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. (No need to handle fancy punctuation.)
- 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"]
```
```if:java
For java users, the calls will actually be in the form: `TopWords.top3(String s)`, expecting you to return a `List`.
```
Bonus points (not really, but just for fun):
------------
1. Avoid creating an array whose memory footprint is roughly as big as the input text.
2. Avoid sorting the entire array of unique words.
from string import punctuation

def top_3_words(text):
    for p in punctuation: 
        if p != "'": text = text.replace(p, " ")
    count = {}
    for w in text.lower().split():
        if w.strip(punctuation) == '':
            pass
        elif w in count.keys():
            count[w] += 1
        else:
            count[w] = 1
    lcount = [(w,c) for w,c in count.items()]
    lcount.sort(key=lambda x: x[1], reverse=True)
    return [x[0] for x in lcount[:3]]
from string import punctuation

def top_3_words(text):
    for p in punctuation: 
        if p != "'": text = text.replace(p, " ")
    count = {}
    for w in text.lower().split():
        if w.strip(punctuation) == '':
            pass
        elif w in count.keys():
            count[w] += 1
        else:
            count[w] = 1
    lcount = [(w,c) for w,c in count.items()]
    lcount.sort(key=lambda x: x[1], reverse=True)
    return [x[0] for x in lcount[:3]]