Exercise 8: Manual Word Count Without .split() — Possible Solution
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sentence = "The quick brown fox jumps over the lazy dog"

word_count = 0
previous_was_space = True   # treat the very start as "just after a space"

for ch in sentence:
    if ch.isspace():
        previous_was_space = True
    else:
        if previous_was_space:
            word_count += 1
        previous_was_space = False

print(f"Word count: {word_count}")

Output:
Word count: 9

WHY THIS WORKS AS AN ANSWER
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Without .split(), the only way to know a new word has started is to
notice the exact moment the loop crosses from "on a space" to "on a
non-space character" — a boundary, not a character type on its own.
previous_was_space is what remembers what kind of character came
immediately before the current one, updated on every single pass
regardless of which branch ran.

word_count only increments inside the else branch, and only when
previous_was_space was True going into that character — meaning
this non-space character is the very first one of a new word, not
the second, third, or later character of a word already being
counted. Starting previous_was_space as True (rather than False)
is what correctly counts the very first word in the sentence, which
has no space before it to trigger the transition naturally.