introduction.md

Introduction

There are at least a couple of idiomatic approaches to solving Isogram. One is to use filter() and map() with allMatch(). Another is to use filter() and mapToObj() with distinct().

General guidance

The key to solving Isogram is to determine if any of the letters in the String being checked are repeated. A repeated letter means the String is not an Isogram. The occurrence of the letter a and the letter A count as a repeated letter, so Alpha would not be an isogram.

Approach: filter() and map() with allMatch()

import java.util.HashSet;

final class IsogramChecker {

    boolean isIsogram(final String phrase) {
        return phrase.chars()
            .filter(Character::isLetter)
            .map(Character::toLowerCase)
            .allMatch(new HashSet < > ()::add);
    }
}

For more information, check the filter() and map() with allMatch() approach.

Approach: filter() and mapToObj() with distinct()

import java.util.stream.Collectors;

public class IsogramChecker {

    public boolean isIsogram(String input) {
        final var scrubbed = input.chars()
            .filter(Character::isLetter)
            .mapToObj(Character::toLowerCase)
            .collect(Collectors.toList());

        return scrubbed.size() == scrubbed.stream().distinct().count();
    }
}

For more information, check the filter() and mapToObj() with distinct() approach.

Which approach to use?

Since benchmarking with the Java Microbenchmark Harness is currently outside the scope of this document, the choice between the various approaches can be made by perceived readability.

Due to the short-circuiting of the allMatch() method, it is probably more performant than the distinct() method, unless constructing the HashSet takes more time than constructing a short list and a short stream.