Develop Reference

Java regex: Replace all characters with `+` except instances of a given string

I have the following problem which states
Replace all characters in a string with + symbol except instances of the given string in the method
so for example if the string given was abc123efg and they want me to replace every character except every instance of 123 then it would become +++123+++.
I figured a regular expression is probably the best for this and I came up with this.
str.replaceAll("[^str]","+")
where str is a variable, but its not letting me use the method without putting it in quotations. If I just want to replace the variable string str how can I do that? I ran it with the string manually typed and it worked on the method, but can I just input a variable?
as of right now I believe its looking for the string "str" and not the variable string.
Here is the output its right for so many cases except for two :(
List of open test cases:
plusOut("12xy34", "xy") → "++xy++"
plusOut("12xy34", "1") → "1+++++"
plusOut("12xy34xyabcxy", "xy") → "++xy++xy+++xy"
plusOut("abXYabcXYZ", "ab") → "ab++ab++++"
plusOut("abXYabcXYZ", "abc") → "++++abc+++"
plusOut("abXYabcXYZ", "XY") → "++XY+++XY+"
plusOut("abXYxyzXYZ", "XYZ") → "+++++++XYZ"
plusOut("--++ab", "++") → "++++++"
plusOut("aaxxxxbb", "xx") → "++xxxx++"
plusOut("123123", "3") → "++3++3"

Looks like this is the plusOut problem on CodingBat.
I had 3 solutions to this problem, and wrote a new streaming solution just for fun.
Solution 1: Loop and check
Create a StringBuilder out of the input string, and check for the word at every position. Replace the character if doesn't match, and skip the length of the word if found.
public String plusOut(String str, String word) {
StringBuilder out = new StringBuilder(str);
for (int i = 0; i < out.length(); ) {
if (!str.startsWith(word, i))
out.setCharAt(i++, '+');
else
i += word.length();
}
return out.toString();
}
This is probably the expected answer for a beginner programmer, though there is an assumption that the string doesn't contain any astral plane character, which would be represented by 2 char instead of 1.
Solution 2: Replace the word with a marker, replace the rest, then restore the word
public String plusOut(String str, String word) {
return str.replaceAll(java.util.regex.Pattern.quote(word), "#").replaceAll("[^#]", "+").replaceAll("#", word);
}
Not a proper solution since it assumes that a certain character or sequence of character doesn't appear in the string.
Note the use of Pattern.quote to prevent the word being interpreted as regex syntax by replaceAll method.
Solution 3: Regex with \G
public String plusOut(String str, String word) {
word = java.util.regex.Pattern.quote(word);
return str.replaceAll("\\G((?:" + word + ")*+).", "$1+");
}
Construct regex \G((?:word)*+)., which does more or less what solution 1 is doing:
\G makes sure the match starts from where the previous match leaves off
((?:word)*+) picks out 0 or more instance of word - if any, so that we can keep them in the replacement with $1. The key here is the possessive quantifier *+, which forces the regex to keep any instance of the word it finds. Otherwise, the regex will not work correctly when the word appear at the end of the string, as the regex backtracks to match .
. will not be part of any word, since the previous part already picks out all consecutive appearances of word and disallow backtrack. We will replace this with +
Solution 4: Streaming
public String plusOut(String str, String word) {
return String.join(word,
Arrays.stream(str.split(java.util.regex.Pattern.quote(word), -1))
.map((String s) -> s.replaceAll("(?s:.)", "+"))
.collect(Collectors.toList()));
}
The idea is to split the string by word, do the replacement on the rest, and join them back with word using String.join method.
Same as above, we need Pattern.quote to avoid split interpreting the word as regex. Since split by default removes empty string at the end of the array, we need to use -1 in the second parameter to make split leave those empty strings alone.
Then we create a stream out of the array and replace the rest as strings of +. In Java 11, we can use s -> String.repeat(s.length()) instead.
The rest is just converting the Stream to an Iterable (List in this case) and joining them for the result

This is a bit trickier than you might initially think because you don't just need to match characters, but the absence of specific phrase - a negated character set is not enough. If the string is 123, you would need:
(?<=^|123)(?!123).*?(?=123|$)
https://regex101.com/r/EZWMqM/1/
That is - lookbehind for the start of the string or "123", make sure the current position is not followed by 123, then lazy-repeat any character until lookahead matches "123" or the end of the string. This will match all characters which are not in a "123" substring. Then, you need to replace each character with a +, after which you can use appendReplacement and a StringBuffer to create the result string:
String inputPhrase = "123";
String inputStr = "abc123efg123123hij";
StringBuffer resultString = new StringBuffer();
Pattern regex = Pattern.compile("(?<=^|" + inputPhrase + ")(?!" + inputPhrase + ").*?(?=" + inputPhrase + "|$)");
Matcher m = regex.matcher(inputStr);
while (m.find()) {
String replacement = m.group(0).replaceAll(".", "+");
m.appendReplacement(resultString, replacement);
}
m.appendTail(resultString);
System.out.println(resultString.toString());
Output:
+++123+++123123+++
Note that if the inputPhrase can contain character with a special meaning in a regular expression, you'll have to escape them first before concatenating into the pattern.

You can do it in one line:
input = input.replaceAll("((?:" + str + ")+)?(?!" + str + ").((?:" + str + ")+)?", "$1+$2");
This optionally captures "123" either side of each character and puts them back (a blank if there's no "123"):

So instead of coming up with a regular expression that matches the absence of a string. We might as well just match the selected phrase and append + the number of skipped characters.
StringBuilder sb = new StringBuilder();
Matcher m = Pattern.compile(Pattern.quote(str)).matcher(input);
while (m.find()) {
for (int i = 0; i < m.start(); i++) sb.append('+');
sb.append(str);
}
int remaining = input.length() - sb.length();
for (int i = 0; i < remaining; i++) {
sb.append('+');
}

Absolutely just for the fun of it, a solution using CharBuffer (unexpectedly it took a lot more that I initially hoped for):
private static String plusOutCharBuffer(String input, String match) {
int size = match.length();
CharBuffer cb = CharBuffer.wrap(input.toCharArray());
CharBuffer word = CharBuffer.wrap(match);
int x = 0;
for (; cb.remaining() > 0;) {
if (!cb.subSequence(0, size < cb.remaining() ? size : cb.remaining()).equals(word)) {
cb.put(x, '+');
cb.clear().position(++x);
} else {
cb.clear().position(x = x + size);
}
}
return cb.clear().toString();
}

To make this work you need a beast of a pattern. Let's say you you are operating on the following test case as an example:
plusOut("abXYxyzXYZ", "XYZ") → "+++++++XYZ"
What you need to do is build a series of clauses in your pattern to match a single character at a time:
Any character that is NOT "X", "Y" or "Z" -- [^XYZ]
Any "X" not followed by "YZ" -- X(?!YZ)
Any "Y" not preceded by "X" -- (?<!X)Y
Any "Y" not followed by "Z" -- Y(?!Z)
Any "Z" not preceded by "XY" -- (?<!XY)Z
An example of this replacement can be found here: https://regex101.com/r/jK5wU3/4
Here is an example of how this might work (most certainly not optimized, but it works):
import java.util.regex.Pattern;
public class Test {
public static void plusOut(String text, String exclude) {
StringBuilder pattern = new StringBuilder("");
for (int i=0; i<exclude.length(); i++) {
Character target = exclude.charAt(i);
String prefix = (i > 0) ? exclude.substring(0, i) : "";
String postfix = (i < exclude.length() - 1) ? exclude.substring(i+1) : "";
// add the look-behind (?<!X)Y
if (!prefix.isEmpty()) {
pattern.append("(?<!").append(Pattern.quote(prefix)).append(")")
.append(Pattern.quote(target.toString())).append("|");
}
// add the look-ahead X(?!YZ)
if (!postfix.isEmpty()) {
pattern.append(Pattern.quote(target.toString()))
.append("(?!").append(Pattern.quote(postfix)).append(")|");
}
}
// add in the other character exclusion
pattern.append("[^" + Pattern.quote(exclude) + "]");
System.out.println(text.replaceAll(pattern.toString(), "+"));
}
public static void main(String [] args) {
plusOut("12xy34", "xy");
plusOut("12xy34", "1");
plusOut("12xy34xyabcxy", "xy");
plusOut("abXYabcXYZ", "ab");
plusOut("abXYabcXYZ", "abc");
plusOut("abXYabcXYZ", "XY");
plusOut("abXYxyzXYZ", "XYZ");
plusOut("--++ab", "++");
plusOut("aaxxxxbb", "xx");
plusOut("123123", "3");
}
}
UPDATE: Even this doesn't quite work because it can't deal with exclusions that are just repeated characters, like "xx". Regular expressions are most definitely not the right tool for this, but I thought it might be possible. After poking around, I'm not so sure a pattern even exists that might make this work.

The problem in your solution that you put a set of instance string str.replaceAll("[^str]","+") which it will exclude any character from the variable str and that will not solve your problem
EX: when you try str.replaceAll("[^XYZ]","+") it will exclude any combination of character X , character Y and character Z from your replacing method so you will get "++XY+++XYZ".
Actually you should exclude a sequence of characters instead in str.replaceAll.
You can do it by using capture group of characters like (XYZ) then use a negative lookahead to match a string which does not contain characters sequence : ^((?!XYZ).)*$
Check this solution for more info about this problem but you should know that it may be complicated to find regular expression to do that directly.
I have found two simple solutions for this problem :
Solution 1:
You can implement a method to replace all characters with '+' except the instance of given string:
String exWord = "XYZ";
String str = "abXYxyzXYZ";
for(int i = 0; i < str.length(); i++){
// exclude any instance string of exWord from replacing process in str
if(str.substring(i, str.length()).indexOf(exWord) + i == i){
i = i + exWord.length()-1;
}
else{
str = str.substring(0,i) + "+" + str.substring(i+1);//replace each character with '+' symbol
}
}
Note : str.substring(i, str.length()).indexOf(exWord) + i this if statement will exclude any instance string of exWord from replacing process in str.
Output:
+++++++XYZ
Solution 2:
You can try this Approach using ReplaceAll method and it doesn't need any complex regular expression:
String exWord = "XYZ";
String str = "abXYxyzXYZ";
str = str.replaceAll(exWord,"*"); // replace instance string with * symbol
str = str.replaceAll("[^*]","+"); // replace all characters with + symbol except *
str = str.replaceAll("\\*",exWord); // replace * symbol with instance string
Note : This solution will work only if your input string str doesn't contain any * symbol.
Also you should escape any character with a special meaning in a regular expression in phrase instance string exWord like : exWord = "++".

Split string in latest occurrence of substring

I need help to split string when substring is found, only on latest occurrence
Substring to search(example):123
String: 123hello123boy123guy123girl
Res: 123hello123boy123guy (result on split[0])
Ex: hello123boy
Res:hello
I'm trying with this function, but this split only on first occource.
public static String[] getSplitArray(String toSplitString, String spltiChar) {
return toSplitString.split("(?<=" + spltiChar + ")");
}

If you want to do it in single split(), this should work:
string.split("123((?!123).)*?$")
E.g.:
String s = "foo123bar123hhh123tillHere123Bang";
System.out.println(s.split("123((?!123).)*?$")[0]);
Outputs:
foo123bar123hhh123tillHere
Another approach would be, just split("123") then join the elements you required by 123 as delimiter.

Improvement on Kent's answer: this will actually give you a split array rather than a single-element array with the first part of your String:
String[] text = {
"123hello123boy123guy123girl", // 123hello123boy123guy
"hello123boy"// hello
};
for (String s: text) {
System.out.println(
Arrays.toString(
s.split("123(?=((?!123).)*?$)")
)
);
}
Output
[123hello123boy123guy, girl]
[hello, boy]

You don't need a regex for that, you can simply use String#lastIndexOf:
-> String s = "123hello123boy123guy123girl"
-> s.substring(0, s.lastIndexOf("123"));
| Expression value is: "123hello123boy123guy"
-> s.substring(s.lastIndexOf("123"));
| Expression value is: "123girl"

Remove trailing substring from String in Java

I am looking to remove parts of a string if it ends in a certain string.
An example would be to take this string: "am.sunrise.ios#2x.png"
And remove the #2x.png so it looks like: "am.sunrise.ios"
How would I go about checking to see if the end of a string contains "#2x.png" and remove it?

You could check the lastIndexOf, and if it exists in the string, use substring to remove it:
String str = "am.sunrise.ios#2x.png";
String search = "#2x.png";
int index = str.lastIndexOf(search);
if (index > 0) {
str = str.substring(0, index);
}

Assuming you have a string initialized as String file = "am.sunrise.ios#2x.png";.
if(file.endsWith("#2x.png"))
file = file.substring(0, file.lastIndexOf("#2x.png"));
The endsWith(String) method returns a boolean determining if the string has a certain suffix. Depending on that you can replace the string with a substring of itself starting with the first character and ending before the index of the character that you are trying to remove.

private static String removeSuffixIfExists(String key, String suffix) {
return key.endswith(suffix)
? key.substring(0, key.length() - suffix.length())
: key;
}
}
String suffix = "#2x.png";
String key = "am.sunrise.ios#2x.png";
String output = removeSuffixIfExists(key, suffix);

public static void main(String [] args){
String word = "am.sunrise.ios#2x.png";
word = word.replace("#2x.png", "");
System.out.println(word);
}

If you want to generally remove entire content of string from # till end you can use
yourString = yourString.replaceAll("#.*","");
where #.* is regex (regular expression) representing substring starting with # and having any character after it (represented by .) zero or more times (represented by *).
In case there will be no #xxx part your string will be unchanged.
If you want to change only this particular substring #2x.png (and not substirng like #3x.png) while making sure that it is placed at end of your string you can use
yourString = yourString.replaceAll("#2x\\.png$","");
where
$ represents end of string
\\. represents . literal (we need to escape it since like shown earlier . is metacharacter representing any character)

Since I was trying to do this on an ArrayList of items similarly styled I ended up using the following code:
for (int image = 0; image < IBImages.size(); image++) {
IBImages.set(image, IBImages.get(image).split("~")[0].split("#")[0].split(".png")[0]);
}
If I have a list of images with the names
[am.sunrise.ios.png, am.sunrise.ios#2x.png, am.sunrise.ios#3x.png, am.sunrise.ios~ipad.png, am.sunrise.ios~ipad#2x.png]
This allows me to split the string into 2 parts.
For example, "am.sunrise.ios~ipad.png" will be split into "am.sunrise.ios" and "~ipad.png" if I split on "~". I can just get the first part back by referencing [0]. Therefore I get what I'm looking for in one line of code.
Note that image is "am.sunrise.ios~ipad.png"

You could use String.split():
public static void main(String [] args){
String word = "am.sunrise.ios#2x.png";
String[] parts = word.split("#");
if (parts.length == 2) {
System.out.println("looks like user#host...");
System.out.println("User: " + parts[0]);
System.out.println("Host: " + parts[1]);
}
}
Then you haven an array of Strings, where the first element contains the part before "#" and the second element the part after the "#".

Combining the answers 1 and 2:
String str = "am.sunrise.ios#2x.png";
String search = "#2x.png";
if (str.endsWith(search)) {
str = str.substring(0, str.lastIndexOf(search));
}

java split () method

I've got a string '123' (yes, it's a string in my program). Could anyone explain, when I use this method:
String[] str1Array = str2.split(" ");
Why I got str1Array[0]='123' rather than str1Array[0]=1?

str2 does not contain any spaces, therefore split copies the entire contents of str2 to the first index of str1Array.
You would have to do:
String str2 = "1 2 3";
String[] str1Array = str2.split(" ");
Alternatively, to find every character in str2 you could do:
for (char ch : str2.toCharArray()){
System.out.println(ch);
}
You could also assign it to the array in the loop.

str2.split("") ;
Try this:to split each character in a string .
Output:
[, 1, 2, 3]
but it will return an empty first value.
str2.split("(?!^)");
Output :
[1, 2, 3]

the regular expression that you pass to the split() should have a match in the string so that it will split the string in places where there is a match found in the string. Here you are passing " " which is not found in '123' hence there is no split happening.

Because there's no space in your String.
If you want single chars, try char[] characters = str2.toCharArray()

Simple...You are trying to split string by space and in your string "123", there is no space

This is because the split() method literally splits the string based on the characters given as a parameter.
We remove the splitting characters and form a new String every time we find the splitting characters.
String[] strs = "123".split(" ");
The String "123" does not have the character " " (space) and therefore cannot be split apart. So returned is just a single item in the array - { "123" }.

To do the "Split" you must use a delimiter, in this case insert a "," between each number
public static void main(String[] args) {
String[] list = "123456".replaceAll("(\\d)", ",$1").substring(1)
.split(",");
for (String string : list) {
System.out.println(string);
}
}

Try this:
String str = "123";
String res = str.split("");
will return the following result:
1,2,3

Java Split not working as expected

I am trying to use a simple split to break up the following string: 00-00000
My expression is: ^([0-9][0-9])(-)([0-9])([0-9])([0-9])([0-9])([0-9])
And my usage is:
String s = "00-00000";
String pattern = "^([0-9][0-9])(-)([0-9])([0-9])([0-9])([0-9])([0-9])";
String[] parts = s.split(pattern);
If I play around with the Pattern and Matcher classes I can see that my pattern does match and the matcher tells me my groupCount is 7 which is correct. But when I try and split them I have no luck.

String.split does not use capturing groups as its result. It finds whatever matches and uses that as the delimiter. So the resulting String[] are substrings in between what the regex matches. As it is the regex matches the whole string, and with the whole string as a delimiter there is nothing else left so it returns an empty array.
If you want to use regex capturing groups you will have to use Matcher.group(), String.split() will not do.

for your example, you could simply do this:
String s = "00-00000";
String pattern = "-";
String[] parts = s.split(pattern);

I can not be sure, but I think what you are trying to do is to get each matched group into an array.
Matcher matcher = Pattern.compile(pattern).matcher();
if (matcher.matches()) {
String s[] = new String[matcher.groupCount()) {
for (int i=0;i<matches.groupCount();i++) {
s[i] = matcher.group(i);
}
}
}

From the documentation:
String[] split(String regex) -- Returns: the array of strings computed by splitting this string around matches of the given regular expression
Essentially the regular expression is used to define delimiters in the input string. You can use capturing groups and backreferences in your pattern (e.g. for lookarounds), but ultimately what matters is what and where the pattern matches, because that defines what goes into the returned array.
If you want to split your original string into 7 parts using regular expression, then you can do something like this:
String s = "12-3456";
String[] parts = s.split("(?!^)");
System.out.println(parts.length); // prints "7"
for (String part : parts) {
System.out.println("Part [" + part + "]");
} // prints "[1] [2] [-] [3] [4] [5] [6] "
This splits on zero-length matching assertion (?!^), which is anywhere except before the first character in the string. This prevents the empty string to be the first element in the array, and trailing empty string is already discarded because we use the default limit parameter to split.
Using regular expression to get individual character of a string like this is an overkill, though. If you have only a few characters, then the most concise option is to use foreach on the toCharArray():
for (char ch : "12-3456".toCharArray()) {
System.out.print("[" + ch + "] ");
}
This is not the most efficient option if you have a longer string.
Splitting on -
This may also be what you're looking for:
String s = "12-3456";
String[] parts = s.split("-");
System.out.println(parts.length); // prints "2"
for (String part : parts) {
System.out.print("[" + part + "] ");
} // prints "[12] [3456] "