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Solution: Letter Combinations of a Phone Number

Fledgling software developer; the struggle is a Rational Approximation.
・4 min read

This is part of a series of Leetcode solution explanations (index). If you liked this solution or found it useful, please like this post and/or upvote my solution post on Leetcode's forums.

Leetcode Problem #17 (Medium): Letter Combinations of a Phone Number


(Jump to: Solution Idea || Code: JavaScript | Python | Java | C++)

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

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

Description Visual 1


Example 1:
Input: digits = "23"
Output: ["ad","ae","af","bd","be","bf","cd","ce","cf"]
Example 2:
Input: digits = ""
Output: []
Example 3:
Input: digits = "2"
Output: ["a","b","c"]


  • 0 <= digits.length <= 4
  • digits[i] is a digit in the range ['2', '9'].


(Jump to: Problem Description || Code: JavaScript | Python | Java | C++)

Since each digit can possibly mean one of several characters, we'll need to create code that branches down the different paths as we iterate through the input digit string (D).

This quite obviously calls for a depth-first search (DFS) approach as we will check each permutation of characters and store them in our answer array (ans). For a DFS approach we can use one of several options, but a recursive solution is generally the cleanest.

But first, we'll need to set up a lookup table (L) to convert a digit to its possible characters. Since the digits are actually low-indexed integers, we can actually choose between an array or map/dictionary here with little difference.

For our DFS function (dfs), we'll have to feed it the current position (pos) in D as well as the string (str) being built. The function will also need to have access to D, L, and ans.

The DFS function itself is fairly simple. It will push a completed str onto ans, otherwise it will look up the characters that match the current pos, and then fire off new recursive functions down each of those paths.

Once we're done, we should be ready to return ans.


Javascript and Python will have scoped access to D, L, and ans inside dfs, so won't need to pass in references via arguments.

Java should make sure to use a char[][] and a StringBuilder for better performance here.

Javascript Code:

(Jump to: Problem Description || Solution Idea)

const L = {'2':"abc",'3':"def",'4':"ghi",'5':"jkl",

var letterCombinations = function(D) {
    let len = D.length, ans = []
    if (!len) return []
    const bfs = (pos, str) => {
        if (pos === len) ans.push(str)
        else {
            let letters = L[D[pos]]
            for (let i = 0; i < letters.length; i++)
    return ans
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Python Code:

(Jump to: Problem Description || Solution Idea)

L = {'2':"abc",'3':"def",'4':"ghi",'5':"jkl",

class Solution:
    def letterCombinations(self, D: str) -> List[str]:
        lenD, ans = len(D), []
        if D == "": return []
        def bfs(pos: int, st: str):
            if pos == lenD: ans.append(st)
                letters = L[D[pos]]
                for letter in letters:
        return ans
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Java Code:

(Jump to: Problem Description || Solution Idea)

class Solution {
    final char[][] L = {{},{},{'a','b','c'},{'d','e','f'},{'g','h','i'},{'j','k','l'},

    public List<String> letterCombinations(String D) {
        int len = D.length();
        List<String> ans = new ArrayList<>();
        if (len == 0) return ans;
        bfs(0, len, new StringBuilder(), ans, D);
        return ans;

    public void bfs(int pos, int len, StringBuilder sb, List<String> ans, String D) {
        if (pos == len) ans.add(sb.toString());
        else {
            char[] letters = L[Character.getNumericValue(D.charAt(pos))];
            for (int i = 0; i < letters.length; i++)
                bfs(pos+1, len, new StringBuilder(sb).append(letters[i]), ans, D);
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C++ Code:

(Jump to: Problem Description || Solution Idea)

unordered_map<char, string> L({{'2',"abc"},{'3',"def"},{'4',"ghi"},

class Solution {
    vector<string> letterCombinations(string D) {
        int len = D.size();
        vector<string> ans;
        if (!len) return ans;
        bfs(0, len, "", ans, D);
        return ans;

    void bfs(int pos, int &len, string str, vector<string> &ans, string &D) {
        if (pos == len) ans.push_back(str);
        else {
            string letters = L[D[pos]];
            for (int i = 0; i < letters.size(); i++)
                bfs(pos+1, len, str+letters[i], ans, D);
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Discussion (1)

inhuofficial profile image
InHuOfficial • Edited

I have been enjoying this series for a while, always helps me with my imposter syndrome (and by help I mean makes me realise I am rubbish 🤣)

Could I make a tiny tiny suggestion. For the JavaScript examples throw them into an inline fiddle.

I always enjoy picking apart the solutions and the logic you implement but as I am a bit thick playing with the code rather than reading it is far more educational.

It isn't like it is a big task to copy into a fiddle for me, it would just be a nice touch.

Either way I look forward to the next instalment! ❤🦄

p.s. finally remembered to hit the bloody follow button, I always seem to forget on your posts as I am too busy crying into my spaghetti code! 😋