17. Letter Combinations of a Phone Number - Leetcode

solution.py

# https://leetcode.com/problems/letter-combinations-of-a-phone-number/#/description
class Solution(object):
    DIGIT_DICT = {
        "2" : ['a', 'b', 'c'],
        "3" : ['d', 'e', 'f'],
        "4" : ['g', 'h', 'i'],
        "5" : ['j', 'k', 'l'],
        "6" : ['m', 'n', 'o'],
        "7" : ['p', 'q', 'r', 's'],
        "8" : ['t', 'u', 'v'],
        "9" : ['w', 'x', 'y', 'z']
    }
    
    def letterCombinations(self, digits):
        """
        :type digits: str
        :rtype: List[str]
        """
        pointer = 0
        letters = ""
        if len(digits) == 0:
            return []
        else:
            return (self.create_strings(letters, digits, pointer))
    
    def create_strings(self, letters, digits, pointer):
        if pointer == len(digits):
            return [letters]
        next_chars = self.DIGIT_DICT[digits[pointer]]
        new_strings = []
        for x in next_chars:
            for s in self.create_strings(letters + x, digits, pointer + 1):
                new_strings.append(s)
        return new_strings
        
    
    

return (self.create_strings(letters, digits, pointer))

Try out keyword args in python.
e.g.

return (self.create_strings(letters="", digits=digits, pointer=0))

you also don't have to declare the 1 time use letters + pointer variables.

class Solution(object):
    DIGIT_DICT = {
        "2" : ['a', 'b', 'c'],
        "3" : ['d', 'e', 'f'],
        "4" : ['g', 'h', 'i'],
        "5" : ['j', 'k', 'l'],
        "6" : ['m', 'n', 'o'],
        "7" : ['p', 'q', 'r', 's'],
        "8" : ['t', 'u', 'v'],
        "9" : ['w', 'x', 'y', 'z']
    }
    
    def letterCombinations(self, digits, letters="", pointer=0):
        if len(digits) == 0:
            return []
        if pointer == len(digits):
            return [letters]
        next_chars = self.DIGIT_DICT[digits[pointer]]
        new_strings = []
        for x in next_chars:
            for s in self.letterCombinations(letters=letters + x, digits=digits, pointer=pointer + 1):
                new_strings.append(s)
        return new_strings

If leetcode accepted [""] as an answer for empty (they want [] instead of [""], then you can condense it to:

class Solution(object):
    DIGIT_DICT = {
        "2" : ['a', 'b', 'c'],
        "3" : ['d', 'e', 'f'],
        "4" : ['g', 'h', 'i'],
        "5" : ['j', 'k', 'l'],
        "6" : ['m', 'n', 'o'],
        "7" : ['p', 'q', 'r', 's'],
        "8" : ['t', 'u', 'v'],
        "9" : ['w', 'x', 'y', 'z']
    }
    
    def letterCombinations(self, digits, letters="", pointer=0):
        if pointer == len(digits):
            return [letters]
        next_chars = self.DIGIT_DICT[digits[pointer]]
        new_strings = []
        for x in next_chars:
            for s in self.letterCombinations(letters=letters + x, digits=digits, pointer=pointer + 1):
                new_strings.append(s)
        return new_strings