Java regular expression word match

Question

I have 3 values IU, PRI and RET. if my input string contains any one or more value(s),
the Java regular expression should return true.

Answer 1

Ok here's a crazy solution with anagrams of each given String, built into a Pattern just for fun:

public static void main(String[] args) {
    try {
        Pattern pattern = makePattern("IU", "PRI", "RET");
        System.out.println(pattern.pattern());
        String test = "put returns UI between paragraphs, also IRP and TER";
        Matcher matcher = pattern.matcher(test);
        while (matcher.find()) {
            System.out.println(matcher.group());
        }
    }
    catch (Exception e) {
        e.printStackTrace();
    }
}
public static Pattern makePattern(String... words) throws Exception {
    if (words == null || words.length == 0) {
        throw new Exception("TODO handle invalid argument");
    }
    StringBuilder patternBuilder = new StringBuilder("(");
    for (String word : words) {
        if (word == null || word.isEmpty()) {
            throw new Exception("TODO invalid word");
        }
        for (String anagram: doAnagrams(word, null)) {
            patternBuilder.append("\\b").append(anagram).append("\\b").append("|");
        }
    }
    patternBuilder.deleteCharAt(patternBuilder.length() - 1);
    patternBuilder.append(")");
    return Pattern.compile(patternBuilder.toString());
}
public static Set doAnagrams(String original, Set processed) {
    if (original == null || original.isEmpty()) {
        return new LinkedHashSet();
    }
    Set result;
    if (processed == null) {
        result = new LinkedHashSet();
        result.add(original);
    } else {
        result = processed;
    }
    if (original.length() <= 1) {
        return result;
    }
    String sub = original.substring(1);
    String subStart = original.substring(0, 1);
    for (String subAnagram : doAnagrams(sub, null)) {
        result.add(subAnagram.concat(subStart));
    }
    if (sub.concat(original.substring(0, 1)).equals(result.iterator().next())) {
        return result;
    } 
    else {
        return doAnagrams(sub.concat(subStart), result);
    }
}

Output:

(\bIU\b|\bUI\b|\bPRI\b|\bRIP\b|\bIRP\b|\bIPR\b|\bPIR\b|\bRPI\b|\bRET\b|\bETR\b|\bTER\b|\bTRE\b|\bRTE\b|\bERT\b)
UI
IRP
TER