Challenge: phone keypad to letters - return all possible strings

Question

The challenge:

Given a mapping of digits to letters (as in a phone number), and a digit string, return all possible letters the number could represent. You can assume each valid number in the mapping is a single digit.

Example:

If {"2": ["a", "b", "c"], 3: ["d", "e", "f"], …} then "23" should return ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

My iterative solution:

const digitMap = (digiStr) => { const digitMapping = { 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'] } const length = digiStr.length; if (length === 1) return digitMapping[digiStr]; // in case of single digit strings const indexes = []; // index addresses of charsets, in order for (let i = 0; i < length; i++) { indexes.push(0); } let nextI = length === 2 ? 0 : 1; // current spot of nextI in indexes const possibleStrs = []; // for final answer of possible permutations let str = ''; let endCount = 0; // if all addresses in indexes are at their end, then break while loop while (true) { console.log(indexes); for (let i = 0; i < length; i++) { str += digitMapping[digiStr[i]][indexes[i]]; if (indexes[i] === digitMapping[digiStr[i]].length - 1) { endCount += 1; } } possibleStrs.push(str); str = ''; if (endCount === length) { break; } endCount = 0; // increment last item in indexes as long as it < length of corresponding inner char array if (indexes[indexes.length - 1] < digitMapping[digiStr[digiStr.length - 1]].length - 1) { indexes[indexes.length - 1] += 1; } else { indexes[indexes.length - 1] = 0; // increment nextI as long as < length of corresponding inner array if (indexes[nextI] < digitMapping[digiStr[nextI]].length - 1) { indexes[nextI] += 1; } else { // otherwise shift position of nextI in indexes array // if next position of nextI not end of indexes array // shift nextI 1 spot forward in indexes if (nextI + 1 < length - 1) { nextI += 1; indexes[nextI] += 1; } else { // else reset all indexes to 0, starting from position 1 for (let i = 1; i < indexes.length; i++) { indexes[i] = 0; } // increment first item in indexes indexes[0] += 1; nextI = 1; // place nextI at position 1 } } } } return possibleStrs; } console.log(digitMap("2")); console.log(digitMap("23")); console.log(digitMap("234")); console.log(digitMap("2345")); console.log(digitMap("23456")); 

Produces the correct output for all test cases. Any alternative approaches--recursive or iterative--to cleaning up the conditional logic would be greatly appreciated.

Answer 1

Interesting question;

High Level Overview

The code is close to good, I definitely would not go for a recursive solution. These challenges often have at least 1 test with a really long string.

Naming

• Some of your naming choices are unfortunate:
  • possibleStrs -> I would go for possibleStrings, or combinations, or out
  • I prefer <verb><thing> over <thing><verb>; so mapDigits over digitMap or even generatePossiblePhoneWordCombinations

Comments

• I like your approach to commenting

JSHint.com

• Your code is only missing 2 semicolons, consider using http://jshint.com/ to perfect your code

Production Code

• Remove all references to console.log()
• You are mixing variable declarations and logic, I would group the variable declarations up front

Alternatives

• The below could use a built-in function

  const indexes = []; // index addresses of charsets, in order for (let i = 0; i < length; i++) { indexes.push(0); } 

  could be

  const indexes = Array(length).fill(0); 

Counter Proposal

The logic you are using afterwards is a little convoluted, and is hard to read. In essence, you can convert the input string to an array, and use that array to keep track of your state.

// "23" should return ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]. function mapDigits(s){ const digitMapping = { 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'] } //Get the digits, ignore ones and zeroes let digits = s.split('').filter(i => i > 1); let out = [], tmp = []; //Some shortcuts if(!digits.length){ return out; } if(digits.length == 1){ return digitMapping[digits[0]]; } //We're still here, prep out and digits (shift modifies digits) out = digitMapping[digits.shift()]; while(digits.length){ const nextLetters = digitMapping[digits.shift()]; tmp = out; out = []; tmp.forEach(s => nextLetters.forEach(c => out.push(s+c))); } return out; } console.log(mapDigits("23")); 

Answer 2

Too long for a comment...

The problem as presented seems to have much in common with the string permutations problem in terms of solution structure.

You should be able to do something similar to: https://stackoverflow.com/questions/4240080/generating-all-permutations-of-a-given-string with slight modification to get a compact recursive formulation.

Pseudo code below:

solution(string digits, string prefix="", array<string> outputs) if digits empty then add prefix to outputs else currentDigit := digits.get(0) remainingDigits := digits.remove(0); for each character in listOfCharsForDigit[currentDigit] do solution(remainingDigits, prefix+character, outputs) endfor endif 

Hope that helps...