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How to construct regular expression to balance characters in a string?

I have come across regular expressions for different problems but I could not find out regex s to balance characters in a string.
I came across a problem, to find if a string is balanced.
ex: aabbccdd is a balanced one, as a characters are repeated in even numbers
but aabbccddd is not a balanced one since ddd is repeated in odd number mode. This is applicable for all characters give an input not to specific a,b,c and d. If i give input as 12344321 or 123454321, it should return balanced and unbalanced result respectively.
How to find the balance using regex. What type of regular expression we should use to find if the string is balanced?
Edit:
I tried to find solution using regex only as the problem demands answer in regex pattern. I would implemented using any other solution if regex was not mentioned explicitly

I don't think you can do it with regex. Why do you need to use them?
I tried this: it works and it's pretty simple
static boolean isBalanced(String str) {
ArrayList<Character> odds = new ArrayList<>(); //Will contain the characters read until now an odd number of times
for (char x : str.toCharArray()) { //Reads each char of the string
if (odds.contains(x)) { //If x was in the arraylist we found x an even number of times so let's remove it
odds.remove(odds.indexOf(x));
}
else {
odds.add(x);
}
}
return odds.isEmpty();
}

Regular expression for this problem exists, but doesn't speed up anythings and will be totally messy. It's easier to prepare NFA, and then switch to REGEX. Still, it's not proper tool.
public static void main(String args[]) {
String s = args[0];
int[] counter = new int[256];
for (int i = 0; i < s.length(); i++) {
counter[s.charAt(i)]++;
}
if (validate(counter)) {
System.out.println("valid");
} else {
System.out.println("invalid");
}
}
public static boolean validate(int[] tab) {
for (int i : tab) {
if (i%2 == 1) {
return false;
}
}
return true;
}
Edit: for pointing the regex existance
Reference for a finite automate for just two characters. Start on the very left, win with double circle. Each state named by the set of characters that have odd count so far.

Alternative to successive String.replace

I want to replace some strings in a String input :
string=string.replace("<h1>","<big><big><big><b>");
string=string.replace("</h1>","</b></big></big></big>");
string=string.replace("<h2>","<big><big>");
string=string.replace("</h2>","</big></big>");
string=string.replace("<h3>","<big>");
string=string.replace("</h3>","</big>");
string=string.replace("<h4>","<b>");
string=string.replace("</h4>","</b>");
string=string.replace("<h5>","<small><b>");
string=string.replace("</h5>","</b><small>");
string=string.replace("<h6>","<small>");
string=string.replace("</h6>","</small>");
As you can see this approach is not the best, because each time I have to search for the portion to replace etc, and Strings are immutable... Also the input is large, which means that some performance issues are to be considered.
Is there any better approach to reduce the complexity of this code ?

Although StringBuilder.replace() is a huge improvement compared to String.replace(), it is still very far from being optimal.
The problem with StringBuilder.replace() is that if the replacement has different length than the replaceable part (applies to our case), a bigger internal char array might have to be allocated, and the content has to be copied, and then the replace will occur (which also involves copying).
Imagine this: You have a text with 10.000 characters. If you want to replace the "XY" substring found at position 1 (2nd character) to "ABC", the implementation has to reallocate a char buffer which is at least larger by 1, has to copy the old content to the new array, and it has to copy 9.997 characters (starting at position 3) to the right by 1 to fit "ABC" into the place of "XY", and finally characters of "ABC" are copied to the starter position 1. This has to be done for every replace! This is slow.
Faster Solution: Building Output On-The-Fly
We can build the output on-the-fly: parts that don't contain replaceable texts can simply be appended to the output, and if we find a replaceable fragment, we append the replacement instead of it. Theoretically it's enough to loop over the input only once to generate the output. Sounds simple, and it's not that hard to implement it.
Implementation:
We will use a Map preloaded with mappings of the replaceable-replacement strings:
Map<String, String> map = new HashMap<>();
map.put("<h1>", "<big><big><big><b>");
map.put("</h1>", "</b></big></big></big>");
map.put("<h2>", "<big><big>");
map.put("</h2>", "</big></big>");
map.put("<h3>", "<big>");
map.put("</h3>", "</big>");
map.put("<h4>", "<b>");
map.put("</h4>", "</b>");
map.put("<h5>", "<small><b>");
map.put("</h5>", "</b></small>");
map.put("<h6>", "<small>");
map.put("</h6>", "</small>");
And using this, here is the replacer code: (more explanation after the code)
public static String replaceTags(String src, Map<String, String> map) {
StringBuilder sb = new StringBuilder(src.length() + src.length() / 2);
for (int pos = 0;;) {
int ltIdx = src.indexOf('<', pos);
if (ltIdx < 0) {
// No more '<', we're done:
sb.append(src, pos, src.length());
return sb.toString();
}
sb.append(src, pos, ltIdx); // Copy chars before '<'
// Check if our hit is replaceable:
boolean mismatch = true;
for (Entry<String, String> e : map.entrySet()) {
String key = e.getKey();
if (src.regionMatches(ltIdx, key, 0, key.length())) {
// Match, append the replacement:
sb.append(e.getValue());
pos = ltIdx + key.length();
mismatch = false;
break;
}
}
if (mismatch) {
sb.append('<');
pos = ltIdx + 1;
}
}
}
Testing it:
String in = "Yo<h1>TITLE</h1><h3>Hi!</h3>Nice day.<h6>Hi back!</h6>End";
System.out.println(in);
System.out.println(replaceTags(in, map));
Output: (wrapped to avoid scroll bar)
Yo<h1>TITLE</h1><h3>Hi!</h3>Nice day.<h6>Hi back!</h6>End
Yo<big><big><big><b>TITLE</b></big></big></big><big>Hi!</big>Nice day.
<small>Hi back!</small>End
This solution is faster than using regular expressions as that involves much overhead, like compiling a Pattern, creating a Matcher etc. and regexp is also much more general. It also creates many temporary objects under the hood which are thrown away after the replace. Here I only use a StringBuilder (plus char array under its hood) and the code iterates over the input String only once. Also this solution is much faster that using StringBuilder.replace() as detailed at the top of this answer.
Notes and Explanation
I initialized the StringBuilder in the replaceTags() method like this:
StringBuilder sb = new StringBuilder(src.length() + src.length() / 2);
So basically I created it with an initial capacity of 150% of the length of the original String. This is because our replacements are longer than the replaceable texts, so if replacing occurs, the output will obviously be longer than the input. Giving a larger initial capacity to StringBuilder will result in no internal char[] reallocation at all (of course the required initial capacity depends on the replaceable-replacement pairs and their frequency/occurrence in the input, but this +50% is a good upper estimation).
I also utilized the fact that all replaceable strings start with a '<' character, so finding the next potential replaceable position becomes blazing-fast:
int ltIdx = src.indexOf('<', pos);
It's just a simple loop and char comparisons inside String, and since it always starts searching from pos (and not from the start of the input), overall the code iterates over the input String only once.
And finally to tell if a replaceable String does occur at the potential position, we use the String.regionMatches() method to check the replaceable stings which is also blazing-fast as all it does is just compares char values in a loop and returns at the very first mismatching character.
And a PLUS:
The question doesn't mention it, but our input is an HTML document. HTML tags are case-insensitive which means the input might contain <H1> instead of <h1>.
To this algorithm this is not a problem. The regionMatches() in the String class has an overload which supports case-insensitive comparison:
boolean regionMatches(boolean ignoreCase, int toffset, String other,
int ooffset, int len);
So if we want to modify our algorithm to also find and replace input tags which are the same but are written using different letter case, all we have to modify is this one line:
if (src.regionMatches(true, ltIdx, key, 0, key.length())) {
Using this modified code, replaceable tags become case-insensitive:
Yo<H1>TITLE</H1><h3>Hi!</h3>Nice day.<H6>Hi back!</H6>End
Yo<big><big><big><b>TITLE</b></big></big></big><big>Hi!</big>Nice day.
<small>Hi back!</small>End

For performance - use StringBuilder.
For convenience you can use Map to store values and replacements.
Map<String, String> map = new HashMap<>();
map.put("<h1>","<big><big><big><b>");
map.put("</h1>","</b></big></big></big>");
map.put("<h2>","<big><big>");
...
StringBuilder builder = new StringBuilder(yourString);
for (String key : map.keySet()) {
replaceAll(builder, key, map.get(key));
}
... To replace all occurences in StringBuilder you can check here:
Replace all occurrences of a String using StringBuilder?
public static void replaceAll(StringBuilder builder, String from, String to)
{
int index = builder.indexOf(from);
while (index != -1)
{
builder.replace(index, index + from.length(), to);
index += to.length(); // Move to the end of the replacement
index = builder.indexOf(from, index);
}
}

Unfortunately StringBuilder doesn't provide a replace(string,string) method, so you might want to consider using Pattern and Matcher in conjunction with StringBuffer:
String input = ...;
StringBuffer sb = new StringBuffer();
Pattern p = Pattern.compile("</?(h1|h2|...)>");
Matcher m = p.matcher( input );
while( m.find() )
{
String match = m.group();
String replacement = ...; //get replacement for match, e.g. by lookup in a map
m.appendReplacement( sb, replacement );
}
m.appendTail( sb );
You could do something similar with StringBuilder but in that case you'd have to implement appendReplacement etc. yourself.
As for the expression you could also just try and match any html tag (although that might cause problems since regex and arbitrary html don't fit very well) and when the lookup doesn't have any result you just replace the match with itself.

The particular example you provide seems to be HTML or XHTML. Trying to edit HTML or XML using regular expressions is frought with problems. For the kind of editing you seem to be interested in doing you should look at using XSLT. Another possibility is to use SAX, the streaming XML parser, and have your back-end write the edited output on the fly. If the text is actually HTML, you might be better using a tolerant HTML parser, such as JSoup, to build a parsed representation of the document (like the DOM), and manipulate that before outputting it.

StringBuilder is backed by a char array. So, unlike String instances, it is mutable. Thus, you can call indexOf() and replace() on the StringBuilder.

I would do something like this
StringBuilder sb = new StringBuilder();
for (int i = 0; i < str.length(); i++) {
if (tagEquals(str, i, "h1")) {
sb.append("<big><big><big><b>");
i += 2;
} else (tagEquals(s, i, "/h1")) {
...
} else {
sb.append(str.charAt(i));
}
}
tagEquals is a func which checks a tag name

Use Apache Commons StringUtils.replaceEach.
String[] searches = new String[]{"<h1>", "</h1>", "<h2>", ...};
String[] replacements = new String[]("<big><big><big><b>", "</b></big></big></big>", "<big><big>" ...};
string = StringUtils.replaceEach(string, searches, replacements);

Simplify & condense multiple editorial operations on an array. Java

I have some raw output that I want to clean up and make presentable but right now I go about it in a very ugly and cumbersome way, I wonder if anyone might know a clean and elegant way in which to perform the same operation.
int size = charOutput.size();
for (int i = size - 1; i >= 1; i--)
{
if(charOutput.get(i).compareTo(charOutput.get(i - 1)) == 0)
{
charOutput.remove(i);
}
}
for(int x = 0; x < charOutput.size(); x++)
{
if(charOutput.get(x) == '?')
{
charOutput.remove(x);
}
}
String firstOne = Arrays.toString(charOutput.toArray());
String secondOne = firstOne.replaceAll(",","");
String thirdOne = secondOne.substring(1, secondOne.length() - 1);
String output = thirdOne.replaceAll(" ","");
return output;

ZouZou has the right code for fixing the final few calls in your code. I have some suggestions for the for loops. I hope I got them right...
These work after you get the String represented by charOutput, using a method such as the one suggested by ZouZou.
Your first block appears to remove all repeated letters. You can use a regular expression for that:
Pattern removeRepeats = Pattern.compile("(.)\\1{1,}");
// "(.)" creates a group that matches any character and puts it into a group
// "\\1" gets converted to "\1" which is a reference to the first group, i.e. the character that "(.)" matched
// "{1,}" means "one or more"
// So the overall effect is "one or more of a single character"
To use:
removeRepeats.matcher(s).replaceAll("$1");
// This creates a Matcher that matches the regex represented by removeRepeats to the contents of s, and replaces the parts of s that match the regex represented by removeRepeats with "$1", which is a reference to the first group captured (i.e. "(.)", which is the first character matched"
To remove the question mark, just do
Pattern removeQuestionMarks = Pattern.compile("\\?");
// Because "?" is a special symbol in regex, you have to escape it with a backslash
// But since backslashes are also a special symbol, you have to escape the backslash too.
And then to use, do the same thing as was done above except with replaceAll("");
And you're done!
If you really wanted to, you can combine a lot of regex into two super-regex expressions (and one normal regex expression):
Pattern p0 = Pattern.compile("(\\[|\\]|\\,| )"); // removes brackets, commas, and spaces
Pattern p1 = Pattern.compile("(.)\\1{1,}"); // Removes duplicate characters
Pattern p2 = Pattern.compile("\\?");
String removeArrayCharacters = p0.matcher(charOutput.toString()).replaceAll("");
String removeDuplicates = p1.matcher(removeArrayCharacters).replaceAll("$1");
return p2.matcher(removeDuplicates).replaceAll("");

Use a StringBuilder and append each character you want, at the end just return myBuilder.toString();
Instead of this:
String firstOne = Arrays.toString(charOutput.toArray());
String secondOne = firstOne.replaceAll(",","");
String thirdOne = secondOne.substring(1, secondOne.length() - 1);
String output = thirdOne.replaceAll(" ","");
return output;
Simply do:
StringBuilder sb = new StringBuilder();
for(Character c : charOutput){
sb.append(c);
}
return sb.toString();
Note that you are doing a lot of unnecessary work (by iterating through the list and removing some elements). What you can actually do is just iterate one time and then if the condition fullfits your requirements (the two adjacent characters are not the same and no question mark) then append it to the StringBuilder directly.
This task could also be a job for a regular expression.

If you don't want to use Regex try this version to remove consecutive characters and '?':
int size = charOutput.size();
if (size == 1) return Character.toString((Character)charOutput.get(0));
else if (size == 0) return null;
StringBuilder sb = new StringBuilder();
for (int i = 0; i < size - 1; i++) {
Character temp = (Character)charOutput.get(i);
if (!temp.equals(charOutput.get(i+1)) && !temp.equals('?'))
sb.append(temp);
}
//for the last element
if (!charOutput.get(size-1).equals(charOutput.get(size-2))
&& !charOutput.get(size-1).equals('?'))
sb.append(charOutput.get(size-1));
return sb.toString();

Efficient way to search for a set of strings in a string in Java

I have a set of elements of size about 100-200. Let a sample element be X.
Each of the elements is a set of strings (number of strings in such a set is between 1 and 4). X = {s1, s2, s3}
For a given input string (about 100 characters), say P, I want to test whether any of the X is present in the string.
X is present in P iff for all s belong to X, s is a substring of P.
The set of elements is available for pre-processing.
I want this to be as fast as possible within Java. Possible approaches which do not fit my requirements:
Checking whether all the strings s are substring of P seems like a costly operation
Because s can be any substring of P (not necessarily a word), I cannot use a hash of words
I cannot directly use regex as s1, s2, s3 can be present in any order and all of the strings need to be present as substring
Right now my approach is to construct a huge regex out of each X with all possible permutations of the order of strings. Because number of elements in X <= 4, this is still feasible. It would be great if somebody can point me to a better (faster/more elegant) approach for the same.
Please note that the set of elements is available for pre-processing and I want the solution in java.

You can use regex directly:
Pattern regex = Pattern.compile(
"^ # Anchor search to start of string\n" +
"(?=.*s1) # Check if string contains s1\n" +
"(?=.*s2) # Check if string contains s2\n" +
"(?=.*s3) # Check if string contains s3",
Pattern.DOTALL | Pattern.COMMENTS);
Matcher regexMatcher = regex.matcher(subjectString);
foundMatch = regexMatcher.find();
foundMatch is true if all three substrings are present in the string.
Note that you might need to escape your "needle strings" if they could contain regex metacharacters.

It sounds like you're prematurely optimising your code before you've actually discovered a particular approach is actually too slow.
The nice property about your set of strings is that the string must contain all elements of X as a substring -- meaning we can fail fast if we find one element of X that is not contained within P. This might turn out a better time saving approach than others, especially if the elements of X are typically longer than a few characters and contain no or only a few repeating characters. For instance, a regex engine need only check 20 characters in 100 length string when checking for the presence of a 5 length string with non-repeating characters (eg. coast). And since X has 100-200 elements you really, really want to fail fast if you can.
My suggestion would be to sort the strings in order of length and check for each string in turn, stopping early if one string is not found.

Looks like a perfect case for the Rabin–Karp algorithm:
Rabin–Karp is inferior for single pattern searching to Knuth–Morris–Pratt algorithm, Boyer–Moore string search algorithm and other faster single pattern string searching algorithms because of its slow worst case behavior. However, Rabin–Karp is an algorithm of choice for multiple pattern search.

When the preprocessing time doesn't matter, you could create a hash table which maps every one-letter, two-letter, three-letter etc. combination which occurs in at least one string to a list of strings in which it occurs.
The algorithm to index a string would look like that (untested):
HashMap<String, Set<String>> indexes = new HashMap<String, Set<String>>();
for (int pos = 0; pos < string.length(); pos++) {
for (int sublen=0; sublen < string.length-pos; sublen++) {
String substring = string.substr(pos, sublen);
Set<String> stringsForThisKey = indexes.get(substring);
if (stringsForThisKey == null) {
stringsForThisKey = new HashSet<String>();
indexes.put(substring, stringsForThisKey);
}
stringsForThisKey.add(string);
}
}
Indexing each string that way would be quadratic to the length of the string, but it only needs to be done once for each string.
But the result would be constant-speed access to the list of strings in which a specific string occurs.

You are probably looking for Aho-Corasick algorithm, which constructs an automata (trie-like) from the set of strings (dictionary), and try to match the input string to the dictionary using this automata.

You might want to consider using a "Suffix Tree" as well. I haven't used this code, but there is one described here
I have used proprietary implementations (that I no longer even have access to) and they are very fast.

One way is to generate every possible substring and add this to a set. This is pretty inefficient.
Instead you can create all the strings from any point to the end into a NavigableSet and search for the closest match. If the closest match starts with the string you are looking for, you have a substring match.
static class SubstringMatcher {
final NavigableSet<String> set = new TreeSet<String>();
SubstringMatcher(Set<String> strings) {
for (String string : strings) {
for (int i = 0; i < string.length(); i++)
set.add(string.substring(i));
}
// remove duplicates.
String last = "";
for (String string : set.toArray(new String[set.size()])) {
if (string.startsWith(last))
set.remove(last);
last = string;
}
}
public boolean findIn(String s) {
String s1 = set.ceiling(s);
return s1 != null && s1.startsWith(s);
}
}
public static void main(String... args) {
Set<String> strings = new HashSet<String>();
strings.add("hello");
strings.add("there");
strings.add("old");
strings.add("world");
SubstringMatcher sm = new SubstringMatcher(strings);
System.out.println(sm.set);
for (String s : "ell,he,ow,lol".split(","))
System.out.println(s + ": " + sm.findIn(s));
}
prints
[d, ello, ere, hello, here, ld, llo, lo, old, orld, re, rld, there, world]
ell: true
he: true
ow: false
lol: false

String Match How to implement

I have a class Matcher() as follows. find method accepts two strings: pattern (string to be looked for) and source (string where to look for). Example if pattern = "abc" and source = "abc cda abc" is passed to find method. It returns [0 4], i.e. the pattern abc(exact match) is found at index 0 and index 4 of source. Whatever you pass to pattern, it will consider that string as one pattern. Without modifying the following Matcher class, if I want to search for more than one pattern. What is the best way to do it? For example I want to call the find method from other class and if i have two patterns stored in arraylist then i want to pass one pattern first and return the result and again pass the 2nd pattern and return the result in one time. I want to stop processing only after both the patterns or the patterns in arraylist are looked for in source. Need your idea.
public class Matcher {
public static List<Integer> find(String pattern, String source) {
char[] x = pattern.toCharArray(), y = source.toCharArray();
int i, j, m = x.length, n = y.length;
List<Integer> result = new ArrayList<Integer>();
/* Searching */
for (j = 0; j <= n - m; ++j) {
for (i = 0; i < m && x[i] == y[i + j]; ++i)
;
if (i >= m)
result.add(j);
}
return result;
}
}

So, you want your method find to return a list of indexes? Did you have a look at String.indexOf? That might do exactly what you want.

It seems to me that you answered your own question. You'll need to use a loop in your client code to make repeated calls to the find() method. You won't be able to do it in a single call unless you rewrite your find method, which you don't want to do. Your client code in mangled pseudocode:
declare a Matcher object
for (each pattern I want to match)
call the find method with the pattern and the source string
store the result
end loop
How you deal with the result will depend on what you need to do with it. You could create an ArrayList object and append the List objects to it. Or you could create a HashMap and use the pattern as a key to the List objects if you need to know which pattern is where.
Let me know if I completely missed your point.
Cheers,
dean