I have been given an infinite wrap around of the string str="abcdefghijklmnopqrstuvwxyz" so it looks like
"..zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd...." and another string p.
I need to find out how many unique non-empty substrings of p are present in the infinite wraparound string str?
For example: "zab"
There are 6 substrings "z", "a", "b", "za", "ab", "zab" of string "zab" in str.
I tried finding all suffixes of p in a particular concatenation of the string str say for example: "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz"
and as soon as i get a suffix which is a part of the above i add all its substrings to my result, as:
for (int i=0;i<length;i++) {
String suffix = p.substring(i,length);
if(isPresent(suffix)) {
sum += (suffix.length()*(suffix.length()+1))/2;
break;
} else {
sum++;
}
}
And my isPresent function is:
private boolean isPresent(String s) {
if(s.length()==1) {
return true;
}
String main = "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcde
fghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz";
return main.contains(s);
}
If the length of p is greater than my assumed concatenated string assumed in isPresent function, my algorithm fails!!
So how should i find the substrings irrespective of the the wrap around string str? Is there a better approach for this problem?
Some ideas/suggestions (not a full algo)
you don't need to consider an infinite repetition of the wrap around string but only len(p)/len(repeating-fragment) + 1 (integral division) repetitions. Let's denote this string with S **
if a substring sp of p is a substring of S, than any substrings of sp will be substrings of S
So the problem seems to reduce to:
find sp (substring of both p and S) with the maximal length. This is called longest common substring and admits a dynamic programming solution with the complexity of O(n*m) (lengths of the two strings). The cited has a pseudo-code algo.
repeat the above recursively with the 'remnants' of p after eliminating the longest common substring.
Now, you have a sequence of "longest common substrings". How many do you need to retain? I feel that the "longest common substring" may be used to trim down the need of brute-forcing every substring of any and all the above, but I'd need more time than I have available now.
I hope the sketch above helps.
** I might be wrong on the number of repetitions which need to be considered. If I am, then in any case there will be a maximal number of repetitions to be considered and there will be an S of minimal length that is sufficient for the purpose.
Suppose we are given two strings s1 and s2(both lowercase). We have two find the minimal lexographic string that can be formed by merging two strings.
At the beginning , it looks prettty simple as merge of the mergesort algorithm. But let us see what can go wrong.
s1: zyy
s2: zy
Now if we perform merge on these two we must decide which z to pick as they are equal, clearly if we pick z of s2 first then the string formed will be:
zyzyy
If we pick z of s1 first, the string formed will be:
zyyzy which is correct.
As we can see the merge of mergesort can lead to wrong answer.
Here's another example:
s1:zyy
s2:zyb
Now the correct answer will be zybzyy which will be got only if pick z of s2 first.
There are plenty of other cases in which the simple merge will fail. My question is Is there any standard algorithm out there used to perform merge for such output.
You could use dynamic programming. In f[x][y] store the minimal lexicographical string such that you've taken x charecters from the first string s1 and y characters from the second s2. You can calculate f in bottom-top manner using the update:
f[x][y] = min(f[x-1][y] + s1[x], f[x][y-1] + s2[y]) \\ the '+' here represents
\\ the concatenation of a
\\ string and a character
You start with f[0][0] = "" (empty string).
For efficiency you can store the strings in f as references. That is, you can store in f the objects
class StringRef {
StringRef prev;
char c;
}
To extract what string you have at certain f[x][y] you just follow the references. To udapate you point back to either f[x-1][y] or f[x][y-1] depending on what your update step says.
It seems that the solution can be almost the same as you described (the "mergesort"-like approach), except that with special handling of equality. So long as the first characters of both strings are equal, you look ahead at the second character, 3rd, etc. If the end is reached for some string, consider the first character of the other string as the next character in the string for which the end is reached, etc. for the 2nd character, etc. If the ends for both strings are reached, then it doesn't matter from which string to take the first character. Note that this algorithm is O(N) because after a look-ahead on equal prefixes you know the whole look-ahead sequence (i.e. string prefix) to include, not just one first character.
EDIT: you look ahead so long as the current i-th characters from both strings are equal and alphabetically not larger than the first character in the current prefix.
Suppose a string is like this "abaabaabaabaaba", the palindrome cycle here is 3, because you can find the string aba at every 3rd position and you can augment the palindrome by concatenating any number of "aba"s to the string.
I think it's possible to detect this efficiently using Manacher's Algorithm but how?
You can find it easily by searching the string S in S+S. The first index you find is the cycle number you want (may be the entire string). In python it would be something like:
In [1]: s = "abaabaabaabaaba"
In [2]: print (s+s).index(s, 1)
3
The 1 is there to ignore the index 0, that would be a trivial match.
Let's say I have a three character string "ABC". I want to generate all permutations of that string where a single letter can be replaced with his lower-case equivalent. For example, "aBC", "abC", "abc", "AbC", "Abc", etc. In other words, given a regexp like [Aa][Bb][Cc] generate every string that can be matched by it.
The problem can be trivially reduced to generating all binary sequences of length n. This has been previously addressed, for example in Fastest way to generate all binary strings of size n into a boolean array? and all permutations of a binary sequence x bits long.
This is a coding exercise. Suppose I have to decide if one string is created by a cyclic shift of another. For example: cab is a cyclic shift of abc but cba is not.
Given two strings s1 and s2 we can do that as follows:
if (s1.length != s2.length)
return false
for(int i = 0; i < s1.length(); i++)
if ((s1.substring(i) + s1.substring(0, i)).equals(s2))
return true
return false
Now what if I have an array of strings and want to find all strings that are cyclic shift of one another? For example: ["abc", "xyz", "yzx", "cab", "xxx"] -> ["abc", "cab"], ["xyz", "yzx"], ["xxx"]
It looks like I have to check all pairs of the strings. Is there a "better" (more efficient) way to do that?
As a start, you can know if a string s1 is a rotation of a string s2 with a single call to contains(), like this:
public boolean isRotation(String s1, String s2){
String s2twice = s2+s2;
return s2twice.contains(s1);
}
Namely, if s1 is "rotation" and s2 is "otationr", the concat gives you "otationrotationr", which contains s1 indeed.
Now, even if we assume this is linear, or close to it (which is not impossible using Rabin-Karp, for instance), you are still left with O(n^2) pair comparisons, which may be too much.
What you could do is build an hashtable where the sorted word is the key, and the posting list contains all the words from your list that, if sorted, give the key (ie. key("bca") and key("cab") both should return "abc"):
private Map<String, List<String>> index;
/* ... */
public void buildIndex(String[] words){
for(String word : words){
String sortedWord = sortWord(word);
if(!index.containsKey(sortedWord)){
index.put(sortedWord, new ArrayList<String>());
}
index.get(sortedWord).add(word);
}
}
CAVEAT: The hashtable will contain, for each key, all the words that have exactly the same letters occurring the same amount of times (not just the rotations, ie. "abba" and "baba" will have the same key but isRotation("abba", "baba") will return false).
But once you have built this index, you can significantly reduce the number of pairs you need to consider: if you want all the rotations for "bca" you just need to sort("bca"), look it up in the hashtable, and check (using the isRotation method above, if you want) if the words in the posting list are the result of a rotation or not.
If strings are short compared to the number of strings in the list, you can do significantly better by rotating all strings to some normal form (lexicographic minimum, for example). Then sort lexicographically and find runs of the same string. That's O(n log n), I think... neglecting string lengths. Something to try, maybe.
Concerning the way to find the pairs in the table, there could be many better way, but what I came up as a first thought is to sort the table and apply the check per adjacent pair.
This is much better and simpler that checking every string with every other string in the table
Consider building an automaton for each string against which you wish to test.
Each automaton should have one entry point for each possible character in the string, and transitions for each character, plus an extra transition from the end to the start.
You could improve performance even further if you amalgated the automata.
I think a combination of the answers by Patrick87 and savinos would make a fair amount of sense. Specifically, in a Java-esque pseudo-code:
List<String> inputs = ["abc", "xyz", "yzx", "cab", "xxx"];
Map<String,List<String>> uniques = new Map<String,List<String>>();
for(String value : inputs) {
String normalized = normalize(value);
if(!uniques.contains(normalized)) {
unqiues.put(normalized, new List<String>());
}
uniques.get(normalized).add(value);
}
// you now have a Map of normalized strings to every string in the input
// that is "equal to" that normalized version
Normalizing the string, as stated by Patrick87 might be best done by picking the rotation of the string that results in the lowest lexographic ordering.
It's worth noting, however, that the "best" algorithm probably relies heavily on the inputs... the number of strings, the length of those string, how many duplicates there are, etc.
You can rotate all the strings to a normalized form using Booth's algorithm (https://en.wikipedia.org/wiki/Lexicographically_minimal_string_rotation) in O(s) time, where s is the length of the string.
You can then use the normalized form as a key in a HashMap (where the value is the set of rotations seen in the input). You can populate this HashMap in a single pass over the data. i.e., for each string
calculate the normalized form
check if the HashMap contains the normalized form as a key - if not insert the empty Set at this key
add the string to the Set in the HashMap
You then just need to output the values of the HashMap. This makes the total runtime of the algorithm O(n * s) - where n is the number of words and s is the average word length. The total space usage is also O(n * s).