Set vs filter to sort uniques

Benchmark created on

Description

are sets a good way to filter out unique values?

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Testing in
TestOps/sec
filter naive
const longest = (s1, s2) => Array.from(s1+s2).sort().filter((x,i,a) => a.indexOf(x) === i).join('')

let a ='asbavdasfsafa', b='daghjkasfashfjasfsvnx8ewrw';

for(let i = 0; i < 1000; i++) longest(a,b);
ready
sort set
const longest = (s1, s2) => [...new Set(s1.concat(s2).split(''))].sort((a,b) => a-b)

let a ='asbavdasfsafa', b='daghjkasfashfjasfsvnx8ewrw';

for(let i = 0; i < 1000; i++) longest(a,b);
ready
filter hashmap
const longest = (s1, s2) => {
  const arr = Array.from(s1+s2).sort();
  const map = arr.reduce((a,c,i) => { a[c] ||= i;return a } , {});
  return arr.filter((x,i) => map[x] === i).join('')
}
let a ='asbavdasfsafa', b='daghjkasfashfjasfsvnx8ewrw';

for(let i = 0; i < 1000; i++) longest(a,b);
ready
filter Object.values
const longest = (s1, s2) => {
  const arr = Array.from(s1+s2).sort();
  const map = arr.reduce((a,c,i) => { a[c] ||= i;return a } , {});
  return Object.keys(map).join('')
}
let a ='asbavdasfsafa', b='daghjkasfashfjasfsvnx8ewrw';

for(let i = 0; i < 1000; i++) longest(a,b);
ready

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