Кодовые войны - Сортировать лишнее (7кю)

Задача

You have an array of numbers.
Your task is to sort ascending odd numbers but even numbers must be on their places.
Zero isn’t an odd number and you don’t need to move it. If you have an empty array, you need to return it.

Пример

sortArray([5, 3, 2, 8, 1, 4]) == [1, 3, 2, 8, 5, 4]

Мое решение

function sortArray(array) {
  const oddArr = [];
  const evenArr = [];
  const result = [];
for (let i = 0; i < array.length; i += 1) {
    if (array[i]%2 === 0) {
      evenArr.push(array[i]);
    } else {
      oddArr.push(array[i]);
    }
  }
  oddArr.sort((a, b) => a - b);
  for (let i = 0; i < array.length; i += 1) {
    if (array[i]%2 === 0) {
      result.push(evenArr.shift());
    } else {
      result.push(oddArr.shift());
    }
  }
  return result;
}

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function sortArray(array) {
   const odd = array.filter((x) => x % 2).sort((a,b) => a - b);
   return array.map((x) => x % 2 ? odd.shift() : x);
}