count_long_words(words, min_length) that counts how many strings in list words, after stripping surrounding whitespace, have length >= min_length. Implement the counting with a list comprehension that uses the walrus operator := to both compute the stripped word and test its length in the same expression, avoiding calling .strip() twice per word.count_long_words([" hi ", "hello", " world "], 5)
2
"hello" and "world" (after stripping) both have length 5.
count_long_words([], 3)
0
No words at all.
Inside the comprehension condition: `if (stripped := w.strip()) and len(stripped) >= min_length` — but here stripped is never falsy-empty relevant, so simply `if len((s := w.strip())) >= min_length`.
The walrus lets you bind s = w.strip() right inside the filter condition, avoiding a second .strip() call when building the kept list.