Discussion on: A classic interview question

[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.