Describe how your solution for the first task (finding unique elements while preserving the last occurrence) is built and its complexity.

To find unique elements while preserving the last occurrence, you can iterate through the list from end to start, keeping track of seen elements in a set. Add elements to the result only if they haven't been seen before. After the iteration, reverse the result to restore the original order with the last occurrences.

Example in Dart:

List<T> uniqueLast<T>(List<T> list) {
  final seen = <T>{};
  final result = <T>[];
  for (var i = list.length - 1; i >= 0; i--) {
    if (!seen.contains(list[i])) {
      seen.add(list[i]);
      result.add(list[i]);
    }
  }
  return result.reversed.toList();
}

The algorithm complexity is O(n), where n is the length of the list, since we traverse the list once and set operations are performed in amortized O(1) time.