Is JavaScript's array sorting stable?

The ECMAScript specification does not require JavaScript's built-in sorting algorithm to be stable. Instead, it's up to JavaScript engines to implement their own sorting algorithms, which may or may not be stable. This means that the order of equal elements is not guaranteed to be preserved after sorting. This is even more of a headache when dealing with more complex data across different environments.
Luckily, implementing a stable sort is pretty trivial. All you need to do is fall back to comparing the two elements' indexes when their values are equal. However, Array.prototype.sort() does not provide a way to access the indexes of the elements being compared.
To work around this, you can first use Array.prototype.map()to pair each element with its index. Then, using Array.prototype.sort(), you can apply the comparator function and, if the values are equal, compare the indexes. Finally, you can use Array.prototype.map() again to extract the original elements from the sorted array.
> [!IMPORTANT]
>

const stableSort = (arr, compare) =>
  arr
    .map((item, index) => ({ item, index }))
    .sort((a, b) => compare(a.item, b.item) || a.index - b.index)
    .map(({ item }) => item);

const arr = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
const stable = stableSort(arr, () => 0); // [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

To work around this, you can first use Array.prototype.map()to pair each element with its index. Then, using Array.prototype.sort(), you can apply the comparator function and, if the values are equal, compare the indexes. Finally, you can use Array.prototype.map() again to extract the original elements from the sorted array.
> [!IMPORTANT]
>
> This implementation doesn't sort in-place, but rather creates a new array. It's also significantly slower than the built-in sorting algorithm, due to the additional mapping steps.

From 30-seconds-of-code: array-stable-sort