17. Letter Combinations of a Phone Number

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)
  }
};