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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.

Capture group multiple times

Lately I have being playing around with regex in Java, and I find myself into a problem which (theoretically) is easy to solve, but I was wandering if there is any easier way to do it (Yes, yes I am lazy), the problem is capture a group multiple times, this is:
public static void main(String[] args) {
Pattern p = Pattern.compile("A (IvI(.*?)IvI)*? A");
Matcher m = p.matcher("A IvI asd IvI IvI qwe IvI A"); //ANY NUMBER of IvI x IvI
//Matcher m = p.matcher("A A");
int loi = 0; //last Occurrence Index
String storage;
while (loi >= 0 && m.find(loi)) {
System.out.println(m.group(1));
if ((storage = m.group(2)) != null) {
System.out.println(storage);
}
//System.out.println(m.group(1));
loi = m.end(1);
}
m.find();
System.out.println("2 opt");
Pattern p2 = Pattern.compile("IvI(.*?)IvI");
Matcher m2 = p2.matcher(m.group(1)); //m.group(1) = "IvI asd IvI IvI qwe IvI"
loi = 0;
while (loi >= 0 && m2.find(loi)) {
if ((storage = m2.group(1)) != null) {
System.out.println(storage);
}
loi = m2.end(0);
}
}
Using ONLY Pattern p is there any way to get what is inside IvI's? (in the test string would be "asd" and "qwe") considering that there could be any number of IvI's sections, something alike of what I am trying to do in the first while which is, finding the first occurrence of the group, then moving the index and search for the next group and so on and so on...
Using the code I wrote in that while it returns asd IvI IvI qwe as the group 2, not just asd and then qwe, in part I suppose it could be because of the (.*?) part, is is not supposed to be greedy but still it goes up to the qwe consuming two of the IvI's, I mention this because otherwise I may be able to use the end index of those with the matcher.find(anInt) method, but it does not work either; I don't think it is anything wrong with the regex, since the next code works without consuming the IvI.
public static void main(String[] args) {
Pattern p = Pattern.compile("(.*?)IvI");
Matcher m = p.matcher("bla bla blaIvI");
m.find();
System.out.println(m.group(1));
}
This prints: bla bla bla
THERE IS A SOLUTION I KNOW (but I am lazy remember)
(Also on the first code, bellow "2 opt" message)
The solution is dividing it into sub-groups and use another regex where you process only those sub-groups one at a time...
BTW: I did my homework
In this page it mentions
Since a capture group with a quantifier holds on to its number, what value does the engine return when you inspect the group? All engines return the last value captured. For instance, if you match the string A_B_C_D_ with ([A-Z])+, when you inspect the match, Group 1 will be D. With the exception of the .NET engine, all intermediate values are lost. In essence, Group 1 gets overwritten each time its pattern is matched.
But I am still hoping you to give me good news...

No, unfortunately, as your citation already mentions, the java.util.regex regular expression implementation does not support retrieving any previous values of a repeated capturing group after a single match. The only way to get those, as your code illustrates, is by find()ing multiple matches of the repeated part of your regular expression.
I've also been looking at other implementations of regular expressions in Java, for example:
http://www.brics.dk/automaton/
but I could not find any that supported it (only the Microsoft .NET engine) . If I understood correctly, implementations of regular expressions based on state machines cannot easily implement this feature. java.util.regex does not use state machines, though.
If anyone knows of a Java regular expression library that supports this behaviour, please share it, because it would be a powerful feature.
p.s. it took me quite a while to understand your question. The title is good, but the body confused me about whether I understood you correctly.

Replace text with data & matched group contents

I don't believe I saw this when searching (believe me, I spent a good amount of time searching for this) for a solution to this so here goes.
Goal:
Match regex in a string and replace it with something that contains the matched value.
Regex used currently:
\b(Connor|charries96|Foo|Bar)\b
For the record I suck at regex incase this isn't the best way to do it.
My current code (and several other methods I tried) can only replace the text with the first match it encounters if there are multiple matches.
private Pattern regexFromList(List<String> input) {
if(input.size() < 1) {
return "";
}
StringBuilder builder = new StringBuilder();
builder.append("\\b");
builder.append("(");
for(String s : input) {
builder.append(s);
if(!s.equals(input.get(input.size() - 1)))
{
builder.append("|");
}
}
builder.append(")");
builder.append("\\b");
return Pattern.compile(builder.toString(), Pattern.CASE_INSENSITIVE);
}
Example input:
charries96's name is Connor.
Example result using TEST as the data to prepend the match with
TESTcharries96's name is TESTcharries96.
Desired result using example input:
TESTcharries96's name is TESTConnor.
Here is my current code for replacing the text:
if(highlight) {
StringBuilder builder = new StringBuilder();
Matcher match = pattern.matcher(event.getMessage());
String string = event.getMessage();
if (match.find()) {
string = match.replaceAll("TEST" + match.group());
// I do realise I'm using #replaceAll but that's mainly given it gives me the same result as other methods so why not just cut to the chase.
}
builder.append(string);
return builder.toString();
}
EDIT:
Working example of desired result on RegExr

There are a few problems here:
You are taking the user input as is and build the regex:
builder.append(s);
If there are special character in the user input, it might be recognized as meta character and cause unexpected behavior.
Always use Pattern.quote if you want to match a string as it is passed in.
builder.append(Pattern.quote(s));
Matcher.replaceAll is a high level function which resets the Matcher (start the match all over again), and search for all the matches and perform the replacement. In your case, it can be as simple as:
String result = match.replaceAll("TEST$1");
The StringBuilder should be thrown away along with the if statement.
Matcher.find, Matcher.group are lower level functions for fine grain control on what you want to do with a match.
When you perform replacement, you need to build the result with Matcher.appendReplacement and Matcher.appendTail.
A while loop (instead of if statement) should be used with Matcher.find to search for and perform replacement for all matched.

Check special arrangement of specific signs in a string in Java

I need to check a string whether it includes a specific arrangements of letters and numbers.
Valid arrangements are for example:
X
X-Y
A-H-K-L-J-Y
A-H-J-Y
123
12?
12*
12-17
Invalid are for example:
-X-Y
-XY
*12
?12
I have written this method in java to solve this problem (but i don´t have some experiences with regular expressions):
public boolean checkPatternMatching(String sourceToScan, String searchPattern) {
boolean patternFounded;
if (sourceToScan == null) {
patternFounded = false;
} else {
Pattern pattern = Pattern.compile(Pattern.quote(searchPattern),
Pattern.CASE_INSENSITIVE);
Matcher matcher = pattern.matcher(sourceToScan);
patternFounded = matcher.find();
}
return patternFounded;
}
How can i implemented this requirement with regular expressions?
By the way: It is a good solution to check a string, whether it includes numeric content by using the method isNumeric from the java class StringUtils?
//EDIT
The link, which was edited by the admins includes not specific arrangements of characters but only an appearance of characters with regular expressions in general !

After a good while trying to help, answering to constantly changing questions, just found out that the same was asked yesterday, and that the OP doesn't accept answers to his questions...all I have left to say is good night sir, good luck
n-th answer follows:
First pattern: [a-z](-[a-z])* : a letter, possibly followed by more letters, separated by -.
Second pattern: \d+(-\d+)*[?*]* : a number, possibly followed by more numbers, separated by -, and possibly ending with ? or *.
So join them together: ^([a-z](-[a-z])*)|(\d+(-\d+)*[?*]*)$. ^ and $ mark the beginning and the end of the string.
Few more comments on the code: you don't need to use Pattern.quote, and you should use matches() instead of find(), because find() returns true if any part of the string matches the pattern, and you want the whole string:
public static boolean checkPatternMatching(String sourceToScan, String searchPattern) {
boolean patternFounded;
if (sourceToScan == null) {
patternFounded = false;
} else {
Pattern pattern = Pattern.compile(searchPattern, Pattern.CASE_INSENSITIVE);
Matcher matcher = pattern.matcher(sourceToScan);
patternFounded = matcher.matches();
}
return patternFounded;
}
Called like this: checkPatternMatching(s, "^([a-z](-[a-z])*)|(\\d+(-\\d+)*[?*]*)$")
About the second question, this is the current implementation of StringUtils.isNumeric:
public static boolean isNumeric(final CharSequence cs) {
if (isEmpty(cs)) {
return false;
}
final int sz = cs.length();
for (int i = 0; i < sz; i++) {
if (Character.isDigit(cs.charAt(i)) == false) {
return false;
}
}
return true;
}
So no, there is nothing wrong about it, that is as simple as it gets. But you need to include an external JAR in your program, which I find unnecessary if you just want to use such a simple method.

I believe that you should first remove the Pattern.quote() method because that would turn the inputting patterns into string literals; and those are not really useful in your context.
To match the valid arrangements with letters, something like this should work:
^[a-z](?:-[a-z])*$
For the numbers (if I understood the rules correctly):
^\\d+(?:[?*]|-\\d+)*$
And if you want to combine them:
^(?:[a-z](?:-[a-z])*|\\d+(?:[?*]|-\\d+)*)$
I'm not familiar with Java itself, nor the isNumeric method, sorry.
As per your comment, if you want to accept *12 or 1?2 or 12*456, you can use:
^\\*?\\d+(?:[?*]\\d*|-\\d+)*$
Then add it to the previous regex like so:
^(?:[a-z](?:-[a-z])*|\\*?\\d+(?:[?*]\\d*|-\\d+)*)$

String splitting

I have a string in what is the best way to put the things in between $ inside a list in java?
String temp = $abc$and$xyz$;
how can i get all the variables within $ sign as a list in java
[abc, xyz]
i can do using stringtokenizer but want to avoid using it if possible.
thx

Maybe you could think about calling String.split(String regex) ...

The pattern is simple enough that String.split should work here, but in the more general case, one alternative for StringTokenizer is the much more powerful java.util.Scanner.
String text = "$abc$and$xyz$";
Scanner sc = new Scanner(text);
while (sc.findInLine("\\$([^$]*)\\$") != null) {
System.out.println(sc.match().group(1));
} // abc, xyz
The pattern to find is:
\$([^$]*)\$
\_____/ i.e. literal $, a sequence of anything but $ (captured in group 1)
1 and another literal $
The […] is a character class. Something like [aeiou] matches one of any of the lowercase vowels. [^…] is a negated character class. [^aeiou] matches one of anything but the lowercase vowels.
(…) is used for grouping. (pattern) is a capturing group and creates a backreference.
The backslash preceding the $ (outside of character class definition) is used to escape the $, which has a special meaning as the end of line anchor. That backslash is doubled in a String literal: "\\" is a String of length one containing a backslash).
This is not a typical usage of Scanner (usually the delimiter pattern is set, and tokens are extracted using next), but it does show how'd you use findInLine to find an arbitrary pattern (ignoring delimiters), and then using match() to access the MatchResult, from which you can get individual group captures.
You can also use this Pattern in a Matcher find() loop directly.
Matcher m = Pattern.compile("\\$([^$]*)\\$").matcher(text);
while (m.find()) {
System.out.println(m.group(1));
} // abc, xyz
Related questions
Validating input using java.util.Scanner
Scanner vs. StringTokenizer vs. String.Split

Just try this one:temp.split("\\$");

I would go for a regex myself, like Riduidel said.
This special case is, however, simple enough that you can just treat the String as a character sequence, and iterate over it char by char, and detect the $ sign. And so grab the strings yourself.
On a side node, I would try to go for different demarkation characters, to make it more readable to humans. Use $ as start-of-sequence and something else as end-of-sequence for instance. Or something like I think the Bash shell uses: ${some_value}. As said, the computer doesn't care but you debugging your string just might :)
As for an appropriate regex, something like (\\$.*\\$)* or so should do. Though I'm no expert on regexes (see http://www.regular-expressions.info for nice info on regexes).

Basically I'd ditto Khotyn as the easiest solution. I see you post on his answer that you don't want zero-length tokens at beginning and end.
That brings up the question: What happens if the string does not begin and end with $'s? Is that an error, or are they optional?
If it's an error, then just start with:
if (!text.startsWith("$") || !text.endsWith("$"))
return "Missing $'s"; // or whatever you do on error
If that passes, fall into the split.
If the $'s are optional, I'd just strip them out before splitting. i.e.:
if (text.startsWith("$"))
text=text.substring(1);
if (text.endsWith("$"))
text=text.substring(0,text.length()-1);
Then do the split.
Sure, you could make more sophisticated regex's or use StringTokenizer or no doubt come up with dozens of other complicated solutions. But why bother? When there's a simple solution, use it.
PS There's also the question of what result you want to see if there are two $'s in a row, e.g. "$foo$$bar$". Should that give ["foo","bar"], or ["foo","","bar"] ? Khotyn's split will give the second result, with zero-length strings. If you want the first result, you should split("\$+").

If you want a simple split function then use Apache Commons Lang which has StringUtils.split. The java one uses a regex which can be overkill/confusing.

You can do it in simple manner writing your own code.
Just use the following code and it will do the job for you
import java.util.ArrayList;
import java.util.List;
public class MyStringTokenizer {
/**
* #param args
*/
public static void main(String[] args) {
List <String> result = getTokenizedStringsList("$abc$efg$hij$");
for(String token : result)
{
System.out.println(token);
}
}
private static List<String> getTokenizedStringsList(String string) {
List <String> tokenList = new ArrayList <String> ();
char [] in = string.toCharArray();
StringBuilder myBuilder = null;
int stringLength = in.length;
int start = -1;
int end = -1;
{
for(int i=0; i<stringLength;)
{
myBuilder = new StringBuilder();
while(i<stringLength && in[i] != '$')
i++;
i++;
while((i)<stringLength && in[i] != '$')
{
myBuilder.append(in[i]);
i++;
}
tokenList.add(myBuilder.toString());
}
}
return tokenList;
}
}

You can use
String temp = $abc$and$xyz$;
String array[]=temp.split(Pattern.quote("$"));
List<String> list=new ArrayList<String>();
for(int i=0;i<array.length;i++){
list.add(array[i]);
}
Now the list has what you want.