# Question

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

A mapping of digit to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.

{
"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"]
}


Example:

Input: "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].


# Solution

class Solution:
def __init__(self):
self.lookup_table = {
"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]
"""
if len(digits) == 0:
return []

if len(digits) == 1 or digits in self.lookup_table:
return self.lookup_table[digits]

left, right = digits[:len(digits) // 2], digits[len(digits) // 2:]
comb_left = self.letterCombinations(left)
comb_right = self.letterCombinations(right)
res = []
if len(comb_left) == 0:
res = comb_right
elif len(comb_right) == 0:
res = comb_left
else:
for l in comb_left:
for r in comb_right:
res.append(l + r)

self.lookup_table[digits] = res
return res