givecampus_2.py

"""
If a sticker is the collection of
the letters of "givecampusinc", return the
number of stickers needed to complete an input
string. "mice span" would require one sticker,
"give mice span" would require 2 because
the 2nd 'e' requires a new sticker. Treat it as case-insensitive.
"""

def how_many_sticks(sticker, letters_input):
    letters_input = letters_input.replace(' ', '')
    abcs = [chr(c) for c in range(ord('a'), ord('z') + 1)]
    sticker_letters_count = {letter: 0 for letter in abcs}  # {'a': 0, 'b': 0, ..., 'z': 0}
    input_letters_count = {letter: 0 for letter in abcs}
    for letter in sticker:
        sticker_letters_count[letter.lower()] += 1
    for letter in letters_input:
        input_letters_count[letter.lower()] += 1
    max_stickers = 1
    for letter in letters_input:
        try:
            max_stickers = max(max_stickers, division_round_up(input_letters_count[letter], sticker_letters_count[
                letter]))
        except ZeroDivisionError:
            return -1
    return max_stickers


def division_round_up(top, bottom):
    return top / bottom + (1 if top % bottom != 0 else 0)


print how_many_sticks('givecampusinc', 'mice span')  # 1
print how_many_sticks('givecampusinc', 'give mice span')  # 2

And I was remembered back from my python hacking days of the wonderful defaultdict that will let me skip that abcs initialization, you can set by default both sticker_letters_count and input_letters_count to return 0 for every letter:

from collections import defaultdict
sticker_letters_count = defaultdict(lambda: 0)
input_letters_count = defaultdict(lambda: 0)