I am given N numbers and for them apply M rules about their order. The rules are represented in a pairs of indexes and every pair (A, B) is telling that the number with index A (A-th number) must be AFTER the B-th number - it doesn't have to be next to him.
Ex: N = 4
1 2 3 4
M = 2
3 2
3 1
Output: 1234, 4213, 4123, 2134, 2143, 2413, 1423 ...Maybe there are even more:)
The algorithm should give me all the permutations available that don't break the rules, like in the example - 3 must always be after 2 and after 1.
I tried bruteforcing but it didn't work (although bruteforce should work in here, N is in the range (1,8). )
Any ideas ?
Just as a hint.
You can treat your set of rules as a graph. Each index is a vertex, each rule is a directed edge.
Any proper ordering of the numbers (i.e. a permutation that satisfies the rules) corresponds to so called topological ordering of the above graph. In order to generate all valid orderings of your numbers you need to generate all possible topological orderings of that graph.
P.S. The first algorithm for topological ordering given at the linked Wikipedia page already allows for a fairly straightforward solution that would enumerate all valid permutations. It will take some effort and some care to implement, but it is not rocket science.
Brute forcing would be going through every permutation, which is O(N!), and for each permutation simply looping through every rule to confirm that they aplpy, which is O(M). This ends up O(N!M) which is kind of ridiculous, but it shouldn't "not work" for such a small set.
Honestly, your best bet is to go back and get the brute force solution working. Once that is done (and if you still have time, etc) you can look for a better algorithm.
EDIT to the down voter. The student is (should be) trying to get his homework done on time. By the sounds of it, his homework is a programming exercise where a brute-force solution would be adequate. Helping him to figure out an efficient algorithm is not addressing his REAL problem.
In this case he has tried the simple brute-force approach (which everyone agrees ought to work for small N values) and given up on it prematurely to try something that is probably more difficult. Any experienced developer will tell you that this is a bad idea. The student needs and deserves to be told so, and if he is sensible he will pay attention. But obviously, it his choice ...
Related
I'm implementing a matrix reduction algorithm, I'm a math student.
Obviously I've searched and read around internet but didn't find exactly what I was looking for (I list at the end what I've found and the papers that I've read.)
Quick overview of the problem:
The bitvector b has FIXED LENGTH N.
b changes at every step (could be only at a couple of indexes (most of the times) or at considerably more indexes (from 1/10 to 1/3), this only in ~10% of the cases).
I already have a sparse implementation, now I'd like to code it using some smart implementation of the bitvector.
//initialize to 0
b=bitvector(0, n=N)
for i in 1 to N
{some operations on the bitvector b}
get I= { j | b[j] == 1 }
{save I}
What I need is:
quickly set b[i]=1 or =0 (possibly O(1))
quickly get the set of indexes I at each step (definitely not more than O(logN), ideally O(1))
a C++ library that allows it
papers/documentation
What would be nice to have:
a fast way to get the "lowest one" (the last index set to 1, namely select(rank(b)), if both operations are fast (O(1)))
What I do not need is:
save space
compress the data
I have been using the library Sdsl 2.0 of Simon Gog et al. (https://github.com/simongog/sdsl-lite) but the select structure
bit_vector::select_1_type
costs O(n) to be initialized, O(1) for every query but does not "follow" the changes in b (right?? I haven't found anything very specific about it), meaning that it needs to be initialized at every step after the modifications.
Papers that I've read are:
"Fast, Small, Simple Rank/Select on Bitmaps" (G. Navarro and E. Providel) and "Practical Entropy-Compressed Rank/Select Dictionary" (D. Okanohara
K. Sadakane) and I would appreciate any link to solid implementations in C++ (if the structure fulfills my requirements)
Things that I've found here on stackexchange about similar topics that didn't help:
Dynamic bit vector in C/C++
Bit vector and bitset
Sorry for the lengthy question, I hope I explained what I need and my determination to finding it. I'm still very confused about various things related with bitvectors, it's definitely not my field of expertise, so any clarification is appreciated.
Thanks in advance.
The structure described here is the closest thing I am aware of to the properties you want.
Specifically:
initialisation is constant time
setting/clearing entries is constant time
testing for membership is constant time
retrieving the set of entries is O(N) in the number of entries (assuming you don't need them sorted - you actually end up walking them in order of insertion; you're not going to do better than O(N) overall if you need to walk all of them for whatever happens next, of course)
This question already has answers here:
How to find the only number in an array that doesn't occur twice [duplicate]
(5 answers)
Closed 7 years ago.
What would be the best algorithm for finding a number that occurs only once in a list which has all other numbers occurring exactly twice.
So, in the list of integers (lets take it as an array) each integer repeats exactly twice, except one. To find that one, what is the best algorithm.
The fastest (O(n)) and most memory efficient (O(1)) way is with the XOR operation.
In C:
int arr[] = {3, 2, 5, 2, 1, 5, 3};
int num = 0, i;
for (i=0; i < 7; i++)
num ^= arr[i];
printf("%i\n", num);
This prints "1", which is the only one that occurs once.
This works because the first time you hit a number it marks the num variable with itself, and the second time it unmarks num with itself (more or less). The only one that remains unmarked is your non-duplicate.
By the way, you can expand on this idea to very quickly find two unique numbers among a list of duplicates.
Let's call the unique numbers a and b. First take the XOR of everything, as Kyle suggested. What we get is a^b. We know a^b != 0, since a != b. Choose any 1 bit of a^b, and use that as a mask -- in more detail: choose x as a power of 2 so that x & (a^b) is nonzero.
Now split the list into two sublists -- one sublist contains all numbers y with y&x == 0, and the rest go in the other sublist. By the way we chose x, we know that a and b are in different buckets. We also know that each pair of duplicates is still in the same bucket. So we can now apply ye olde "XOR-em-all" trick to each bucket independently, and discover what a and b are completely.
Bam.
O(N) time, O(N) memory
HT= Hash Table
HT.clear()
go over the list in order
for each item you see
if(HT.Contains(item)) -> HT.Remove(item)
else
ht.add(item)
at the end, the item in the HT is the item you are looking for.
Note (credit #Jared Updike): This system will find all Odd instances of items.
comment: I don't see how can people vote up solutions that give you NLogN performance. in which universe is that "better" ?
I am even more shocked you marked the accepted answer s NLogN solution...
I do agree however that if memory is required to be constant, then NLogN would be (so far) the best solution.
Kyle's solution would obviously not catch situations were the data set does not follow the rules. If all numbers were in pairs the algorithm would give a result of zero, the exact same value as if zero would be the only value with single occurance.
If there were multiple single occurance values or triples, the result would be errouness as well.
Testing the data set might well end up with a more costly algorithm, either in memory or time.
Csmba's solution does show some errouness data (no or more then one single occurence value), but not other (quadrouples). Regarding his solution, depending on the implementation of HT, either memory and/or time is more then O(n).
If we cannot be sure about the correctness of the input set, sorting and counting or using a hashtable counting occurances with the integer itself being the hash key would both be feasible.
I would say that using a sorting algorithm and then going through the sorted list to find the number is a good way to do it.
And now the problem is finding "the best" sorting algorithm. There are a lot of sorting algorithms, each of them with its strong and weak points, so this is quite a complicated question. The Wikipedia entry seems like a nice source of info on that.
Implementation in Ruby:
a = [1,2,3,4,123,1,2,.........]
t = a.length-1
for i in 0..t
s = a.index(a[i])+1
b = a[s..t]
w = b.include?a[i]
if w == false
puts a[i]
end
end
You need to specify what you mean by "best" - to some, speed is all that matters and would qualify an answer as "best" - for others, they might forgive a few hundred milliseconds if the solution was more readable.
"Best" is subjective unless you are more specific.
That said:
Iterate through the numbers, for each number search the list for that number and when you reach the number that returns only a 1 for the number of search results, you are done.
Seems like the best you could do is to iterate through the list, for every item add it to a list of "seen" items or else remove it from the "seen" if it's already there, and at the end your list of "seen" items will include the singular element. This is O(n) in regards to time and n in regards to space (in the worst case, it will be much better if the list is sorted).
The fact that they're integers doesn't really factor in, since there's nothing special you can do with adding them up... is there?
Question
I don't understand why the selected answer is "best" by any standard. O(N*lgN) > O(N), and it changes the list (or else creates a copy of it, which is still more expensive in space and time). Am I missing something?
Depends on how large/small/diverse the numbers are though. A radix sort might be applicable which would reduce the sorting time of the O(N log N) solution by a large degree.
The sorting method and the XOR method have the same time complexity. The XOR method is only O(n) if you assume that bitwise XOR of two strings is a constant time operation. This is equivalent to saying that the size of the integers in the array is bounded by a constant. In that case you can use Radix sort to sort the array in O(n).
If the numbers are not bounded, then bitwise XOR takes time O(k) where k is the length of the bit string, and the XOR method takes O(nk). Now again Radix sort will sort the array in time O(nk).
You could simply put the elements in the set into a hash until you find a collision. In ruby, this is a one-liner.
def find_dupe(array)
h={}
array.detect { |e| h[e]||(h[e]=true; false) }
end
So, find_dupe([1,2,3,4,5,1]) would return 1.
This is actually a common "trick" interview question though. It is normally about a list of consecutive integers with one duplicate. In this case the interviewer is often looking for you to use the Gaussian sum of n-integers trick e.g. n*(n+1)/2 subtracted from the actual sum. The textbook answer is something like this.
def find_dupe_for_consecutive_integers(array)
n=array.size-1 # subtract one from array.size because of the dupe
array.sum - n*(n+1)/2
end
The problem:
N nodes are related to each other by a 'closeness' factor ranging from 0 to 1, where a factor of 1 means that the two nodes have nothing in common and 0 means the two nodes are exactly alike.
If two nodes are both close to another node (i.e. they have a factor close to 0) then this doesn't mean that they will be close together, although probabilistically they do have a much higher chance of being close together.
-
The question:
If another node is placed in the set, find the node that it is closest to in the shortest possible amount of time.
This isn't a homework question, this is a real world problem that I need to solve - but I've never taken any algorithm courses etc so I don't have a clue what sort of algorithm I should be researching.
I can index all of the nodes before another one is added and gather closeness data between each node, but short of comparing all nodes to the new node I haven't been able to come up with an efficient solution. Any ideas or help would be much appreciated :)
Because your 'closeness' metric obeys the triangle inequality, you should be able to use a variant of BK-Trees to organize your elements. Adapting them to real numbers should simply be a matter of choosing an interval to quantize your number on, and otherwise using the standard Bk-Tree procedure. Some experimentation may be required - you might want to increase the resolution of the quantization as you progress down the tree, for instance.
but short of comparing all nodes to
the new node I haven't been able to
come up with an efficient solution
Without any other information about the relationships between nodes, this is the only way you can do it since you have to figure out the closeness factor between the new node and each existing node. A O(n) algorithm can be a perfectly decent solution.
One addition you might consider - keep in mind we have no idea what data structure you are using for your objects - is to organize all present nodes into a graph, where nodes with factors below a certain threshold can be considered connected, so you can first check nodes that are more likely to be similar/related.
If you want the optimal algorithm in terms of speed, but O(n^2) space, then for each node create a sorted list of other nodes (ordered by closeness).
When you get a new node, you have to add it to the indexed list of all the other nodes, and all the other nodes need to be added to its list.
To find the closest node, just find the first node on any node's list.
Since you already need O(n^2) space (in order to store all the closeness information you need basically an NxN matrix where A[i,j] represents the closeness between i and j) you might as well sort it and get O(1) retrieval.
If this closeness forms a linear spectrum (such that closeness to something implies closeness to other things that are close to it, and not being close implies not being close to those close), then you can simply do a binary or interpolation sort on insertion for closeness, handling one extra complexity: at each point you have to see if closeness increases or decreases below or above.
For example, if we consider letters - A is close to B but far from Z - then the pre-existing elements can be kept sorted, say: A, B, E, G, K, M, Q, Z. To insert say 'F', you start by comparing with the middle element, [3] G, and the one following that: [4] K. You establish that F is closer to G than K, so the best match is either at G or to the left, and we move halfway into the unexplored region to the left... 3/2=[1] B, followed by E, and we find E's closer to F, so the match is either at E or to its right. Halving the space between our earlier checks at [3] and [1], we test at [2] and find it equally-distant, so insert it in between.
EDIT: it may work better in probabilistic situations, and require less comparisons, to start at the ends of the spectrum and work your way in (e.g. compare F to A and Z, decide it's closer to A, see if A's closer or the halfway point [3] G). Also, it might be good to finish with a comparison to the closest few points either side of where the binary/interpolation led you.
ACM Surveys September 2001 carried two papers that might be relevant, at least for background. "Searching in Metric Spaces", lead author Chavez, and "Searching in High Dimensional Spaces - Index Structures for Improving the Performance of Multimedia Databases", lead author Bohm. From memory, if all you have is the triangle inequality, you can use it to some effect, but if you can trim your data down to a sensible number of dimensions, you can do better by using a search structure that knows about this dimensional structure.
Facebook has this thing where it puts you and all of your friends in a graph, then slowly moves everyone around until people are grouped together based on mutual friends and so on.
It looked to me like they just made anything <0.5 an attractive force, anything >0.5 a repulsive force, and moved people with every iteration based on the net force. After a couple hundred iterations, it was looking pretty darn good.
Note: this is not an algorithm it is a heuristic. In the facebook implementation I saw, two people were not able to reach equilibrium and kept dancing around each other. It turns out they were actually the same person with two different accounts.
Also, it took about 15 minutes on a decent computer and ~100 nodes. YMMV.
It looks suspiciously like a Nearest Neighbor Search problem (also called a similarity search)
In one of the applications I work on, it is necessary to have a function like this:
bool IsInList(int iTest)
{
//Return if iTest appears in a set of numbers.
}
The number list is known at app load up (But are not always the same between two instances of the same application) and will not change (or added to) throughout the whole of the program. The integers themselves maybe large and have a large range so it is not efficient to have a vector<bool>. Performance is a issue as the function sits in a hot spot. I have heard about Perfect hashing but could not find out any good advice. Any pointers would be helpful. Thanks.
p.s. I'd ideally like if the solution isn't a third party library because I can't use them here. Something simple enough to be understood and manually implemented would be great if it were possible.
I would suggest using Bloom Filters in conjunction with a simple std::map.
Unfortunately the bloom filter is not part of the standard library, so you'll have to implement it yourself. However it turns out to be quite a simple structure!
A Bloom Filter is a data structure that is specialized in the question: Is this element part of the set, but does so with an incredibly tight memory requirement, and quite fast too.
The slight catch is that the answer is... special: Is this element part of the set ?
No
Maybe (with a given probability depending on the properties of the Bloom Filter)
This looks strange until you look at the implementation, and it may require some tuning (there are several properties) to lower the probability but...
What is really interesting for you, is that for all the cases it answers No, you have the guarantee that it isn't part of the set.
As such a Bloom Filter is ideal as a doorman for a Binary Tree or a Hash Map. Carefully tuned it will only let very few false positive pass. For example, gcc uses one.
What comes to my mind is gperf. However, it is based in strings and not in numbers. However, part of the calculation can be tweaked to use numbers as input for the hash generator.
integers, strings, doesn't matter
http://videolectures.net/mit6046jf05_leiserson_lec08/
After the intro, at 49:38, you'll learn how to do this. The Dot Product hash function is demonstrated since it has an elegant proof. Most hash functions are like voodoo black magic. Don't waste time here, find something that is FAST for your datatype and that offers some adjustable SEED for hashing. A good combo there is better than the alternative of growing the hash table.
#54:30 The Prof. draws picture of a standard way of doing perfect hash. The perfect minimal hash is beyond this lecture. (good luck!)
It really all depends on what you mod by.
Keep in mind, the analysis he shows can be further optimized by knowing the hardware you are running on.
The std::map you get very good performance in 99.9% scenarios. If your hot spot has the same iTest(s) multiple times, combine the map result with a temporary hash cache.
Int is one of the datatypes where it is possible to just do:
bool hash[UINT_MAX]; // stackoverflow ;)
And fill it up. If you don't care about negative numbers, then it's twice as easy.
A perfect hash function maps a set of inputs onto the integers with no collisions. Given that your input is a set of integers, the values themselves are a perfect hash function. That really has nothing to do with the problem at hand.
The most obvious and easy to implement solution for testing existence would be a sorted list or balanced binary tree. Then you could decide existence in log(N) time. I doubt it'll get much better than that.
For this problem I would use a binary search, assuming it's possible to keep the list of numbers sorted.
Wikipedia has example implementations that should be simple enough to translate to C++.
It's not necessary or practical to aim for mapping N distinct randomly dispersed integers to N contiguous buckets - i.e. a perfect minimal hash - the important thing is to identify an acceptable ratio. To do this at run-time, you can start by configuring a worst-acceptible ratio (say 1 to 20) and a no-point-being-better-than-this-ratio (say 1 to 4), then randomly vary (e.g. changing prime numbers used) a fast-to-calculate hash algorithm to see how easily you can meet increasingly difficult ratios. For worst-acceptible you don't time out, or you fall back on something slower but reliable (container or displacement lists to resolve collisions). Then, allow a second or ten (configurable) for each X% better until you can't succeed at that ratio or reach the no-pint-being-better ratio....
Just so everyone's clear, this works for inputs only known at run time with no useful patterns known beforehand, which is why different hash functions have to be trialed or actively derived at run time. It is not acceptible to simple say "integer inputs form a hash", because there are collisions when %-ed into any sane array size. But, you don't need to aim for a perfectly packed array either. Remember too that you can have a sparse array of pointers to a packed array, so there's little memory wasted for large objects.
Original Question
After working with it for a while, I came up with a number of hash functions that seemed to work reasonably well on strings, resulting in a unique - perfect hashing.
Let's say the values ranged from L to H in the array. This yields a Range R = H - L + 1.
Generally it was pretty big.
I then applied the modulus operator from H down to L + 1, looking for a mapping that keeps them unique, but has a smaller range.
In you case you are using integers. Technically, they are already hashed, but the range is large.
It may be that you can get what you want, simply by applying the modulus operator.
It may be that you need to put a hash function in front of it first.
It also may be that you can't find a perfect hash for it, in which case your container class should have a fall back position.... binary search, or map or something like that, so that
you can guarantee that the container will work in all cases.
A trie or perhaps a van Emde Boas tree might be a better bet for creating a space efficient set of integers with lookup time bring constant against the number of objects in the data structure, assuming that even std::bitset would be too large.
The common interview problem of determining the missing value in a range from 1 to N has been done a thousand times over. Variations include 2 missing values up to K missing values.
Example problem: Range [1,10] (1 2 4 5 7 8 9 10) = {3,6}
Here is an example of the various solutions:
Easy interview question got harder: given numbers 1..100, find the missing number(s)
My question is that seeing as the simple case of one missing value is of O(n) complexity and that the complexity of the larger cases converge at roughly something larger than O(nlogn):
Couldn't it just be easier to answer the question by saying sort (mergesort) the range and iterate over it observing the missing elements?
This solution should take no more than O(nlogn) and is capable of solving the problem for ranges other than 1-to-N such as 10-to-1000 or -100 to +100 etc...
Is there any reason to believe that the given solutions in the above SO link will be better than the sorting based solution for larger number of missing values?
Note: It seems a lot of the common solutions to this problem, assume an only number theoretic approach. If one is being asked such a question in an S/E interview wouldn't it be prudent to use a more computer science/algorithmic approach, assuming the approach is on par with the number theoretic solution's complexity...
More related links:
https://mathoverflow.net/questions/25374/duplicate-detection-problem
How to tell if an array is a permutation in O(n)?
You are only specifying the time complexity, but the space complexity is also important to consider.
The problem complexity can be specified in term of N (the length of the range) and K (the number of missing elements).
In the question you link, the solution of using equations is O(K) in space (or perhaps a bit more ?), as you need one equation per unknown value.
There is also the preservation point: may you alter the list of known elements ? In a number of cases this is undesirable, in which case any solution involving reordering the elements, or consuming them, must first make a copy, O(N-K) in space.
I cannot see faster than a linear solution: you need to read all known elements (N-K) and output all unknown elements (K). Therefore you cannot get better than O(N) in time.
Let us break down the solutions
Destroying, O(N) space, O(N log N) time: in-place sort
Preserving, O(K) space ?, O(N log N) time: equation system
Preserving, O(N) space, O(N) time: counting sort
Personally, though I find the equation system solution clever, I would probably use either of the sorting solutions. Let's face it: they are much simpler to code, especially the counting sort one!
And as far as time goes, in a real execution, I think the "counting sort" would beat all other solutions hands down.
Note: the counting sort does not require the range to be [0, X), any range will do, as any finite range can be transposed to the [0, X) form by a simple translation.
EDIT:
Changed the sort to O(N), one needs to have all the elements available to sort them.
Having had some time to think about the problem, I also have another solution to propose. As noted, when N grows (dramatically) the space required might explode. However, if K is small, then we could change our representation of the list, using intervals:
{4, 5, 3, 1, 7}
can be represented as
[1,1] U [3,5] U [7,7]
In the average case, maintaining a sorted list of intervals is much less costly than maintaining a sorted list of elements, and it's as easy to deduce the missing numbers too.
The time complexity is easy: O(N log N), after all it's basically an insertion sort.
Of course what's really interesting is that there is no need to actually store the list, thus you can feed it with a stream to the algorithm.
On the other hand, I have quite a hard time figuring out the average space complexity. The "final" space occupied is O(K) (at most K+1 intervals), but during the construction there will be much more missing intervals as we introduce the elements in no particular order.
The worst case is easy enough: N/2 intervals (think odd vs even numbers). I cannot however figure out the average case though. My gut feeling is telling me it should be better than O(N), but I am not that trusting.
Whether the given solution is theoretically better than the sorting one depends on N and K. While your solution has complexity of O(N*log(N)), the given solution is O(N*K). I think that the given solution is (same as the sorting solution) able to solve any range [A, B] just by transforming the range [A, B] to [1, N].
What about this?
create your own set containing all the numbers
remove the given set of numbers from your set (no need to sort)
What's left in your set are the missing numbers.
My question is that seeing as the [...] cases converge at roughly
something larger than O(nlogn) [...]
In 2011 (after you posted this question) Caf posted a simple answer that solves the problem in O(n) time and O(k) space [where the array size is n - k].
Importantly, unlike in other solutions, Caf's answer has no hidden memory requirements (using bit array's, adding numbers to elements, multiplying elements by -1 - these would all require O(log(n)) space).
Note: The question here (and the original question) didn't ask about the streaming version of the problem, and the answer here doesn't handle that case.
Regarding the other answers: I agree that many of the proposed "solutions" to this problem have dubious complexity claims, and if their time complexities aren't better in some way than either:
count sort (O(n) time and space)
compare (heap) sort (O(n*log(n)) time, O(1) space)
...then you may as well just solve the problem by sorting.
However, we can get better complexities (and more importantly, genuinely faster solutions):
Because the numbers are taken from a small, finite range, they can be 'sorted' in linear time.
All we do is initialize an array of 100 booleans, and for each input, set the boolean corresponding to each number in the input, and then step through reporting the unset booleans.
If there are total N elements where each number x is such that 1 <= x <= N then we can solve this in O(nlogn) time complexity and O(1) space complexity.
First sort the array using quicksort or mergesort.
Scan through the sorted array and if the difference between previously scanned number, a and current number, b is equal to 2 (b - a = 2), then the missing number is a+1. This can be extended to condition where (b - a > 2).
Time complexity is O(nlogn)+O(n) almost equal to O(nlogn) when N > 100.
I already answered it HERE
You can also create an array of boolean of the size last_element_in_the_existing_array + 1.
In a for loop mark all the element true that are present in the existing array.
In another for loop print the index of the elements which contains false AKA The missing ones.
Time Complexity: O(last_element_in_the_existing_array)
Space Complexity: O(array.length)
If the range is given to you well ahead, in this case range is [1,10] you can perform XOR operation with your range and the numbers given to you. Since XOR is commutative operation. You will be left with {3,6}
(1 2 3 4 5 6 7 8 9 10) XOR (1 2 4 5 7 8 9 10) ={3,6}