Solution for ACM SER 2016 Division 2 "Simplicity"

simplicity.py

# A class for holding the unique letters from a string, and their number of occurences
class LetterSet:

    # constructor, adds the words to a dictionary with letters as keys and letter counts as values
    # ala: { 'a': 4, 'b' : 2, ... } etc.
    def __init__(self, word):
        self.letter_counts = {}

        for letter in word:
            self.add_letter(letter)

    # adds a single letter to the letter set. If the letter is already there, the count is incremented
    # otherwise, it's added to the set
    def add_letter(self, letter):
        # if letter is not in the set yet, add it
        if letter not in self.letter_counts:
            self.letter_counts[letter] = 0

        # count this occurrence of the letter
        self.letter_counts[letter] += 1

        # return the new value, just in case we want to use it
        return self.letter_counts[letter]

    # A quick "to string" function
    def __str__(self):
        return repr(self.letter_counts)

    # gets the distinct letters sorted ascendingly by frequency
    def letters_by_frequency(self):
        sorted_keys_with_vals = sorted(self.letter_counts.items(), key=lambda x : x[1]) # sort by dict vals (counts)
        return list(map(lambda x : x[0], sorted_keys_with_vals))

    # gets the frequency of a particular letter
    def frequency_of_letter(self, letter):
        if letter in self.letter_counts:
            return self.letter_counts[letter]
        else:
            return 0

# read the word from stdin, and convert to a letter set
word = raw_input().strip()
letter_set = LetterSet(word)

# Get a list of the distinct letters in the letter set
distinct_letters = letter_set.letters_by_frequency()

# assume that we don't have to delete any characters, i.e. assume there are 2 or less distinct characters in the word
deletion_count = 0

# if there are more than two distinct letters get all of the distinct letters but 2 (these are the least frequent
# letters), and count how many deletions it would take to delete those characters
if len(distinct_letters) > 2:
    letters_to_remove = distinct_letters[:-2]
    deletion_count = sum(map(lambda letter : letter_set.frequency_of_letter(letter), letters_to_remove))

print(deletion_count)