Problem:
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.
Example:
Input: "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
Note:
Although the above answer is in lexicographical order, your answer could be in any order you want.
Solution:
ONE
JavaScript specific optimization.
Array.prototype.push accepts arbitrary arguments which enables tighter loops.
Also, appending string is faster than prepending.
/**
* @param {string} digits
* @return {string[]}
*/
let letterCombinations = function (digits) {
if (digits.length <= 0) {
return [];
}
const letters = [
,
,
["a", "b", "c"],
["d", "e", "f"],
["g", "h", "i"],
["j", "k", "l"],
["m", "n", "o"],
["p", "q", "r", "s"],
["t", "u", "v"],
["w", "x", "y", "z"],
];
let result = [""];
for (let i = 0; i < digits.length; i++) {
const arr = letters[digits[i]];
let newResult = [];
arr.forEach((c) => newResult.push(...result.map((r) => r + c)));
result = newResult;
}
return result;
};
TWO
General recursive DFS solution.
/**
* @param {string} digits
* @return {string[]}
*/
let letterCombinations = function (digits) {
const letters = [
,
,
"abc",
"def",
"ghi",
"jkl",
"mno",
"pqrs",
"tuv",
"wxyz",
];
const result = [];
if (digits.length > 0) {
dfs(digits, 0, "", letters, result);
}
return result;
};
function dfs(digits, idigit, path, letters, result) {
if (idigit >= digits.length) {
result.push(path);
return;
}
const str = letters[digits[idigit]];
for (let i = 0; i < str.length; i++) {
dfs(digits, idigit + 1, path + str[i], letters, result);
}
}
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