Develop Reference

Using a recursive method to determine if a word is elf-ish

public static boolean Xish
This method should take in two parameters, in the following order: A String of the word to check and a String made up of the letters to check for. For example, a word is considered elf-ish, if it contains the letters e, l, and f, in any order (“waffle”, “rainleaf”) and a true return of the method would be Xish(“waffle”, ”elf”). If there are multiple occurrences of a letter to check for, it must occur multiple times in the search word. Return true if the word contains all the needed characters and false if it does not contain all the characters.
This is what I have so far, but I am lost how I would recall the method and check to see if there are multiple occurrences (2nd part).
public static boolean Xish(String check, String letters) {
String word = check;
String contains= letters;
if(word.indexOf(contains) >= 0)
return true;
else
return false;
}

Actually, doing this recursively will also take care of the multiple occurrences issue.
First, your own method is not really correct - it looks for the whole letters in the word. That is, if letters is elf, then true will be returned for self, but not for heartfelt, and that's wrong. You are supposed to look for the individual letters, because the order is not important.
For recursion:
If the letters is an empty string - return true. You can say that any word is fine if there are no restrictions.
If the check is an empty string - return false. An empty string does not contain the letters in letters (and we already know that letters is not empty).
Take the first letter in letters. Look for it in check. If it's not there, return false.
If it was there, then call the same method, but pass only what remains of check and letters. For example, if check was selfish and letters was elf, you found that e exists. Return the result of Xish("slfish","lf"). This will take care of the multiple occurrences. You do that by using substring and concatenating the applicable parts.
If multiple occurrences weren't an issue, you could pass the check as-is to the next level of the recursion. But since they matter, we need to remove one letter for each letter requested, to make sure that we don't match the same position again for the next occurrenc.

The title mentions a recursive function so I will propose a recursive solution.
For each character in your check string, compare it against the first character in your letters string.
If the compared characters are equivalent, remove the first character from your letters string and pass both strings back into your function.
If the check string is fully iterated without finding a character in the letters string, return false
If letters is empty at any point, return true
This is a brute force approach, and there are several other ways to accomplish what you are looking for. Maybe think about how you could check every character in your in you check string a single time?

public static boolean Xish(String check, String letters) {
boolean ish = true;
String word = check;
char[] contains= letters.toCharArray();
for(int i = 0; i < contains.length; i++){
if(word.indexOf(contains[i]) < 0){
ish = false;
}else {
StringBuilder sb = new StringBuilder(word);
sb.deleteCharAt(word.indexOf(contains[i]));
word = sb.toString();
// System.out.println(word);
}
}
return ish;
}
This could be one way, but it is not recursive.
Xish("Waffle", "elff") returns true, but
Xish("Waffle", "elfff") returns false.

Not sure whether it solves your question 100 %. But i tried a recursive method. See if this helps.
package com.company;
public class Selfish {
public static void main(String args[]) {
String check = "waffle";
String letters = "elf"; // "eof"
int xishCount = xish(check, letters, 0);
if(letters.length()== xishCount) {
System.out.println("TRUE");
}else{
System.out.println("FALSE");
}
}
static int xish(String check, String letters, int xishCount) {
if(letters.length() < 1) {
return 0;
}
if(check.contains(letters.substring(0, 1))) {
xishCount = 1;
}
return xishCount + xish(check, letters.substring(1, letters.length()), 0);
}
}

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.

String.split() returning a "" unexpectedly

I have a simple method splitting a string into an array. It splits it where there are non-letter characters. The line I am using right now is as follows:
String[] words = str.split("[^a-zA-Z]");
So this should split the string where there are only alphabetical characters. But the problem is that when it splits it works for some, but not all. For example:
String str = "!!day--yaz!!";
String[] words = str.split("[^a-zA-Z]");
String result = "";
for (int i = 0; i < words.length; i++) {
result += words[i] + "1 ";
}
return result;
I added the 1 in there to see where the split takes place, becuase i was getting errors on null values. Anyway, when I run this code I get an output of:
1 1 day1 1 yaz1
Why is it splitting between the first two !'s and after one of the -'s, but not after the last two !'s? Why is it even splitting there at all? Any help on this would be great!

It doesn't split before or after it splits ON the matches, therefore you get an empty String between the dashes and the bangs.
This doesn't apply to the trailing bangs, because trailing empty Strings are omitted as described in the javadoc
Trailing empty strings are therefore not included in the resulting
array.

This happens because it indeed uses every non-letter character as a delimiter. It means that string "!" will be splitted into array of 2 empty strings to the left and to the right of the exclamation sign.
Your problem can be solved withing 2 steps.
use "[^a-zA-Z]+" instead of "[^a-zA-Z]". The + will help you to avoid empty string between 2 dashes.
Remove starting and trailing non-letter characters before splitting. This will remove leading and trailing empty strings: str.replaceFirst("[^a-zA-Z]+").replaceFirst("[^a-zA-Z]+$")
Finally your split will look like:
String[] words = str..replaceFirst("[^a-zA-Z]+").replaceFirst("[^a-zA-Z]+$")split("[^a-zA-Z]");

If you want to get rid of some of the extra splits, use split("[^a-zA-Z]+") instead of split("[^a-zA-Z]"). This will match a continuous part of the String that matches the pattern.

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.

Java - removing first character of a string

In Java, I have a String:
Jamaica
I would like to remove the first character of the string and then return amaica
How would I do this?

const str = "Jamaica".substring(1)
console.log(str)
Use the substring() function with an argument of 1 to get the substring from position 1 (after the first character) to the end of the string (leaving the second argument out defaults to the full length of the string).

public String removeFirstChar(String s){
return s.substring(1);
}

In Java, remove leading character only if it is a certain character
Use the Java ternary operator to quickly check if your character is there before removing it. This strips the leading character only if it exists, if passed a blank string, return blankstring.
String header = "";
header = header.startsWith("#") ? header.substring(1) : header;
System.out.println(header);
header = "foobar";
header = header.startsWith("#") ? header.substring(1) : header;
System.out.println(header);
header = "#moobar";
header = header.startsWith("#") ? header.substring(1) : header;
System.out.println(header);
Prints:
blankstring
foobar
moobar
Java, remove all the instances of a character anywhere in a string:
String a = "Cool";
a = a.replace("o","");
//variable 'a' contains the string "Cl"
Java, remove the first instance of a character anywhere in a string:
String b = "Cool";
b = b.replaceFirst("o","");
//variable 'b' contains the string "Col"

Use substring() and give the number of characters that you want to trim from front.
String value = "Jamaica";
value = value.substring(1);
Answer: "amaica"

You can use the substring method of the String class that takes only the beginning index and returns the substring that begins with the character at the specified index and extending to the end of the string.
String str = "Jamaica";
str = str.substring(1);

substring() method returns a new String that contains a subsequence of characters currently contained in this sequence.
The substring begins at the specified start and extends to the character at index end - 1.
It has two forms. The first is
String substring(int FirstIndex)
Here, FirstIndex specifies the index at which the substring will
begin. This form returns a copy of the substring that begins at
FirstIndex and runs to the end of the invoking string.
String substring(int FirstIndex, int endIndex)
Here, FirstIndex specifies the beginning index, and endIndex specifies
the stopping point. The string returned contains all the characters
from the beginning index, up to, but not including, the ending index.
Example
String str = "Amiyo";
// prints substring from index 3
System.out.println("substring is = " + str.substring(3)); // Output 'yo'

you can do like this:
String str = "Jamaica";
str = str.substring(1, title.length());
return str;
or in general:
public String removeFirstChar(String str){
return str.substring(1, title.length());
}

public String removeFirst(String input)
{
return input.substring(1);
}

The key thing to understand in Java is that Strings are immutable -- you can't change them. So it makes no sense to speak of 'removing a character from a string'. Instead, you make a NEW string with just the characters you want. The other posts in this question give you a variety of ways of doing that, but its important to understand that these don't change the original string in any way. Any references you have to the old string will continue to refer to the old string (unless you change them to refer to a different string) and will not be affected by the newly created string.
This has a number of implications for performance. Each time you are 'modifying' a string, you are actually creating a new string with all the overhead implied (memory allocation and garbage collection). So if you want to make a series of modifications to a string and care only about the final result (the intermediate strings will be dead as soon as you 'modify' them), it may make more sense to use a StringBuilder or StringBuffer instead.

I came across a situation where I had to remove not only the first character (if it was a #, but the first set of characters.
String myString = ###Hello World could be the starting point, but I would only want to keep the Hello World. this could be done as following.
while (myString.charAt(0) == '#') { // Remove all the # chars in front of the real string
myString = myString.substring(1, myString.length());
}
For OP's case, replace while with if and it works aswell.

You can simply use substring().
String myString = "Jamaica"
String myStringWithoutJ = myString.substring(1)
The index in the method indicates from where we are getting the result string, in this case we are getting it after the first position because we dont want that "J" in "Jamaica".

Another solution, you can solve your problem using replaceAll with some regex ^.{1} (regex demo) for example :
String str = "Jamaica";
int nbr = 1;
str = str.replaceAll("^.{" + nbr + "}", "");//Output = amaica

My version of removing leading chars, one or multiple. For example, String str1 = "01234", when removing leading '0', result will be "1234". For a String str2 = "000123" result will be again "123". And for String str3 = "000" result will be empty string: "". Such functionality is often useful when converting numeric strings into numbers.The advantage of this solution compared with regex (replaceAll(...)) is that this one is much faster. This is important when processing large number of Strings.
public static String removeLeadingChar(String str, char ch) {
int idx = 0;
while ((idx < str.length()) && (str.charAt(idx) == ch))
idx++;
return str.substring(idx);
}

##KOTLIN
#Its working fine.
tv.doOnTextChanged { text: CharSequence?, start, count, after ->
val length = text.toString().length
if (length==1 && text!!.startsWith(" ")) {
tv?.setText("")
}
}