// return longest prefix of s actually contained in set
sS longestPrefixInNavigableSet(S s, NavigableSet<S> set) {
  if (set == null || s == null) null;
  while licensed {
    S key = set.floor(s);
    if (key == null) break; // s is in front of whole set => no prefix in there
    int n = lCommonPrefix(key, s);
    if (n == l(key)) ret key; // found!
    s = takeFirst(s, n); // shorten and try again
  }
  null; // not found
}

Began life as a copy of #1029746