is a lot slower. Going from 6*N to 8*N bytes would not be nice, but still acceptable if my find and update go to O(log N) both.
If the only operations are (1) check if value 'a' belongs to A and (2) update values in A, why don't you use a hash table in place of the sorted array B? Especially if A does not grow or shrink in size and the values only change this would be a much better solution. A hash table does not require significantly more memory than an array. (Alternatively, B should be changed not to a heap but to a binary search tree, that could be self-balancing, e.g. a splay tree or a red-black tree. However, trees require extra memory because of the left- and right-pointers.)
A practical solution that grows memory use from 6N to 8N bytes is to aim for exactly 50% filled hash table, i.e. use a hash table that consists of an array of 2N shorts. I would recommend implementing the Cuckoo Hashing mechanism (see http://en.wikipedia.org/wiki/Cuckoo_hashing). Read the article further and you find that you can get load factors above 50% (i.e. push memory consumption down from 8N towards, say, 7N) by using more hash functions. "Using just three hash functions increases the load to 91%."
From Wikipedia:
A study by Zukowski et al. has shown that cuckoo hashing is much
faster than chained hashing for small, cache-resident hash tables on
modern processors. Kenneth Ross has shown bucketized versions of
cuckoo hashing (variants that use buckets that contain more than one
key) to be faster than conventional methods also for large hash
tables, when space utilization is high. The performance of the
bucketized cuckoo hash table was investigated further by Askitis,
with its performance compared against alternative hashing schemes.
std::set usually provides the O(log(n)) insert and delete by using a binary search tree. This unfortunately uses 3*N space for most pointer based implementations. Assuming word sized data, 1 for data, 2 for pointers to left and right child on each node.
If you have some constant N and can guarantee that ceil(log2(N)) is less than half the word size you can use a fixed length array of tree nodes each 2*N size. Use 1 for data, 1 for the indexes of the two child nodes, stored as the upper and lower half of the word. Whether this would let you use a self balancing binary search tree of some manner depends on your N and word size. For a 16 bit system you only get N = 256, but for 32 its 65k.
Since you have limited N, can't you use std::set<short, cmp, pool_allocator> B with Boost's pool_allocator?
I have a list of items; I want to sort them, but I want a small element of randomness so they are not strictly in order, only on average ordered.
How can I do this most efficiently?
I don't mind if the quality of the random is not especially good, e.g. it simply based on the chance ordering of the input, e.g. an early-terminated incomplete sort.
The context is implementing a nearly-greedy search by introducing a very slight element of inexactness; this is in a tight loop and so the speed of sorting and calling random() are to be considered
My current code is to do a std::sort (this being C++) and then do a very short shuffle just in the early part of the array:
for(int i=0; i<3; i++) // I know I have more than 6 elements
std::swap(order[i],order[i+rand()%3]);
Use first two passes of JSort. Build heap twice, but do not perform insertion sort. If element of randomness is not small enough, repeat.
There is an approach that (unlike incomplete JSort) allows finer control over the resulting randomness and has time complexity dependent on randomness (the more random result is needed, the less time complexity). Use heapsort with Soft heap. For detailed description of the soft heap, see pdf 1 or pdf 2.
You could use a standard sort algorithm (is a standard library available?) and pass a predicate that "knows", given two elements, which is less than the other, or if they are equal (returning -1, 0 or 1). In the predicate then introduce a rare (configurable) case where the answer is random, by using a random number:
pseudocode:
if random(1000) == 0 then
return = random(2)-1 <-- -1,0,-1 randomly choosen
Here we have 1/1000 chances to "scamble" two elements, but that number strictly depends on the size of your container to sort.
Another thing to add in the 1000 case, could be to remove the "right" answer because that would not scramble the result!
Edit:
if random(100 * container_size) == 0 then <-- here I consider the container size
{
if element_1 < element_2
return random(1); <-- do not return the "correct" value of -1
else if element_1 > element_2
return random(1)-1; <-- do not return the "correct" value of 1
else
return random(1)==0 ? -1 : 1; <-- do not return 0
}
in my pseudocode:
random(x) = y where 0 <= y <=x
One possibility that requires a bit more space but would guarantee that existing sort algorithms could be used without modification would be to create a copy of the sort value(s) and then modify those in some fashion prior to sorting (and then use the modified value(s) for the sort).
For example, if the data to be sorted is a simple character field Name[N] then add a field (assuming data is in a structure or class) called NameMod[N]. Fill in the NameMod with a copy of Name but add some randomization. Then 3% of the time (or some appropriate amount) change the first character of the name (e.g., change it by +/- one or two characters). And then 10% of the time change the second character +/- a few characters.
Then run it through whatever sort algorithm you prefer. The benefit is that you could easily change those percentages and randomness. And the sort algorithm will still work (e.g., it would not have problems with the compare function returning inconsistent results).
If you are sure that element is at most k far away from where they should be, you can reduce quicksort N log(N) sorting time complexity down to N log(k)....
edit
More specifically, you would create k buckets, each containing N/k elements.
You can do quick sort for each bucket, which takes k * log(k) times, and then sort N/k buckets, which takes N/k log(N/k) time. Multiplying these two, you can do sorting in N log(max(N/k,k))
This can be useful because you can run sorting for each bucket in parallel, reducing total running time.
This works if you are sure that any element in the list is at most k indices away from their correct position after the sorting.
but I do not think you meant any restriction.
Split the list into two equally-sized parts. Sort each part separately, using any usual algorithm. Then merge these parts. Perform some merge iterations as usual, comparing merged elements. For other merge iterations, do not compare the elements, but instead select element from the same part, as in the previous step. It is not necessary to use RNG to decide, how to treat each element. Just ignore sorting order for every N-th element.
Other variant of this approach nearly sorts an array nearly in-place. Split the array into two parts with odd/even indexes. Sort them. (It is even possible to use standard C++ algorithm with appropriately modified iterator, like boost::permutation_iterator). Reserve some limited space at the end of the array. Merge parts, starting from the end. If merged part is going to overwrite one of the non-merged elements, just select this element. Otherwise select element in sorted order. Level of randomness is determined by the amount of reserved space.
Assuming you want the array sorted in ascending order, I would do the following:
for M iterations
pick a random index i
pick a random index k
if (i<k)!=(array[i]<array[k]) then swap(array[i],array[k])
M controls the "sortedness" of the array - as M increases the array becomes more and more sorted. I would say a reasonable value for M is n^2 where n is the length of the array. If it is too slow to pick random elements then you can precompute their indices beforehand. If the method is still too slow then you can always decrease M at the cost of getting a poorer sort.
Take a small random subset of the data and sort it. You can use this as a map to provide an estimate of where every element should appear in the final nearly-sorted list. You can scan through the full list now and move/swap elements that are not in a good position.
This is basically O(n), assuming the small initial sorting of the subset doesn't take a long time. Hopefully you can build the map such that the estimate can be extracted quickly.
Bubblesort to the rescue!
For a unsorted array, you could pick a few random elements and bubble them up or down. (maybe by rotation, which is a bit more efficient) It will be hard to control the amount of (dis)order, even if you pick all N elements, you are not sure that the whole array will be sorted, because elements are moved and you cannot ensure that you touched every element only once.
BTW: this kind of problem tends to occur in game playing engines, where the list with candidate moves is kept more-or-less sorted (because of weighted sampling), and sorting after each iteration is too expensive, and only one or a few elements are expected to move.
I have a 2D array (an image actually) that is size N x N. I need to find the indices of the M largest values in the array ( M << N x N) . Linearized index or the 2D coords are both fine. The array must remain intact (since it's an image). I can make a copy for scratch, but sorting the array will bugger up the indices.
I'm fine with doing a full pass over the array (ie. O(N^2) is fine). Anyone have a good algorithm for doing this as efficiently as possible?
Selection is sorting's austere sister (repeat this ten times in a row). Selection algorithms are less known than sort algorithms, but nonetheless useful.
You can't do better than O(N^2) (in N) here, since nothing indicates that you must not visit each element of the array.
A good approach is to keep a priority queue made of the M largest elements. This makes something O(N x N x log M).
You traverse the array, enqueuing pairs (elements, index) as you go. The queue keeps its elements sorted by first component.
Once the queue has M elements, instead of enqueuing you now:
Query the min element of the queue
If the current element of the array is greater, insert it into the queue and discard the min element of the queue
Else do nothing.
If M is bigger, sorting the array is preferable.
NOTE: #Andy Finkenstadt makes a good point (in the comments to your question) : you definitely should traverse your array in the "direction of data locality": make sure that you read memory contiguously.
Also, this is trivially parallelizable, the only non parallelizable part is when you merge the queues when joining the sub processes.
You could copy the array into a single dimensioned array of tuples (value, original X, original Y ) and build a basic heap out of it in (O(n) time), provided you implement the heap as an array.
You could then retrieve the M largest tuples in O(M lg n) time and reference their original x and y from the tuple.
If you are going to make a copy of the input array in order to do a sort, that's way worse than just walking linearly through the whole thing to pick out numbers.
So the question is how big is your M? If it is small, you can store results (i.e. structs with 2D indexes and values) in a simple array or a vector. That'll minimize heap operations but when you find a larger value than what's in your vector, you'll have to shift things around.
If you expect M to get really large, then you may need a better data structure like a binary tree (std::set) or use sorted std::deque. std::set will reduce number of times elements must be shifted in memory, while if you use std::deque, it'll do some shifting, but it'll reduce number of times you have to go to the heap significantly, which may give you better performance.
Your problem doesn't use the 2 dimensions in any interesting way, it is easier to consiger the equivalent problem in a 2d array.
There are 2 main ways to solve this problem:
Mantain a set of M largest elements, and iterate through the array. (Using a heap allows you to do this efficiently).
This is simple and is probably better in your case (M << N)
Use selection, (the following algorithm is an adaptation of quicksort):
Create an auxiliary array, containing the indexes [1..N].
Choose an arbritary index (and corresponding value), and partition the index array so that indexes corresponding to elements less go to the left, and bigger elements go to the right.
Repeat the process, binary search style until you narrow down the M largest elements.
This is good for cases with large M. If you want to avoid worst case issues (the same quicksort has) then look at more advanced algorithms, (like median of medians selection)
How many times do you search for the largest value from the array?
If you only search 1 time, then just scan through it keeping the M largest ones.
If you do it many times, just insert the values into a sorted list (probably best implemented as a balanced tree).