Simple Isogram Check - Could it be as simple as this?

[SibochCZ]

Hello,

I’m wondering if the check could be as simple as this one I’ve written?

bool FBullCowGame::IsWordIsogram(FString word) const{
	for (int i = 0; i < word.length(); i++) {
		for (int j = i + 1; j < word.length(); j++) {
			if (word[i] == word[j]) {
				return false;
			}
		}
	}
	return true;
}

Cheers,
Robert

[SibochCZ]

Here, you can check it’s function:

https://codepen.io/B3hindL0gic/pen/yrGrez

[DanM]

Yes, see here

Using Regex for Isogram checl

@ben To work out if the the word provided is an isogram I use a regular expression. const std::regex isogram("^(?:([a-z])(?!.*\\1))*$", std::regex_constants::ECMAScript | std::regex_constants::icase); ... else if(!std::regex_match(guess, isogram)) return GuessValidity::NOT_ISOGRAM; Are there any disadvantages to doing it this way instead of the way you showed in lecture 47?