prize18-4  Exercise 2: Counting UTF-16 Units
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Idea: a code point at or below U+FFFF is one 16-bit unit; anything above needs a surrogate pair, two units. len() counts code points, so the two
numbers differ exactly when the string contains characters above U+FFFF. The family emoji is five code points: three people (each above U+FFFF, so 2 units)
plus two joiners (U+200D, below U+FFFF, so 1 unit each): 3 x 2 + 2 = 8.

ex2.py:

  def utf16_units(s):
      return sum(2 if ord(c) > 0xFFFF else 1 for c in s)

  samples = ["hello", "caf\u00e9", "\u20ac5", "\U0001F600", "a\U00010437b", "\U0001F468\u200D\U0001F469\u200D\U0001F467"]
  for s in samples:
      check = len(s.encode("utf-16-le")) // 2
      print(f"{s!r:12} code points={len(s)}  utf-16 units={utf16_units(s)}  agrees={utf16_units(s) == check}")

Real output (Python 3.14):

  'hello'      code points=5  utf-16 units=5  agrees=True
  'café'       code points=4  utf-16 units=4  agrees=True
  '€5'         code points=2  utf-16 units=2  agrees=True
  '😀'          code points=1  utf-16 units=2  agrees=True
  'a𐐷b'        code points=3  utf-16 units=4  agrees=True
  '👨\u200d👩\u200d👧' code points=5  utf-16 units=8  agrees=True

Why it matters: in a language that reports UTF-16 units as the string length, a single emoji has length 2 and this family has length 8, while a person
would say the family is one character. (Which languages do this is from general knowledge: check.)