Front end developer specialising in JavaScript and React. Experienced in all aspects of modern front end development. Passionate about making accessible, secure and performant software.
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.
Front end developer specialising in JavaScript and React. Experienced in all aspects of modern front end development. Passionate about making accessible, secure and performant software.
aaaaa
andabbbb
is not an anagram in the first place, is it?That's right. That's what we're testing for: whether they are anagrams of each other. So in the
str1 = aaaaa
andstr2 = abbbb
case it should returnfalse
because as you say they're not anagrams of each other.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.
Nice :)
The new method returns
true
for"abb", "aab"
, I don't think it should.You're right,
every
just won't work. My method works for a few cases but not all.