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Get all combinations of Strings

I have requirement that need to get all combinations of given Strings.
For example
I have a String of digits and some special charcters
String chars="0123456789##$%&";
String guessedPw="authentic";
So I want to get combinations like this
$authentic%4
authentic##
5&authentic
authentic8
How should I improve my method to get all combinations?
This is the code that I have written.
but it doesn't give me all combinations.
private static String combination(String prefix, String s, String pw, String md5Pw) {
String pwFound = "";
if (s.length() > 0) {
// System.out.println(prefix + s.charAt(0) + pw);
String tempPw1 = prefix + s.charAt(0) + pw;
System.out.println("pw1 : " + tempPw1);
if (md5(tempPw1).equals(md5Pw)) {
// String tempPw1;
pwFound = tempPw1;
return pwFound;
}
String tempPw2 = pw + prefix + s.charAt(0);
if (md5(tempPw2).equals(md5Pw)) {
// String tempPw1;
pwFound = tempPw2;
return pwFound;
}
pwFound = combination(prefix + s.charAt(0), s.substring(1), pw, md5Pw);
pwFound = combination(prefix, s.substring(1), pw, md5Pw);
}
return pwFound;
}

Just if you don't want write own algo, you can use google.
Try to search for: "Generate Permutations".
For example, on this link: Generate all combinations from multiple lists there is a algo which you can use it (but with list).
void GeneratePermutations(List<List<Character>> Lists, List<String> result, int depth, String current)
{
if(depth == Lists.size())
{
result.add(current);
return;
}
for(int i = 0; i < Lists.get(depth).size(); ++i)
{
GeneratePermutations(Lists, result, depth + 1, current + Lists.get(depth).get(i));
}
}
But sure, there are a lot of other ways.

If you want to code everything by your own then this is how you should approach your problem =>
Have a special character say ~ to denote guessedPw, and create another string say str = chars + "~".
Now first you need to look for all possible combinations of str, finding which will take exponential time , and then for each found combination, you should generate all it's permutation , which is again factorial time complex. And then in the final string , you should replace all the occurrence of '~' with guessedPw, and get the answer string.
Here you go for the link for generating permutations and combinations :
Link : http://www.geeksforgeeks.org/write-a-c-program-to-print-all-permutations-of-a-given-string/
To find out combinations , you can use bit-masking if you have at max. 64 characters in the string , combinations from which you want to extract.

To make new combination, you can add one char as prefix or add one char as suffix. To avoid repeats, stop adding prefix when suffix exists
Pseudocode:
GenPass(CurrentString, boolAllowPrefix)
check CurrentString, return it if good result
check length limit, exit if too long
for c in Chars
if (boolAllowPrefix)
GenPass(Chars[i] + CurrentString, True)
GenPass(s + Chars[i], False)
Delphi implementation to check
procedure GenPass(s: string; bPrefix: Boolean);
var
i: integer;
begin
List.Add(s);
if Length(s) = MaxLen then
Exit;
if bPrefix then
for i := 1 to Length(Chars) do
GenPass(Chars[i] + s, True);
for i := 1 to Length(Chars) do
GenPass(s + Chars[i], False);
end;

Mask part of the String data which is of different lengths using Java

I am in need to mask PII data for my application. The PII data will be of String format and of variable lengths, as it may include name, address, mail id's etc.
So i need to mask these data before logging them, it should not be a full mask instead, if the length of string is less than or equal to 8 characters then mask the first half with "XXX etc.."
If the length is more than 8 then mask the first and last portion of the string such that only the mid 5 characters are visible.
I know we can do this using java sub-stringa nd iterating over the string, but want to know if there is any other simple solution to address this.
Thanks in advance

If you are using Apache Commons, you can do like
String maskChar = "*";
//number of characters to be masked
String maskString = StringUtils.repeat( maskChar, 4);
//string to be masked
String str = "FirstName";
//this will mask first 4 characters of the string
System.out.println( StringUtils.overlay(str, maskString, 0, 4) );
You can check the string length before generating maskString using if else statement.

You can use this function; change the logic of half's as per your needs:
public static String maskedVariableString(String original)
{
String maskedString = null;
if(original.length()<9)
{
int half = original.length()/2;
StringBuilder sb =new StringBuilder("");
for(int i=0;i<(original.length()-half);i++)
{
sb.append("X");
}
maskedString = original.replaceAll("\\b.*(\\d{"+half+"})", sb.toString()+"$1");
}
else
{
int maskLength = original.length()-5;
int firstMaskLength = maskLength/2;
int secondMaskLength = maskLength-firstMaskLength;
StringBuilder sb =new StringBuilder("");
for(int i=0;i<firstMaskLength;i++)
{
sb.append("X");
}
String firstMask = sb.toString();
StringBuilder sb1 =new StringBuilder("");
for(int i=0;i<secondMaskLength;i++)
{
sb1.append("X");
}
String secondMask = sb1.toString();
maskedString = original.replaceAll("\\b(\\d{"+firstMaskLength+"})(\\d{5})(\\d{"+secondMaskLength+"})", firstMask+"$2"+secondMask);
}
return maskedString;
}
Explanation:
() groups the regular expression and we can use $ to access this group($1, $2,$3).
The \b boundary helps check that we are the start of the digits (there are other ways to do this, but here this will do).
(\d{+half+}) captures (half) no of digits to Group 1. The same happens in the else part also.

Java: Get Float and String value from a String

I have a string, such as "4.25GB"
I'd like to get the floating part "4.25"
And get the string part "GB"
How to get the two values respectively in Java.
Thanks.

Try
String s = "4.25GB"
Float value = Float.valueOf(s.replaceAll("[^\\d.]", "")); // remove all non-numeric symbols
String f = s.replaceAll("[0-9]",""); // remove all numbers

To get Number Part: String numberPart = "4.25GB".replaceAll("[^0-9.]", "");
To get String part: String stringPart = "4.25GB".replaceAll("[^A-Za-z]", "");

Use String.replaceAll to first replace all non-digits and dot with "" to get the number then otherwise

You can write a function that will be similar to C# int.TryParse method, and use it in loop on your string, it will only work if you alwayes have a (NUM)(STRING) formation :
boolean tryParse(String value)
{
try
{
Integer.parseInt(value);
return true;
} catch(NumberFormatException e)
{
return false;
}
}

Use split/ substring concept. divide the string like below:
String Str = new String("4.25GB");
for (String retval: Str.split("G")){
System.out.println(retval);
}
//or u can use
String[] r = s.split("(?=\\p{Upper})");

You could use public String substring(int beginIndex, int endIndex)
String start = "4.25GB";
String numbers = start.substring(0,4);
String letters = start.substring(4,6);
Read more about substrings and how to use them here

Tested, works:
String str = "4.25GB" ;
String parts[] = str.split("(?i)(?<=\\d)(?=[a-z])|(?<=[a-z])(?=\\d)");
float number = Float.parseFloat(parts[0]) ;
String string = parts[1] ;
System.out.println(number); //4.25
System.out.println(string); //GB

You can use regular expression like this :
String s = "4.25GB";
String num = s.replaceAll("[^0-9.]", "");
System.out.println(num);
String str = s.replaceAll("[0-9.]", "");
System.out.println(str);
wish help you.

That depends on what "such as" means. Are all the strings in the format "x.xxGB"? If that's the case, then you can use substring(), as you know the exact number of 'float' chars and 'suffix' chars.
String theStr = "x.xxGB";
String numStr = theStr.substring(0, 4); // grab first 4 chars: "x.xx"
float numFloat = Float.parseFloat(numStr);
String suffix = theStr.substring(5); // or .substring(5, 7); if you know the exact length
If it's more variable than that, it gets more complicated. If you don't know the length of the leading number string, you'd have to check the first part as a valid float, with perhaps the easiest way to be gathering characters as the start and checking each succession as a valid float, with all the rest being considered a suffix. Maybe something like this (pseudocode-ish):
String theStr = "324.994SUFFIX"; // SomeArbitraryNumberAndSuffixString
String currNumStr = "";
Boolean bHaveFloat = true;
for (int i = 1; i < theStr.length(); i++){
String testStr = theStr.substring(0, i);
try{
float f = Float.parseFloat(testStr);
} catch (NumberFormatException nfe){
// handle the exception, printStackTrace, etc...
// failed so No longer have Valid String...
break;
}
currNumStr = testStr;
}
// currNumStr now has the valid numberString

Java codingbat help - withoutString

I'm using codingbat.com to get some java practice in. One of the String problems, 'withoutString' is as follows:
Given two strings, base and remove, return a version of the base string where all instances of the remove string have been removed (not case sensitive).
You may assume that the remove string is length 1 or more. Remove only non-overlapping instances, so with "xxx" removing "xx" leaves "x".
This problem can be found at: http://codingbat.com/prob/p192570
As you can see from the the dropbox-linked screenshot below, all of the runs pass except for three and a final one called "other tests." The thing is, even though they are marked as incorrect, my output matches exactly the expected output for the correct answer.
Here's a screenshot of my output:
And here's the code I'm using:
public String withoutString(String base, String remove) {
String result = "";
int i = 0;
for(; i < base.length()-remove.length();){
if(!(base.substring(i,i+remove.length()).equalsIgnoreCase(remove))){
result = result + base.substring(i,i+1);
i++;
}
else{
i = i + remove.length();
}
if(result.startsWith(" ")) result = result.substring(1);
if(result.endsWith(" ") && base.substring(i,i+1).equals(" ")) result = result.substring(0,result.length()-1);
}
if(base.length()-i <= remove.length() && !(base.substring(i).equalsIgnoreCase(remove))){
result = result + base.substring(i);
}
return result;
}

Your solution IS failing AND there is a display bug in coding bat.
The correct output should be:
withoutString("This is a FISH", "IS") -> "Th a FH"
Yours is:
withoutString("This is a FISH", "IS") -> "Th a FH"
Yours fails because it is removing spaces, but also, coding bat does not display the correct expected and run output string due to HTML removing extra spaces.
This recursive solution passes all tests:
public String withoutString(String base, String remove) {
int remIdx = base.toLowerCase().indexOf(remove.toLowerCase());
if (remIdx == -1)
return base;
return base.substring(0, remIdx ) +
withoutString(base.substring(remIdx + remove.length()) , remove);
}
Here is an example of an optimal iterative solution. It has more code than the recursive solution but is faster since far fewer function calls are made.
public String withoutString(String base, String remove) {
int remIdx = 0;
int remLen = remove.length();
remove = remove.toLowerCase();
while (true) {
remIdx = base.toLowerCase().indexOf(remove);
if (remIdx == -1)
break;
base = base.substring(0, remIdx) + base.substring(remIdx + remLen);
}
return base;
}

I just ran your code in an IDE. It compiles correctly and matches all tests shown on codingbat. There must be some bug with codingbat's test cases.
If you are curious, this problem can be solved with a single line of code:
public String withoutString(String base, String remove) {
return base.replaceAll("(?i)" + remove, ""); //String#replaceAll(String, String) with case insensitive regex.
}
Regex explaination:
The first argument taken by String#replaceAll(String, String) is what is known as a Regular Expression or "regex" for short.
Regex is a powerful tool to perform pattern matching within Strings. In this case, the regular expression being used is (assuming that remove is equal to IS):
(?i)IS
This particular expression has two parts: (?i) and IS.
IS matches the string "IS" exactly, nothing more, nothing less.
(?i) is simply a flag to tell the regex engine to ignore case.
With (?i)IS, all of: IS, Is, iS and is will be matched.
As an addition, this is (almost) equivalent to the regular expressions: (IS|Is|iS|is), (I|i)(S|s) and [Ii][Ss].
EDIT
Turns out that your output is not correct and is failing as expected. See: dansalmo's answer.

public String withoutString(String base, String remove) {
String temp = base.replaceAll(remove, "");
String temp2 = temp.replaceAll(remove.toLowerCase(), "");
return temp2.replaceAll(remove.toUpperCase(), "");
}

Please find below my solution
public String withoutString(String base, String remove) {
final int rLen=remove.length();
final int bLen=base.length();
String op="";
for(int i = 0; i < bLen;)
{
if(!(i + rLen > bLen) && base.substring(i, i + rLen).equalsIgnoreCase(remove))
{
i +=rLen;
continue;
}
op += base.substring(i, i + 1);
i++;
}
return op;
}
Something things go really weird on codingBat this is just one of them.

I am adding to a previous solution, but using a StringBuilder for better practice. Most credit goes to Anirudh.
public String withoutString(String base, String remove) {
//create a constant integer the size of remove.length();
final int rLen=remove.length();
//create a constant integer the size of base.length();
final int bLen=base.length();
//Create an empty string;
StringBuilder op = new StringBuilder();
//Create the for loop.
for(int i = 0; i < bLen;)
{
//if the remove string lenght we are looking for is not less than the base length
// and the base substring equals the remove string.
if(!(i + rLen > bLen) && base.substring(i, i + rLen).equalsIgnoreCase(remove))
{
//Increment by the remove length, and skip adding it to the string.
i +=rLen;
continue;
}
//else, we add the character at i to the string builder.
op.append(base.charAt(i));
//and increment by one.
i++;
}
//We return the string.
return op.toString();
}

Taylor's solution is the most efficient one, however I have another solution that is a naive one and it works.
public String withoutString(String base, String remove) {
String returnString = base;
while(returnString.toLowerCase().indexOf(remove.toLowerCase())!=-1){
int start = returnString.toLowerCase().indexOf(remove.toLowerCase());
int end = remove.length();
returnString = returnString.substring(0, start) + returnString.substring(start+end);
}
return returnString;
}

#Daemon
your code works. Thanks for the regex explanation. Though dansalmo pointed out that codingbat is displaying the intended output incorrectly, I through in some extra lines to your code to unnecessarily account for the double spaces with the following:
public String withoutString(String base, String remove){
String result = base.replaceAll("(?i)" + remove, "");
for(int i = 0; i < result.length()-1;){
if(result.substring(i,i+2).equals(" ")){
result = result.replace(result.substring(i,i+2), " ");
}
else i++;
}
if(result.startsWith(" ")) result = result.substring(1);
return result;
}

public String withoutString(String base, String remove){
return base.replace(remove,"");
}

Short string format

I've an array with hundreds of string values, Is it possible to use specific formats to make them shorter?
e.g
Normal -> "Hello John, What's up?!"
With short format -> "Hello John..."
After using substring, I got errors.
private String[] finalString;
for (int i = 0; i < arrays.PodcastTitle.length; i++) {
finalString[i] = arrays.PodcastTitle[i].substring(0, 5);
}

Since you haven't given any details, this is the shortest approach :
String little = normalString.substring(0, 10); // use anything 5 or 10 or 15, depending on how short you want to make your String
From your edit:
Please change:
private String[] finalString;
to:
private String[] finalString = new String[whateverSizeYouWant];

String toLitteString(String str, int length) {
if (str.length() > length) return str.substring(0, length) + "...";
return str;
}
Function that will truncate longer strings to length (and add a "...") or return the short string. If you want the length to include the "..." then just change length to length - 3.

Why won't you consider implementing a method that takes a String argument, explodes it by space character and returns the String with required number of words in it?