# Challenge: phone keypad to letters - return all possible strings

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.

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

• 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"));

• As a comment, the approach from @Zefick is very different and superior to our approaches. – konijn Jul 24 '19 at 17:33

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