What is the time complexity of String.contains();
lets say n is the length of the string that is compared against another string of length k.
There is no answer without knowing the actual implementation of the String.contains() that you're interested in; or what algorithm you intend to use.
A completely naive implementation might take (n+1-k)*kcomparisons to decide that a given string of length n does not contain a particular substring of length k. That's O(nk) for the worst case.
Even stopping substring comparisons after the first unequal comparison, while having a smaller coefficient, still is O(nk). Construct a string that's a repetition of many isolated letters, each separated by exactly k-1 spaces, and search that for an occurrence of k consecutive spaces. The search will fail, but each substring comparison will take an amortized k/2 compares to find that out, and you're still at O(nk).
If k is known to be much less than n, you could treat that as O(n).
The average case depends on the actual algorithm used, and also on the distribution of characters in the two strings; and you haven't said what either of those were.
Related
I'm looking at the time complexity for implementations of a method which determines if a String contains all unique characters.
The basic, brute force, approach would be to iterate through the String one character at a time maintaining a HashSet of seen characters. For each character in the iteration we check if the Set already contains it, and if so return false. We return true if the entire String has been searched. This would be O(n) as a worst case complexity. What would be the average case? O(n/2)?
If we try to optimise this by sorting the String into a char array, would it be more or less efficient? Sorting typically takes O(n log n) which is worse than O(n), but a sorted String allows for duplicate characters to be detected much earlier (especially for long strings).
Do we say the worst case is O(n^2 log n) but the average case is better? If so, what is it?
In the un-sorted case, the average case depends entirely on the string! Without knowing/assuming any distribution, it's hard to make any assumption.
A simple case, for a string with randomly-placed characters, where one of the characters repeats once:
the number of possibilities for the repeated characters being arranged is n*(n-1)/2
the probability it is detected repeated in exactly k steps is (k-1)/(n-1)
the probability it is detected in at most k steps is (k*(k-1))/(n*(n-1)), meaning that on average you will detect it (for large n) in about 0.7071*n... [incomplete]
For multiple characters that occur with different frequencies, or you make different assumptions on how characters are distributed in the string, you'll get different probabilities.
Hopefully someone can extend on my answer! :)
If the string is sorted, then you don't need the HashSet.
However, the average case still depends on the distribution of characters in the string: if you get two aa in the beggining, it's pretty efficient; if you get two zz, then you didn't win anything.
The worst case is sorting plus detecting-duplicates, so O(n log n + n), or just O(n log n).
So, it appears it's not advantageous to sort the string beforehand, due to the increased complexity, both in average-case and worst-case.
I already searched for posts on this question. But none of them have clear answers.
Find the occurrence of most common substring with length n in given string.
For example, "deded", we set the length of substring to be 3. "ded" will be the most common substring and its occurrence is 2.
Few post suggest using suffix tree and the time complexity is O(nlgn), space complexity is O(n).
First, I'm not familiar with suffix tree. My idea is to use hashmap store the occurrence of each substring with length of 3. The time is O(n) while space is also O(n). Is this better than suffix tree? Should I take hashmap collison into account?
Extra: if above problem is addressed, how can we solve the problem that length of substring doesn't matter. Just find the most common substring in given string.
If the length of the most common substring doesn't matter (but say, you want it to be greater than 1) then the best solution is to look for the most common substring of length 2. You can do this with a suffix tree in linear time, if you look up suffix trees then it will be clear how to do this. If you want the length M of the most common substring to be an input parameter, then you can hash all substrings of length M in linear time using hashing with multiply-and-add where you multiply the previous string hash value by a constant and then add the value for the next least significant value in the string, and take the modulus modulo a prime P. If you pick your modulus P for the computed string integers to be a randomly chosen prime P such that you can store O(P) memory, then this will do the trick, in linear time if you assume that your hashing has no collisions. If you assume that your hashing might have a lot of collisions, and the substring is of length M and the total string length is N, then the running time would be O(MN) because you have to check all collisions, which in the worst case could be checking all substrings of length M for example if your string is a string of all one character. Suffix trees are better in the worst case, let me know if you want some details (but not completely, because suffix trees are complicated) and I can explain at a high level how to get a faster solution with suffix trees.
I take a input from the user and its a string with a certain substring which repeats itself all through the string. I need to output the substring or its length AKA period.
Say
S1 = AAAA // substring is A
S2 = ABAB // Substring is AB
S3 = ABCAB // Substring is ABC
S4 = EFIEFI // Substring is EFI
I could start with a Single char and check if it is same as its next character if it is not, I could do it with two characters then with three and so on. This would be a O(N^2) algo. I was wondering if there is a more elegant solution to this.
You can do this in linear time and constant additional space by inductively computing the period of each prefix of the string. I can't recall the details (there are several things to get right), but you can find them in Section 13.6 of "Text algorithms" by Crochemore and Rytter under function Per(x).
Let me assume that the length of the string n is at least twice greater than the period p.
Algorithm
Let m = 1, and S the whole string
Take m = m*2
Find the next occurrence of the substring S[:m]
Let k be the start of the next occurrence
Check if S[:k] is the period
if not go to 2.
Example
Suppose we have a string
CDCDFBFCDCDFDFCDCDFBFCDCDFDFCDC
For each power m of 2 we find repetitions of first 2^m characters. Then we extend this sequence to it's second occurrence. Let's start with 2^1 so CD.
CDCDFBFCDCDFDFCDCDFBFCDCDFDFCDC
CDCD CDCD CDCD CDCD CD
We don't extend CD since the next occurrence is just after that. However CD is not the substring we are looking for so let's take the next power: 2^2 = 4 and substring CDCD.
CDCDFBFCDCDFDFCDCDFBFCDCDFDFCDC
CDCD CDCD
Now let's extend our string to the first repetition. We get
CDCDFBF
we check if this is periodic. It is not so we go further. We try 2^3 = 8, so CDCDFBFC
CDCDFBFCDCDFDFCDCDFBFCDCDFDFCDC
CDCDFBFC CDCDFBFC
we try to extend and we get
CDCDFBFCDCDFDF
and this indeed is our period.
I expect this to work in O(n log n) with some KMP-like algorithm for checking where a given string appears. Note that some edge cases still should be worked out here.
Intuitively this should work, but my intuition failed once on this problem already so please correct me if I'm wrong. I will try to figure out a proof.
A very nice problem though.
You can build a suffix tree for the entire string in linear time (suffix tree is easy to look up online), and then recursively compute and store the number of suffix tree leaves (occurences of the suffix prefix) N(v) below each internal node v of the suffix tree. Also recursively compute and store the length of each suffix prefix L(v) at each node of the tree. Then, at an internal node v in the tree, the suffix prefix encoded at v is a repeating subsequence that generates your string if N(v) equals the total length of the string divided by L(v).
We can actually optimise the time complexity by creating a Z Array. We can create Z array in O(n) time and O(n) space. Now, lets say if there is string
S1 = abababab
For this the z array would like
z[]={8,0,6,0,4,0,2,0};
In order to calcutate the period we can iterate over the z array and
use the condition, where i+z[i]=S1.length. Then, that i would be the period.
Well if every character in the input string is part of the repeating substring, then all you have to do is store first character and compare it with rest of the string's characters one by one. If you find a match, string until to matched one is your repeating string.
I too have been looking for the time-space-optimal solution to this problem. The accepted answer by tmyklebu essentially seems to be it, but I would like to offer some explanation of what it's actually about and some further findings.
First, this question by me proposes a seemingly promising but incorrect solution, with notes on why it's incorrect: Is this algorithm correct for finding period of a string?
In general, the problem "find the period" is equivalent to "find the pattern within itself" (in some sense, "strstr(x+1,x)"), but with no constraints matching past its end. This means that you can find the period by taking any left-to-right string matching algorith, and applying it to itself, considering a partial match that hits the end of the haystack/text as a match, and the time and space requirements are the same as those of whatever string matching algorithm you use.
The approach cited in tmyklebu's answer is essentially applying this principle to String Matching on Ordered Alphabets, also explained here. Another time-space-optimal solution should be possible using the GS algorithm.
The fairly well-known and simple Two Way algorithm (also explained here) unfortunately is not a solution because it's not left-to-right. In particular, the advancement after a mismatch in the left factor depends on the right factor having been a match, and the impossibility of another match misaligned with the right factor modulo the right factor's period. When searching for the pattern within itself and disregarding anything past the end, we can't conclude anything about how soon the next right-factor match could occur (part or all of the right factor may have shifted past the end of the pattern), and therefore a shift that preserves linear time cannot be made.
Of course, if working space is available, a number of other algorithms may be used. KMP is linear-time with O(n) space, and it may be possible to adapt it to something still reasonably efficient with only logarithmic space.
I am looking for the most efficient way to store binary strings in a data structure (insert function) and then when getting a string I want to check if some cyclic string of the given string is in my structure.
I thought about storing the input strings in a Trie but then when trying to determine whether some cyclic string of the string I got now was inserted to the Trie means to do |s| searches in the Trie for all the possible cyclic strings.
Is there any way to do that more efficiently while the place complexity will be like in a Trie?
Note: When I say cyclic strings of a string I mean that for example all the cyclic strings of 1011 are: 0111, 1110, 1101, 1011
Can you come up with a canonicalizing function for cyclic strings based on the following:
Find the largest run of zeroes.
Rotate the string so that that run of zeroes is at the front.
For each run of zeroes of equal size, see if rotating that to the front produces a lexicographically lesser string and if so use that.
This would canonicalize everything in the equivalence class (1011, 1101, 1110, 0111) to the lexicographically least value: 0111.
0101010101 is a thorny instance for which this algo will not perform well, but if your bits are roughly randomly distributed, it should work well in practice for long strings.
You can then hash based on the canonical form or use a trie that will include only the empty string and strings that start with 0 and a single trie run will answer your question.
EDIT:
if I have a string of a length |s| it can take a lot of time to find the least lexicographically value..how much time will it actually take?
That's why I said 010101.... is a value for which it performs badly. Let's say the string is of length n and the longest run of 1's is of length r. If the bits are randomly distributed, the length of the longest run is O(log n) according to "Distribution of longest run".
The time to find the longest run is O(n). You can implement shifting using an offset instead of a buffer copy, which should be O(1). The number of runs is worst case O(n / m).
Then, the time to do step 3 should be
Find other long runs: one O(n) pass with O(log n) storage average case, O(n) worst case
For each run: O(log n) average case, O(n) worst case
Shift and compare lexicographically: O(log n) average case since most comparisons of randomly chosen strings fail early, O(n) worst case.
This leads to a worst case of O(n²) but an average case of O(n + log² n) ≅ O(n).
You have n strings s1..sn and given a string t you want to know whether a cyclic permutation of t is a substring of any s1..sn. And you want to store the strings as efficiently as possible. Did I understand your question correctly?
If so, here is a solution, but with a large run-time: for a given input t, let t' = concat(t,t), check t' with every s in s1..sn to see if the longest subsequence of t' and sm is at least |t| If |si| = k, |t| = l it runs in O(n.k.l) time. And you can store s1..sn in any data structure you want. Is that good enough or you?
You are given a string. Develop a function to remove duplicate characters from that string. String could be of any length. Your algorithm must be in space. If you wish you can use constant size extra space which is not dependent any how on string size. Your algorithm must be of complexity of O(n).
My idea was to define an integer array of size of 26 where 0th index would correspond to the letter a and the 25th index for the letter z and initialize all the elements to 0.
Thus we will travel the entire string once and and would increment the value at the desired index as and when we encounter a letter.
and then we will travel the string once again and if the value at the desired index is 1 we print out the letter otherwise we do not.
In this way the time complexity is O(n) and the space used is constant irrespective of the length of the string!!
if anyone can come up with ideas of better efficiency,it will be very helpful!!
Your solution definitely fits the criteria of O(n) time. Instead of an array, which would be very, very large if the allowed alphabet is large (Unicode has over a million characters), you could use a plain hash. Here is your algorithm in (unoptimized!) Ruby:
def undup(s)
seen = Hash.new(0)
s.each_char {|c| seen[c] += 1}
result = ""
s.each_char {|c| result << c if seen[c] == 1}
result
end
puts(undup "")
puts(undup "abc")
puts(undup "Olé")
puts(undup "asdasjhdfasjhdfasbfdasdfaghsfdahgsdfahgsdfhgt")
It makes two passes through the string, and since hash lookup is less than linear, you're good.
You can say the Hashtable (like your array) uses constant space, albeit large, because it is bounded above by the size of the alphabet. Even if the size of the alphabet is larger than that of the string, it still counts as constant space.
There are many variations to this problem, many of which are fun. To do it truly in place, you can sort first; this gives O(n log n). There are variations on merge sort where you ignore dups during the merge. In fact, this "no external hashtable" restriction appears in Algorithm: efficient way to remove duplicate integers from an array (also tagged interview question).
Another common interview question starts with a simple string, then they say, okay now a million character string, okay now a string with 100 billion characters, and so on. Things get very interesting when you start considering Big Data.
Anyway, your idea is pretty good. It can generally be tweaked as follows: Use a set, not a dictionary. Go trough the string. For each character, if it is not in the set, add it. If it is, delete it. Sets take up less space, don't need counters, and can be implemented as bitsets if the alphabet is small, and this algorithm does not need two passes.
Python implementation: http://code.activestate.com/recipes/52560-remove-duplicates-from-a-sequence/
You can also use a bitset instead of the additional array to keep track of found chars. Depending on which characters (a-z or more) are allowed you size the bitset accordingly. This requires less space than an integer array.