Discussion on: A classic interview question

[Spyros Argalias]

Good effort, but unfortunately this solution doesn't always work.
E.g. when str1 is aaaaa and str2 is abbbb, it returns true when it should be false.

It's because .includes checks that the letter is anywhere in the string, not caring about how many times. However the number of times the letter appears also needs to match the other string.

[Ibrahim]

aaaaa and abbbb is not an anagram in the first place, is it?

[Spyros Argalias]

That's right. That's what we're testing for: whether they are anagrams of each other. So in the str1 = aaaaa and str2 = abbbb case it should return false because as you say they're not anagrams of each other.

[Ibrahim]

function anagram(str1, str2) {
return (
str1.length == str2.length &&
str1.split("").every(c => str2.includes(c)) &&
str2.split("").every(c => str1.includes(c))
);
}

... and its still about 70% faster.

[Spyros Argalias]

Nice :)

[DenisPostu]

The new method returns true for "abb", "aab", I don't think it should.

[Ibrahim]

You're right, every just won't work. My method works for a few cases but not all.