Exercise 3 solution

    import re
    tags = "<p>one</p><p>two</p>"

    print(re.findall(r"<p>.*</p>", tags))      # ['<p>one</p><p>two</p>']
    print(re.findall(r"<p>.*?</p>", tags))     # ['<p>one</p>', '<p>two</p>']
    print(re.findall(r"<p>[^<]*</p>", tags))   # ['<p>one</p>', '<p>two</p>']

(All three lines were run and printed exactly the lists shown.)

Why: the greedy .* runs to the LAST </p> and then backs off just enough for
the pattern to finish, so both paragraphs are swallowed as one match. The
lazy .*? stops at the FIRST </p> that lets the pattern succeed. The negated
class [^<]* cannot cross a "<" at all, so it can never run past the end of
the paragraph's text. The negated class is usually the more reliable choice
because it states exactly what is allowed inside the tags.