solution.py

# Write a function that, given a string of text (possibly with punctuation and line-breaks),
# returns an array of the top-3 most occurring words, in descending order of the number of occurrences.

# Assumptions:
# A word is a string of letters (A to Z) optionally containing one or more apostrophes (') in ASCII.
# Apostrophes can appear at the start, middle or end of a word ('abc, abc', 'abc', ab'c are all valid)
# Any other characters (e.g. #, \, / , . ...) are not part of a word and should be treated as whitespace.
# Matches should be case-insensitive, and the words in the result should be lowercased.
# Ties may be broken arbitrarily.
# If a text contains fewer than three unique words, then either the top-2 or top-1 words should be returned,
# or an empty array if a text contains no words.
# Examples:
# top_3_words("In a village of La Mancha, the name of which I have no desire to call to
# mind, there lived not long since one of those gentlemen that keep a lance
# in the lance-rack, an old buckler, a lean hack, and a greyhound for
# coursing. An olla of rather more beef than mutton, a salad on most
# nights, scraps on Saturdays, lentils on Fridays, and a pigeon or so extra
# on Sundays, made away with three-quarters of his income.")
# => ["a", "of", "on"]

# top_3_words("e e e e DDD ddd DdD: ddd ddd aa aA Aa, bb cc cC e e e")
# => ["e", "ddd", "aa"]

# top_3_words("  //wont won't won't")
# => ["won't", "wont"]
# Bonus points (not really, but just for fun):
# Avoid creating an array whose memory footprint is roughly as big as the input text.
# Avoid sorting the entire array of unique words.

import re
from collections import Counter


def top_3_words(text):
    chars_or_words = re.findall(r"[a-z']*[a-z]+[a-z']*", text.lower())
    return [x[0] for x in Counter(chars_or_words).most_common(3)]