How to compare arrays out of element order?

Question

JavaScript

JS language, let's say there are two arrays:
1) [1, 2, 3, 4]
2) [2, 4, 3, 1]

how can I compare them for identity not in the order of the elements?
if through the for of, it will be through the order, but how not?

Answer 1

function haveSameValues(arr1, arr2) {
  if (arr1.length !== arr2.length) {
    return false;
  }

  const count = new Map();
  arr1.forEach(n => count.set(n, (count.get(n) ?? 0) + 1));
  arr2.forEach(n => count.set(n, (count.get(n) ?? 0) - 1));

  return [...count.values()].every(n => n === 0);
}


haveSameValues(
  [ 'hello, world!!', 0, 0, 0, 1, 1, false, false ],
  [ false, false, 1, 1, 0, 0, 0, 'hello, world!!' ]
) // true

haveSameValues(
  [ 1, 2, 3 ],
  [ 3, 2, 2 ]
) // false

haveSameValues(
  [],
  []
) // true

Answer 2

Plan-A : sort before comparison

const areEqual = (arrA, arrB) => {
  if (arrA.length !== arrB.length) return false;
  const a = arrA.slice().sort(), b = arrB.slice().sort();
  return a.every((el, i) => el === b[i]);
}

Plan-B : hash table (value: number of elements with this value). Count in one array, in another, compare: must match. This solution was perfectly implemented in his answer by 0xD34F
Plan-X : remove matches one by one from copies. This option is ineffective. Brought for entertainment.

const areEqual = (arrA, arrB) => {
  if (arrA.length !== arrB.length) return false;
  const a = arrA.slice(), b = arrB.slice();
  while (a.length) {
    const i = b.indexOf(a.pop());
    if (-1 === i) return false;
    b.splice(i, 1);
  }
  return true;
}