Given two lists as input, rearrange the first list to match the order of the second list.

The solution is identical to question 1. Build a mapping from each element in the second list to its position, then sort the first list using that mapping as the key. Complexity remains O(n log n) for sorting and O(n) extra space for the map. Example in JavaScript:

function reorder(l1, l2) {
  const map = new Map(l2.map((v, i) => [v, i]));
  return l1.slice().sort((a, b) => map.get(a) - map.get(b));
}

If the lists contain duplicates, the mapping should store a queue of indices to preserve relative order.