Your solution is definately simple and easy to read. However, technically, it's a lot slower because sorts are slow. Let's say your anagrams were 5000 letters long. This algorithm below will do it in O(2*n) time.

var isAnagram = function(s, t) { const hash = {}; if( s.length != t.length ) return false; // convert first anagram into hashmap with letter counts s.split('').reduce( (acc, val) => { if( val in acc ) acc[val]++; else acc[val] = 1; return acc; }, hash); // iterate over second anagram, subtracting letter counts for( let i = 0; i < t.length; i++ ) { if( t[i] in hash ) { if( --hash[t[i]] < 0 ) return false; } else { // letter doesn't exist, not an anagram return false; } } return true; };