Distant Barcodes LeetCode solution in JavaScript

1
var rearrangeBarcodes = function (barcodes) {
2
  var result = [];
3
  var map = new Map();
4
  barcodes.forEach((n) => map.set(n, map.get(n) + 1 || 1));
5
  let list = [...map.entries()].sort((a, b) => {
6
    return b[1] - a[1];
7
  });
8
  let i = 0; //list[i][0]=>number list[i][1]=>count of this number
9
  while (result.length !== barcodes.length) {
10
    if (list[i][1] > 0) result.push(list[i][0]), list[i][1]--;
11
    i++;
12
    if (list[i] === undefined) i = 0;
13
    if (
14
      list[0][1] - list[i][1] >= 1 &&
15
      result[result.length - 1] !== list[0][0]
16
    )
17
      i = 0;
18
  } //list has sorted, so list[0] appeared most frequent
19
  return result;
20
};