Understanding the Problem

Each dial of the telephone has 3-4 letters on them. It is to have all the possible combination of the dials. For instance, if the dial hits 2, which has letters “abc”, and 3, which has letters “def”, the combination will be [“ad”, “ae”, “af”, “bd”, “be”, “bf”, “cd”, “ce”, “cf”].

The length of the dials could be equal or less than 4.

Approach

It is a recursion problem. Simple as that, but did I know it from the beginning? No, I was again fooled by the multiple iterations and was trying to solve it with the snapshot idea that I had earlier. My snapshot idea would work though, but it will only give me a pain even though I get to solve it, like a victory for nothing.

First, I need a dial dictionary to transform the numbers into letters in the middle of the codes. Then I need to think about how the recursion should work. The idea of slicing and giving the digits[1:] value to the next self function would work, I thought.

Solution

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

class Solution:
    def letterCombinations(self, digits: str) -> List[str]:
        if len(digits) == 0: # exceptions
            return []
        elif len(digits) == 1:
            return [i for i in dials[digits]]

        combination = []
        for i in dials[digits[0]]:
            for j in self.letterCombinations(digits[1:]):
                combination.append(i+j)
        return combination

So, the final code was more simple than I thought. It is important to ascertain the returning codes before running the entire codes. These are the steps I followed to solve the problem.

1. Get dials dictionary

2. Write codes for the exceptions for when the length of the digits are either 0 or 1

3. If the length of the digits is longer than 1, the slice technique will be applied to cut it into smaller list. dials[digits[0]] should be the first for loop to be able to elucidate the operation of the combination with the next slices.

When it is time to go through for j in self.letterCombinations(digits[1:]):, the returned list from elif len(digits) == 1: will be used to have combination.

Reflection

Well, while I was writing the codes down, I found out that I had a elif condition which was useless in the codes but it ran just fine not affecting any runtime ms, but I took it out anyways to make it look much more simple.

It seems that the recursion problem practice somehow augmented my confidence in solving this kind of a problem, that I need to pick up the right spot with the right return codes to make it work like magic.