We Keep Coding

Shortest possible resulting length after iterated string replacement

How would I go about reasonably efficiently finding the shortest possible output given by repeatedly applying replacements to an input sequence? I believe (please correct me if I am wrong) that this is exponential-time in the worst case, but I am not sure due to the second constraint below. The naive method certainly is.
I tried coding the naive method (for all possible replacements, for all valid positions, recurse on a copy of the input after applying the replacement at the position. Return the shortest of all valid recursions and the input, with a cache on the function to catch equivalent replacement sequences), but it is (unworkably) slow, and I'm pretty sure it's an algorithmic issue as opposed to the implementation.
A couple of things that may (or may not) make a difference:
Token is an enumerated type.
The length of the output of each entry in the map is strictly less than the input of the entry.
I do not need what replacements were done and where, just the resulting sequence.
So, as an example where each character is a token (for simplicity's sake), if I have the replacement map as aaba -> a, aaa -> ab, and aba -> bb, and I apply minimalString('aaaaa'), I want to get 'a'.
The actual method signature is something along the following lines:
List<Token> getMinimalAfterReplacements(List<Token> inputList, Map<List<Token>, List<Token>> replacements) {
?
}
Is there a better method than brute-force? If not, is there, for example, a SAT library or similar that could be harnessed? Is there any preprocessing to the map that could be done to make it faster when called multiple times with different token lists but with the same replacement map?

The code below is a Python version to find the shortest possible reduction. It is non-recursive but not too far from the naive algorithm. In every step it tries all possible single reductions, thus, obtaining a set of strings to reduce for the next step.
One optimization that helps in cases when there are "symbol eating" rules like "aa" -> "a" is to check the next set of strings for duplicates.
Another optimization (not implemented in the code below) would be to process the replacement rules into a finite automaton that finds locations of all possible single reductions with a single pass through the input string. This would not help the exponential nature of the main tree search algorithm, though.
class Replacer:
def __init__(self, replacements):
self.replacements = [[tuple(key), tuple(value)] for key, value in replacements.items()]
def get_possible_replacements(self, input):
"Return all possible variations where a single replacement was done to the input"
result = []
for replace_what, replace_with in self.replacements:
#print replace_what, replace_with
for p in range(1 + len(input) - len(replace_what)):
if input[p : p + len(replace_what)] == replace_what:
input_copy = list(input[:])
input_copy[p : p + len(replace_what)] = replace_with
result.append(tuple(input_copy))
return result
def get_minimum_sequence_list(self, input):
"Return the shortest irreducible sequence that can be obtained from the given input"
irreducible = []
to_reduce = [tuple(input)]
to_reduce_new = []
step = 1
while to_reduce:
print "Reduction step", step, ", number of candidates to reduce:", len(to_reduce)
step += 1
for current_input in to_reduce:
reductions = self.get_possible_replacements(current_input)
if not reductions:
irreducible.append(current_input)
else:
to_reduce_new += reductions
to_reduce = set(to_reduce_new[:]) # This dramatically reduces the tree width by removing duplicates
to_reduce_new = []
irreducible_sorted = sorted(set(irreducible), key = lambda x: len(x))
#print "".join(input), "could be reduced to any of", ["".join(x) for x in irreducible_sorted]
return irreducible_sorted[0]
def get_minimum_sequence(self, input):
return "".join(self.get_minimum_sequence_list(list(input)))
input = "aaaaa"
replacements = {
"aaba" : "a",
"aaa" : "ab",
"aba" : "bb",
}
replacer = Replacer(replacements)
replaced = replacer.get_minimum_sequence(input)
print "The shortest string", input, "could be reduced to is", replaced

Just a simple idea which might reduce the branching: With rules like
ba -> c
ca -> b
and a string like
aaabaacaa
^ ^
you can do two substitutions and their order doesn't matter. This is already sort of covered by memoization, however, there's still a considerable overhead for generating the useless string. So I'd suggest the following rule:
After a substitution on position p, consider only substitutions on positions q such that
q + length(lhs_of_the_rule) > p
i.e., such that don't start to the left of the previous substitutions or they overlap.
As a simple low-level optimization I'd suggest to replace the List<Token> by a String or (or an encapsulated byte[] or short[] or whatever). The lower memory footprint should help the cache and you can index an array by a string element (or two) in order to find out what rules may be applicable for it.

Efficient data structure that checks for existence of String

I am writing a program which will add a growing number or unique strings to a data structure. Once this is done, I later need to constantly check for existence of the string in it.
If I were to use an ArrayList I believe checking for the existence of some specified string would iterate through all items until a matching string is found (or reach the end and return false).
However, with a HashMap I know that in constant time I can simply use the key as a String and return any non-null object, making this operation faster. However, I am not keen on filling a HashMap where the value is completely arbitrary. Is there a readily available data structure that uses hash functions, but doesn't require a value to be placed?

If I were to use an ArrayList I believe checking for the existence of some specified string would iterate through all items until a matching string is found
Correct, checking a list for an item is linear in the number of entries of the list.
However, I am not keen on filling a HashMap where the value is completely arbitrary
You don't have to: Java provides a HashSet<T> class, which is very much like a HashMap without the value part.
You can put all your strings there, and then check for presence or absence of other strings in constant time;
Set<String> knownStrings = new HashSet<String>();
... // Fill the set with strings
if (knownString.contains(myString)) {
...
}

It depends on many factors, including the number of strings you have to feed into that data structure (do you know the number by advance, or have a basic idea?), and what you expect the hit/miss ratio to be.
A very efficient data structure to use is a trie or a radix tree; they are basically made for that. For an explanation of how they work, see the wikipedia entry (a followup to the radix tree definition is in this page). There are Java implementations (one of them is here; however I have a fixed set of strings to inject, which is why I use a builder).
If your number of strings is really huge and you don't expect a minimal miss ratio then you might also consider using a bloom filter; the problem however is that it is probabilistic; but you can get very quick answers to "not there". Here also, there are implementations in Java (Guava has an implementation for instance).
Otherwise, well, a HashSet...

A HashSet is probably the right answer, but if you choose (for simplicity, eg) to search a list it's probably more efficient to concatenate your words into a String with separators:
String wordList = "$word1$word2$word3$word4$...";
Then create a search argument with your word between the separators:
String searchArg = "$" + searchWord + "$";
Then search with, say, contains:
bool wordFound = wordList.contains(searchArg);
You can maybe make this a tiny bit more efficient by using StringBuilder to build the searchArg.

As others mentioned HashSet is the way to go. But if the size is going to be large and you are fine with false positives (checking if the username exists) you can use BloomFilters (probabilistic data structure) as well.

Would Java indexOf (brute force method) be more practical for me or some other substring algorithm?

I'm looking at finding very short substrings (pattern, needle) in many short lines of text (haystack). However, I'm not quite sure which method to use outside the naive, brute force method.
Background: I'm doing a side project for fun where I receive text messaging chat logs of multiple users (anywhere from 2000-15000 lines of text and 2-50 users), and I want to find all the various pattern matches in the chat logs based on predetermined words that I've come up with. So far I have about 1600 patterns that I'm looking for, but I may look for more.
So for example, I want to find the number of food related words that are used in an average text message log such as "hamburger", "pizza", "coke", "lunch", "dinner", "restaurant", "McDonalds". While I gave out English examples, I will actually be using Korean for my program. Each of these designated words will have their own respective score, which I put in a hashmap as key and value separately. I then show the top scorers for food related words as well as the most frequent words used by those users for food words.
My current method is to eliminate each line of text by whitespaces, and process each individual word from the haystack by using contains method (which uses the indexOf method and the naive substring search algorithm) of the haystack contains the pattern.
wordFromInput.contains(wordFromPattern);
To give an example, with 17 users in chat, 13000 lines of text, and the 1600 patterns, I've found that this whole program took 12-13 seconds with this method. And on the Android app that I'm developing, it took 2 minutes and 30 seconds to process, which is far too slow.
Originally, I tried to use a hash map and to merely get the pattern instead of searching for it in the ArrayList, but I then realized that is...
not possible with hash table
for what I am trying to do with a substring.
I've looked around through Stackoverflow and found a lot of helpful and related questions, such as these two:
1 and 2. I'm somewhat more familiar with the various string algorithms (Boyer Moore, KMP, etc.)
I initially thought then that the naive method would of course be the worst type of algorithm for my case, but having found this question, I've realized that my case (short pattern, short text), might actually be more effective with the naive method. But I wanted to know if there was something that I was neglecting completely.
Here is a snippet of my code though if anyone wants to see my issue more concretely.
While I removed large parts of the code to simplify it, the primary method that I use to actually match substrings is there in the method matchWords().
I know that's really ugly and bad code (5 for loops...), so if there are any suggestions for that, I'm happy to hear it as well.
So to clean it up:
lines of text from chat logs (2000-10,000+), haystack
1600+ patterns, needle(s)
mostly using Korean characters, although some English is included
Brute force naive method is simply too slow, but debating whether there are other alternatives and even if there are, whether they are practical given the nature of short patterns and text.
I just want some input on my thought process, and possibly some general advice. But additionally, I would like some specific suggestion for a particular algorithm or method if that is possible.

You can replace the hashtable with a Trie.
Split the line of text into words using white space to separate words. Then check if the word is in the Trie. If it is in the Trie, update a counter associated with the word. Ideally, the counter would be integrated into the Trie.
This appraoch is O(C) where C is the number of characters in the text. It's highly unlikely that you can avoid checking each character at least once. Thus this approach should be as good as you can get at least in terms of big O.
However, it sounds like you may not want to list all of the possible words you are searching for. Therefore, you might want to simply use you could build a counting Trie from all of the words. If nothing else that'll probably make it easier for any pattern matching algorithm you use. Although, it might require some modifications to the Trie.

What you're describing sounds like an excellent use case for the Aho-Corasick string-matching algorithm. This algorithm finds all matches of a set of pattern strings inside of a source string and does so in linear time (plus the time to report the matches). If you have a fixed set of strings to search for, you can do linear preprocessing work up front on the patterns to search for all matches very quickly.
There's a Java implementation of Aho-Corasick available here. I haven't tried it out, but it might be a good match.
Hope this helps!

I'm pretty sure string.contains is already highly optimized, so replacing it with something else is not going to do you a lot of good.
So the way to go, I suspect, is not to look for each and every bank-word in your chat words, but rather do multiple comparisons at once.
The first way to do it would be to create one huge regular expression that will match all your bank-words. Compile it and hope the regular expression package is efficient enough (chances are - it is). You will have a rather lengthy setup stage (the regex compilation), but matches should be a lot faster.

You can build an index of the words you need to match and count them as you process them. If you can use a HashMap to lookup the patterns for each word, the cost will be O(n * m)
You can use a HashMap for all the possible words, you can then dissect the words later.
e.g. say you need to match red and apple, you can combine the sum of
redapple = 1
applered = 0
red = 10
apple = 15
This means that red is actually 11 (10 + 1), and apple is 16 (15 + 1)

I don't know Korean so I imagine the same strategies used to tinker with Strings in Korean isn't necessarily possible in the way it is with English, but perhaps this strategy in pseudocode can be applied with your knowledge of Korean to make it work. (Java is of course still the same, but for example, in Korean is it still highly likely for the letters "ough" to be in succession? Are there even letters for "ough"? But with that being said, hopefully the principle can be applied
I would use String.toCharArray to create a two-dimensional array (or ArrayList if variable size needed). The
if (first letter of word matches keyword's first letter)//we have a candidate
skip to last letter of the current word //see comment below
if(last letter of word matches keyword's last letter)//strong candidate
iterate backwards to start+1 checking remainder of letters
The reason I suggest to skip to the last letter is because statistically a "consonant, vowel" for the first two letters of a word is significantly high, especially nouns, which will consist of alot of your keywords since any food is a noun (almost all the keyword examples you gave were matched that structure of consonant, vowel). And since there are only 5 vowels(plus y), the likelihood of the second letter "i" showing up in the keyword "pizza" is inherently highly likely, yet after that point there is still a good chance that the word may turn out to not be a match.
However if you know that the first letter and the last letter match, then you probably have a much stronger candidate and can then iterate in reverse. I think over larger sets of data, this would eliminate candidates much faster than checking letters in order. Basically you'd be letting too many fake candidates past the second iteration, thus increasing your overall conditional operations. It might sound like something small, but in a project like this there's lots of reiterating, so micro-optimizations will accumulate very quickly.
If this approach can be applied in a language that's probably structurally very different from English(I'm speaking from ignorance here though), then I think it might provide some efficiency for you whether you make it happen through iterating a char array or with a scanner, or any other construct.

The trick is to realise that if you can describe the string you are searching for as a regular expression you can also, by definition, describe it with a state machine.
At every character in your message start a state machine for every one of your 1600 patterns and pass the character through it. This sounds scary but believe me most of them will terminate immediately anyway so you aren't really doing a huge amount of work. Bear in mind that a state machine can usually be encoded with a simple switch/case or a ch == s.charAt at each step so they are close to the ultimate in light-weight.
Obviously you know what to do whenever one of your search machines terminates at the end of their search. Any that terminate before full-match can be discarded immediately.
private static class Matcher {
private final int where;
private final String s;
private int i = 0;
public Matcher ( String s, int where ) {
this.s = s;
this.where = where;
}
public boolean match(char ch) {
return s.charAt(i++) == ch;
}
public int matched() {
return i == s.length() ? where: -1;
}
}
// Words I am looking for.
String[] watchFor = new String[] {"flies", "like", "arrow", "banana", "a"};
// Test string to search.
String test = "Time flies like an arrow, fruit flies like a banana";
public void test() {
// Use a LinkedList because it is O(1) to remove anywhere.
List<Matcher> matchers = new LinkedList<> ();
int pos = 0;
for ( char c : test.toCharArray()) {
// Fire off all of the matchers at this point.
for ( String s : watchFor ) {
matchers.add(new Matcher(s, pos));
}
// Discard all matchers that fail here.
for ( Iterator<Matcher> i = matchers.iterator(); i.hasNext(); ) {
Matcher m = i.next();
// Should it be removed?
boolean remove = !m.match(c);
if ( !remove ) {
// Still matches! Is it complete?
int matched = m.matched();
if ( matched >= 0 ) {
// Todo - Should use getters.
System.out.println(" "+m.s +" found at "+m.where+" active matchers "+matchers.size());
// Complete!
remove = true;
}
}
// Remove it where necessary.
if ( remove ) {
i.remove();
}
}
// Step pos to keep track.
pos += 1;
}
}
prints
flies found at 5 active matchers 6
like found at 11 active matchers 6
a found at 16 active matchers 2
a found at 19 active matchers 2
arrow found at 19 active matchers 6
flies found at 32 active matchers 6
like found at 38 active matchers 6
a found at 43 active matchers 2
a found at 46 active matchers 3
a found at 48 active matchers 3
banana found at 45 active matchers 6
a found at 50 active matchers 2
There are several simple optimisations. With some simple pre-processing the most obvious is to use the current character to determine which matchers may be applicable.

This is a pretty broad question, so I won't go into too much detail, but roughly:
Pre-process the haystacks using something like broad lemmatizer to create "topic word only" versions of the messages by noting which topics all words in it cover. For example, any occurrences of "hamburger", "pizza", "coke", "lunch", "dinner", "restaurant", or "McDonalds" would cause the "topic" word "food" to be collected for that message. Some words may have multiple topics, eg "McDonalds" may be in the topics "food" and "business". Most words won't have any topic.
After this process, you'll have haystacks consisting of only "topic" words. Then create a Map<String, Set<Integer>> and populate it with the topic word and the Set of chat message ids that contain it. This is reverse index of topic word to the chat messages that contain it.
The runtime code to find all documents that contain all n words is then trivial and super fast - near O(#terms):
private Map<String, Set<Integer>> index; // pre-populated
Set<Integer> search(String... topics) {
Set<Integer> results = null;
for (String topic : topics) {
Set<Integer> hits = index.get(topic);
if (hits == null)
return Collections.emptySet();
if (results == null)
results = new HashSet<Integer>(hits);
else
results.retainAll(hits);
if (results.isEmpty())
return Collections.emptySet(); // exit early
}
return results;
}
This will perform near O(1), and tell you which messages share all search terms. If you just want the number, use the trivial size() of the returned Set.

Check if a keyword is in a string

I have a list of keywords and I want to be able to find if a string contains any of those keywords. Right now the solution I have takes O(n). Is there a quicker way of doing this search without looping through each keywords and doing a comparison/contains?
i.e.
Keywords = "cat", "hat", "mat", "bat", "fat", "sat", "rat", "pat", "foo bar", "foo-bar"
String = "There is a cat in the box."
The result of this is true because "cat" matches one of the words in the 'keywords'
EDIT:
I guess I wasn't as clear when I said O(n). I mean to say O(n) where n=number of keywords.

You can use Boyer-Moore, which involves preprocessing the string, but you're not going to be able to beat a worst case of O(KN), where K is the sum of the lengths of the keywords, and N is the length of the string. Best case is of course sublinear, but you can't have a worst case sublinear runtime.
Note that the comparisons aren't free. It's not like you can compare two strings in O(1) to see if they're equal, you have to iterate through the characters. Hashing gets you to what you need to compare to in constant time, but doesn't help more than that since two different strings can have the same hash. That's not to say that hashing isn't good, it is, but it does not alter the worst case runtime complexity.
In the end, you need to compare the characters, and Boyer-Moore provides a very good way to do that. Of course if you're using some sort of hash based build, you may be able to rule out certain keywords in amortized constant time, but that doesn't change the fact that in the worst case (and many other cases), you're going to need to compare characters.
Also note that depending on what we assume about the data, and how we construct our indexing structure(s), it's possible to achieve a very good actual runtime. Just because the worst case complexity isn't sublinear doesn't mean that the actual runtime won't be really fast. There is no singleton simple or correct solution, the problem can be approached in myriad ways. There's never a quick and dirty answer to solve all of your problems when it comes to information retrieval.

Could try using contains().
Get the String; String passed = "there is a cat in the box";
use for loop to go through your key words. if keywords is an array.
for(int i = 0; i < keywords.length; i++){
if(passed.toLowerCase().contains(keywords[i]){
//set true;
}else{
//set false;
}
}
Either going through a loop or checking each word individually, i dont think you'll get much better than O(n)

k = # of chars in sentence
n = # of keywords
m = # of words in sentence
You can get O(k + n) time complexity by hashing the words in sentence.
Separating the sentence into words takes O(k). Creating the HashSet also takes O(k). Checking the hash n times takes n*O(1) = O(n), so the overall time complexity is O(k + n).
Edit1: Hashing all n keywords is technically n*O(k/m), where k/m is avg. word length. However, k/m does not scale with the size of the input, so it still gives O(n).
Edit2: FYI, Boyer-Moore will match any substring, not just keywords; E.g. "cat" will match "catepillar". Also, because it is more general it has a worse running time than a simple word match, O(KN) as #SteveP. has in his answer.
So if you only need word matching, as opposed to substring matching, stick with hashing as above.

not sure it will find inO(n).
but the solution to find the element could be like this
List<String> keywords = new ArrayList<String> (Arrays.asList("cat", "hat", "mat", "bat", "fat", "sat", "rat", "pat", "foo bar", "foo-bar"));
String search= "There is a cat in the box." ;
List<String> searchWords = new ArrayList<String> (Arrays.asList(search.split(" ")));
System.out.println(!Collections.disjoint(keywords,searchWords));

You're probably not going to get any better than O(n), since there's a linear component to this piece - you have to trawl the string in some shape, form, or fashion.
Consider using a Set:
Constant-time to add all of the elements (could be amoritized to N for N elements)
Constant-time look up for existence
public boolean inPhrase(String phrase, String searchWord) {
Set<String> phraseSet = new HashSet<>();
// remove the punctuation and split the words on white space.
for(String s: phrase.replaceAll("[.,?!;"'], "").split(" ")) {
phraseSet.add(s);
}
return phraseSet.contains(searchWord);
}

Fastest way to find Strings in String collection that begin with certain chars

I have a large collection of Strings. I want to be able to find the Strings that begin with "Foo" or the Strings that end with "Bar". What would be the best Collection type to get the fastest results? (I am using Java)
I know that a HashSet is very fast for complete matches, but not for partial matches I would think? So, what could I use instead of just looping through a List? Should I look into LinkedList's or similar types? Are there any Collection Types that are optimized for this kind of queries?

The best collection type for this problem is SortedSet. You would need two of them in fact:
Words in regular order.
Words with their characters inverted.
Once these SortedSets have been created, you can use method subSet to find what you are looking for. For example:
Words starting with "Foo":
forwardSortedSet.subSet("Foo","Fop");
Words ending with "Bar":
backwardSortedSet.subSet("raB","raC");
The reason we are "adding" 1 to the last search character is to obtain the whole range. The "ending" word is excluded from the subSet, so there is no problem.
EDIT: Of the two concrete classes that implement SortedSet in the standard Java library, use TreeSet. The other (ConcurrentSkipListSet) is oriented to concurrent programs and thus not optimized for this situation.

It's been a while but I needed to implement this now and did some testing.
I already have a HashSet<String> as source so generation of all other datastructures is included in search time. 100 different sources are used and each time the data structures need to be regenerated. I only need to match a few single Strings each time. These tests ran on Android.
Methods:
Simple loop through HashSet and call endsWith() on
each string
Simple loop through HashSet and perform precompiled
Pattern match (regex) on each string.
Convert HashSet to single String joined by \n and
single match on whole String.
Generate SortedTree with reversed Strings from
HashSet. Then match with subset() as explained by #Mario Rossi.
Results:
Duration for method 1: 173ms (data setup:0ms search:173ms)
Duration for method 2: 6909ms (data setup:0ms search:6909ms)
Duration for method 3: 3026ms (data setup:2377ms search:649ms)
Duration for method 4: 2111ms (data setup:2101ms search:10ms)
Conclusion:
SortedSet/SortedTree is extremely fast in searching. Much faster than just looping through all Strings. However, creating the structure takes a lot of time. Regexes are much slower, but generating a single large String out of hundreds of Strings is more of a bottleneck on Android/Java.
If only a few matches need to be made, then you better loop through your collection. If you have much more matches to make it may be very useful to use a SortedTree!

If the list of words is stable (not many words are added or deleted), a very good second alternative is to create 2 lists:
One with the words in normal order.
The second with the characters in each word reversed.
For speed purposes, make them ArrayLists. Never LinkedLists or other variants which perform extremely bad on random access (the core of binary search; see below).
After the lists are created, they can be sorted with method Collections.sort (only once each) and then searched with Collections.binarySearch. For example:
Collections.sort(forwardList);
Collections.sort(backwardList);
And then to search for words starting in "Foo":
int i= Collections.binarySearch(forwardList,"Foo") ;
while( i < forwardList.size() && forwardList.get(i).startsWith("Foo") ) {
// Process String forwardList.get(i)
i++;
}
And words ending in "Bar":
int i= Collections.binarySearch(backwardList,"raB") ;
while( i < backwardList.size() && backwardList.get(i).startsWith("raB") ) {
// Process String backwardList.get(i)
i++;
}