Exercise 1: An OOP Translation of CountVowels — Possible Solution
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THE PSEUDOCODE (IMPLIED BY THE PROBLEM STATEMENT)
------------------------------
    ALGORITHM CountVowels(word)
        count <- 0
        FOR EACH letter IN word
            IF letter IS A VOWEL THEN
                count <- count + 1
            ENDIF
        ENDFOR
        RETURN count

OOP TRANSLATION, FOLLOWING THIS CHAPTER'S OWN EvenSquareSummer TEMPLATE
------------------------------
    class VowelCounter:
        def __init__(self, word):
            self.word = word

        def compute(self):
            vowels = set('aeiouAEIOU')
            count = 0
            for ch in self.word:
                if ch in vowels:
                    count += 1
            return count

VERIFYING THE COUNT FOR "algorithm"
------------------------------
Running VowelCounter("algorithm").compute() gives 3.

Confirmed by hand: a-l-g-o-r-i-t-h-m
  a -> vowel (count=1)
  l -> not a vowel
  g -> not a vowel
  o -> vowel (count=2)
  r -> not a vowel
  i -> vowel (count=3)
  t -> not a vowel
  h -> not a vowel
  m -> not a vowel

Final count: 3, matching the class's own computed result exactly.

WHY THIS FOLLOWS THIS CHAPTER'S OWN OOP PATTERN
------------------------------
Just like EvenSquareSummer, the class's constructor stores the input
data (the word) as an instance attribute, and a single compute()
method contains the exact same loop-and-condition logic the imperative
version would use - the OOP wrapper changes how the data and the
operation on it are packaged together, not the underlying algorithm
itself, exactly the distinction this chapter's own three-paradigm
comparison established.

WHY THIS WORKS AS AN ANSWER
------------------------------
The translation follows this chapter's own established OOP template
structure precisely (constructor storing input, a compute() method
holding the loop), and the resulting count is verified two independent
ways - by actually running the class's own method, and by a manual
letter-by-letter hand trace - confirming they agree.