Challenge 1: Word Frequency Counter — Possible Solution
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words = ["apple", "banana", "apple", "cherry", "banana", "apple"]

counts = {}

for word in words:
    counts[word] = counts.get(word, 0) + 1

print(counts)

Output:
{'apple': 3, 'banana': 2, 'cherry': 1}

WHY THIS WORKS AS AN ANSWER
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counts starts as an empty dictionary, then the for loop from Chapter 4
visits each word in the list one at a time.

counts.get(word, 0) reuses the chapter's own .get()-with-default
pattern: if word has already been seen and has a running count in
counts, that count is returned; if word has never been seen before,
0 is returned instead of raising a KeyError the way counts[word]
alone would for a first-time key.

Adding 1 to whatever .get() returns and assigning it straight back to
counts[word] handles both cases with the same single line — a brand
new word gets counts[word] = 0 + 1 = 1, and a repeated word gets its
existing count incremented by one. No separate "if word not in counts"
check is needed, because .get()'s default already covers the missing
case.