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Capitalize first letters in words in the string with different separators using java 8 stream

I need to capitalize first letter in every word in the string, BUT it's not so easy as it seems to be as the word is considered to be any sequence of letters, digits, "_" , "-", "`" while all other chars are considered to be separators, i.e. after them the next letter must be capitalized.
Example what program should do:
For input: "#he&llo wo!r^ld"
Output should be: "#He&Llo Wo!R^Ld"
There are questions that sound similar here, but there solutions really don't help.
This one for example:
String output = Arrays.stream(input.split("[\\s&]+"))
.map(t -> t.substring(0, 1).toUpperCase() + t.substring(1))
.collect(Collectors.joining(" "));
As in my task there can be various separators, this solution doesn't work.

It is possible to split a string and keep the delimiters, so taking into account the requirement for delimiters:
word is considered to be any sequence of letters, digits, "_" , "-", "`" while all other chars are considered to be separators
the pattern which keeps the delimiters in the result array would be: "((?<=[^-`\\w])|(?=[^-`\\w]))":
[^-`\\w]: all characters except -, backtick and word characters \w: [A-Za-z0-9_]
Then, the "words" are capitalized, and delimiters are kept as is:
static String capitalize(String input) {
if (null == input || 0 == input.length()) {
return input;
}
return Arrays.stream(input.split("((?<=[^-`\\w])|(?=[^-`\\w]))"))
.map(s -> s.matches("[-`\\w]+") ? Character.toUpperCase(s.charAt(0)) + s.substring(1) : s)
.collect(Collectors.joining(""));
}
Tests:
System.out.println(capitalize("#he&l_lo-wo!r^ld"));
System.out.println(capitalize("#`he`&l+lo wo!r^ld"));
Output:
#He&l_lo-wo!R^Ld
#`he`&L+Lo Wo!R^Ld
Update
If it is needed to process not only ASCII set of characters but apply to other alphabets or character sets (e.g. Cyrillic, Greek, etc.), POSIX class \\p{IsWord} may be used and matching of Unicode characters needs to be enabled using pattern flag (?U):
static String capitalizeUnicode(String input) {
if (null == input || 0 == input.length()) {
return input;
}
return Arrays.stream(input.split("(?U)((?<=[^-`\\p{IsWord}])|(?=[^-`\\p{IsWord}]))")
.map(s -> s.matches("(?U)[-`\\p{IsWord}]+") ? Character.toUpperCase(s.charAt(0)) + s.substring(1) : s)
.collect(Collectors.joining(""));
}
Test:
System.out.println(capitalizeUnicode("#he&l_lo-wo!r^ld"));
System.out.println(capitalizeUnicode("#привет&`ёж`+дос^βιδ/ως"));
Output:
#He&L_lo-wo!R^Ld
#Привет&`ёж`+Дос^Βιδ/Ως

You can't use split that easily - split will eliminate the separators and give you only the things in between. As you need the separators, no can do.
One real dirty trick is to use something called 'lookahead'. That argument you pass to split is a regular expression. Most 'characters' in a regexp have the property that they consume the matching input. If you do input.split("\\s+") then that doesn't 'just' split on whitespace, it also consumes them: The whitespace is no longer part of the individual entries in your string array.
However, consider ^ and $. or \\b. These still match things but don't consume anything. You don't consume 'end of string'. In fact, ^^^hello$$$ matches the string "hello" just as well. You can do this yourself, using lookahead: It matches when the lookahead is there but does not consume it:
String[] args = "Hello World$Huh Weird".split("(?=[\\s_$-]+)");
for (String arg : args) System.out.println("*" + args[i] + "*");
Unfortunately, this 'works', in that it saves your separators, but isn't getting you all that much closer to a solution:
*Hello*
* World*
*$Huh*
* *
* *
* Weird*
You can go with lookbehind as well, but it's limited; they don't do variable length, for example.
The conclusion should rapidly become: Actually, doing this with split is a mistake.
Then, once split is off the table, you should no longer use streams, either: Streams don't do well once you need to know stuff about the previous element in a stream to do the job: A stream of characters doesn't work, as you need to know if the previous character was a non-letter or not.
In general, "I want to do X, and use Y" is a mistake. Keep an open mind. It's akin to asking: "I want to butter my toast, and use a hammer to do it". Oookaaaaayyyy, you can probably do that, but, eh, why? There are butter knives right there in the drawer, just.. put down the hammer, that's toast. Not a nail.
Same here.
A simple loop can take care of this, no problem:
private static final String BREAK_CHARS = "&-_`";
public String toTitleCase(String input) {
StringBuilder out = new StringBuilder();
boolean atBreak = true;
for (char c : input.toCharArray()) {
out.append(atBreak ? Character.toUpperCase(c) : c);
atBreak = Character.isWhitespace(c) || (BREAK_CHARS.indexOf(c) > -1);
}
return out.toString();
}
Simple. Efficient. Easy to read. Easy to modify. For example, if you want to go with 'any non-letter counts', trivial: atBreak = Character.isLetter(c);.
Contrast to the stream solution which is fragile, weird, far less efficient, and requires a regexp that needs half a page's worth of comment for anybody to understand it.
Can you do this with streams? Yes. You can butter toast with a hammer, too. Doesn't make it a good idea though. Put down the hammer!

You can use a simple FSM as you iterate over the characters in the string, with two states, either in a word, or not in a word. If you are not in a word and the next character is a letter, convert it to upper case, otherwise, if it is not a letter or if you are already in a word, simply copy it unmodified.
boolean isWord(int c) {
return c == '`' || c == '_' || c == '-' || Character.isLetter(c) || Character.isDigit(c);
}
String capitalize(String s) {
StringBuilder sb = new StringBuilder();
boolean inWord = false;
for (int c : s.codePoints().toArray()) {
if (!inWord && Character.isLetter(c)) {
sb.appendCodePoint(Character.toUpperCase(c));
} else {
sb.appendCodePoint(c);
}
inWord = isWord(c);
}
return sb.toString();
}
Note: I have used codePoints(), appendCodePoint(int), and int so that characters outside the basic multilingual plane (with code points greater than 64k) are handled correctly.

I need to capitalize first letter in every word
Here is one way to do it. Admittedly this is a might longer but your requirement to change the first letter to upper case (not first digit or first non-letter) required a helper method. Otherwise it would have been easier. Some others seemed to have missed this point.
Establish word pattern, and test data.
String wordPattern = "[\\w_-`]+";
Pattern p = Pattern.compile(wordPattern);
String[] inputData = { "#he&llo wo!r^ld", "0hel`lo-w0rld" };
Now this simply finds each successive word in the string based on the established regular expression. As each word is found, it changes the first letter in the word to upper case and then puts it in a string buffer in the correct position where the match was found.
for (String input : inputData) {
StringBuilder sb = new StringBuilder(input);
Matcher m = p.matcher(input);
while (m.find()) {
sb.replace(m.start(), m.end(),
upperFirstLetter(m.group()));
}
System.out.println(input + " -> " + sb);
}
prints
#he&llo wo!r^ld -> #He&Llo Wo!R^Ld
0hel`lo-w0rld -> 0Hel`lo-W0rld
Since words may start with digits, and the requirement was to convert the first letter (not character) to upper case. This method finds the first letter, converts it to upper case and
returns the new string. So 01_hello would become 01_Hello
public static String upperFirstLetter(String word) {
char[] chs = word.toCharArray();
for (int i = 0; i < chs.length; i++) {
if (Character.isLetter(chs[i])) {
chs[i] = Character.toUpperCase(chs[i]);
break;
}
}
return String.valueOf(chs);
}

How to check if given string can be formed from another string in Java?

I need to check if a given string can be formed from main string or not in Java .
Example :
Input
Main String - lhLoe
HELLO
eel
eL,$H.
Ouput:
HELLO : YES
eel : NO
eL,$H. : YES
I can try with permutation , but if the first string is long , it will take much time . Any other way to do this ?
Note: we can ignore special characters.

You could use regex and define your String as a character group.
Something like this:
String search="lhLoe";
String input="HELLO";
input=input.replaceAll("^[a-zA-Z1-9]","");
Pattern p = Pattern.compile("^["+search+"]+$",Pattern.CASE_INSENSITIVE);
Matcher m = p.matcher(input);
boolean b = m.matches();

First, lowercase both strings, so that:
lowerCasedMainString = "hello"
lowerCasedTargetString = "el,$h"
Next remove special characters, since you say they can be ignored:
lowerCasedTargetStringStripped = "elh"
There are various ways you can tackle the rest of the problem, but I would do a loop like this (pseudocode):
while(length of target > 0) {
char c = target.charAt(1);
if(main contains c) {
main = main with that instance of c removed
} else {
return false
}
target = target.substring(1);
}
return true;
Each individual step of this should be easily achievable. If you're tackling a problem like this, you've probably already seen the methods that will do it. Otherwise, there will be existing answers on Stack Overflow.

You could store the occurrences of each character in the given main string in an array. (Let's call it array a.)
Repeat the same procedure for the given test strings. (Let's call it array b)
If the occurrence of each character in test string (array b) is less than equal to the occurrence of that corresponding character in the main string(array a), then, the test string can be formed using main string.
Runtime complexity will be linear.
Take care of case sensitivity and special symbols.

Making only the first letter of a word uppercase

I have a method that converts all the first letters of the words in a sentence into uppercase.
public static String toTitleCase(String s)
{
String result = "";
String[] words = s.split(" ");
for (int i = 0; i < words.length; i++)
{
result += words[i].replace(words[i].charAt(0)+"", Character.toUpperCase(words[i].charAt(0))+"") + " ";
}
return result;
}
The problem is that the method converts each other letter in a word that is the same letter as the first to uppercase. For example, the string title comes out as TiTle
For the input this is a title this becomes the output This Is A TiTle
I've tried lots of things. A nested loop that checks every letter in each word, and if there is a recurrence, the second is ignored. I used counters, booleans, etc. Nothing works and I keep getting the same result.
What can I do? I only want the first letter in upper case.

Instead of using the replace() method, try replaceFirst().
result += words[i].replaceFirst(words[i].charAt(0)+"", Character.toUpperCase(words[i].charAt(0))+"") + " ";
Will output:
This Is A Title

The problem is that you are using replace method which replaces all occurrences of described character. To solve this problem you can either
use replaceFirst instead
take first letter,
create its uppercase version
concatenate it with rest of string which can be created with a little help of substring method.
since you are using replace(String, String) which uses regex you can add ^ before character you want to replace like replace("^a","A"). ^ means start of input so it will only replace a that is placed after start of input.
I would probably use second approach.
Also currently in each loop your code creates new StringBuilder with data stored in result, append new word to it, and reassigns result of output from toString().
This is infective approach. Instead you should create StringBuilder before loop that will represent your result and append new words created inside loop to it and after loop ends you can get its String version with toString() method.

Doing some Regex-Magic can simplify your task:
public static void main(String[] args) {
final String test = "this is a Test";
final StringBuffer buffer = new StringBuffer(test);
final Pattern patter = Pattern.compile("\\b(\\p{javaLowerCase})");
final Matcher matcher = patter.matcher(buffer);
while (matcher.find()) {
buffer.replace(matcher.start(), matcher.end(), matcher.group().toUpperCase());
}
System.out.println(buffer);
}
The expression \\b(\\p{javaLowerCase}) matches "The beginning of a word followed by a lower-case letter", while matcher.group() is equal to whats inside the () in the part that matches. Example: Applying on "test" matches on "t", so start is 0, end is 1 and group is "t". This can easily run through even a huge amount of text and replace all those letters that need replacement.
In addition: it is always a good idea to use a StringBuffer (or similar) for String manipulation, because each String in Java is unique. That is if you do something like result += stringPart you actually create a new String (equal to result + stringPart) each time this is called. So if you do this with like 10 parts, you will in the end have at least 10 different Strings in memory, while you only need one, which is the final one.
StringBuffer instead uses something like char[] to ensure that if you change only a single character no extra memory needs to be allocated.
Note that a patter only need to be compiled once, so you can keep that as a class variable somewhere.

Create String[] containing only certain characters

I am trying to create a String[] which contains only words that comprise of certain characters. For example I have a dictionary containing a number of words like so:
arm
army
art
as
at
attack
attempt
attention
attraction
authority
automatic
awake
baby
back
bad
bag
balance
I want to narrow the list down so that it only contains words with the characters a, b and g. Therefore the list should only contain the word 'bag' in this example.
Currently I am trying to do this using regexes but having never used them before I can't seem to get it to work.
Here is my code:
public class LetterJugglingMain {
public static void main(String[] args) {
String dictFile = "/Users/simonrhillary/Desktop/Dictionary(3).txt";
fileReader fr = new fileReader();
fr.openFile(dictFile);
String[] dictionary = fr.fileToArray();
String regx = "able";
String[] newDict = createListOfValidWords(dictionary, regx);
printArray(newDict);
}
public static String[] createListOfValidWords(String[] d, String regex){
List<String> narrowed = new ArrayList<String>();
for(int i = 0; i<d.length; i++){
if(d[i].matches(regex)){
narrowed.add(d[i]);
System.out.println("added " + d[i]);
}
}
String[] narrowArray = narrowed.toArray(new String[0]);
return narrowArray;
}
however the array returned is always empty unless the String regex is the exact word! Any ideas? I can post more code if needed...I think I must be trying to initialise the regex wrong.
The narrowed down list must contain ONLY the characters from the regex.

Frankly, I'm not an expert in regexes, but I don't think it's the best tool to do what you want. I would use a method like the following:
public boolean containsAll(String s, Set<Character> chars) {
Set<Character> copy = new HashSet<Character>();
for (int i = 0; i < s.length() && copy.size() < chars.size(); i++) {
char c = s.charAt(i);
if (chars.contains(c)) {
copy.add(c);
}
}
return copy.size() == chars.size();
}

The regex able will match only the string "able". However, if you want a regular expression to match either character of a, b, l or e, the regex you're looking for is [able] (in brackets). If you want words containing several such characters, add a + for repeating the pattern: [able]+.

The OP wants words that contain every character. Not just one of them.
And other characters are not a problem.
If this is the case, I think the simiplest way would be to loop through the entire string, character by character, and check to see if it contains all of the characters you want. Keep flags to check and see if every character has been found.
If this isn't the case.... :
Try using the regex:
^[able]+$
Here's what it does:
^ matches the beginning of the string and $ matches the end of the string. This makes sure that you're not getting a partial match.
[able] matches the characters you want the string to consist of, in this case a, b, l, and e. + Makes sure that there are 1 or more of these characters in the string.
Note: This regex will match a string that contains these 4 letters. For example, it will match:
able, albe, aeble, aaaabbblllleeee
and will not match
qable, treatable, and abled.

A sample regex that filters out words that contains at least one occurrence of all characters in a set. This will match any English word (case-insensitive) that contains at least one occurrence of all the characters a, b, g:
(?i)(?=.*a)(?=.*b)(?=.*g)[a-z]+
Example of strings that match would be bag, baggy, grab.
Example of strings that don't match would be big, argument, nothing.
The (?i) means turns on case-insensitive flag.
You need to append as many (?=.*<character>) as the number of characters in the set, for each of the characters.
I assume a word only contains English alphabet, so I specify [a-z]. Specify more if you need space, hyphen, etc.
I assume matches(String regex) method in String class, so I omitted the ^ and $.
The performance may be bad, since in the worst case (the characters are found at the end of the words), I think that the regex engine may go through the string for around n times where n is the number of characters in the set. It may not be an actual concern at all, since the words are very short, but if it turns out that this is a bottleneck, you may consider doing simple looping.

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