Generate Letter Combinations of a Phone Number

Question

The task: Given a string containing digits from 2-9 inclusive, return all possible letter combinations that the number could represent.

and my solution:

import itertools
letters_stack = list('abcdefghijklmnopqrstuvwxyz')
keypad_dict = {}
for num in range(2, 10):
    size = 3
    if num in [7, 9]:
        size = 4
    keypad_dict[str(num)] = letters_stack[:size]
    letters_stack = letters_stack[size:]

def aux(numbers):
    if len(numbers) == 1:
        return keypad_dict[numbers[0]]
    return itertools.product(keypad_dict[numbers[0]],  iletters(numbers[1:]))    

def iletters(numbers):
    assert len(numbers) > 0
    return [''.join(x) for x in aux(numbers)]



print list(iletters('234'))
# ['adg', 'adh', 'adi', 'aeg', 'aeh', 'aei', 'afg', 'afh', 'afi', 'bdg', 'bdh', 'bdi', 'beg', 'beh', 'bei', 'bfg', 'bfh', 'bfi', 'cdg', 'cdh', 'cdi', 'ceg', 'ceh', 'cei', 'cfg', 'cfh', 'cfi']

Answer 1

You can set the keypad lookup as a constant instead of computing it dynamically. Setting it up as constant makes it clearer that what your intent is.

KEYPAD = {
    '2': 'abc',
    '3': 'def',
    '4': 'ghi',
    '5': 'jkl',
    '6': 'mno',
    '7': 'pqrs',
    '8': 'tuv',
    '9': 'wxyz'}

---

itertools.product can accept multiple lists. And in python, you can use * to unpack an iterable into positional arguments, and ** to unpack as keyword arguments.

To sum up:

def t9_mode(numbers):
    if len(numbers) == 1:
        return KEYPAD[numbers[0]]
    maps = map(KEYPAD.get, numbers)
    return [''.join(_) for _ in itertools.product(*maps)]

does the job.

Answer 2

Python allows indexing of strings, for instance 'abcdefghijklmnopqrstuvwxyz'[3] returns 'd', so you don't need to turn your string into a list. I would just hardcode the keys: keypad = {1:'abc',2:'def' ...]. You spent more code generating keypad_dict than it would take to just declare it.

Having aux and iletters be programs that recursively call each other is making things way more complicated than they need to be. You can just do [''.join(_) for _ in itertools.product(keypad[__] for __ in numbers)]