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Roman Numeral Converter

pickleat profile image Andy Pickle Updated on ・1 min read

My coding education journey has been a bit, disjointed to say the least. I've stop started on freeCodeCamp, CS50, other tutorials, books, and just hanging out with smart people. I made it a goal to go back and get some of the certifications I bounced around on this year, and today solved the "Roman Numeral Converter" on freeCodeCamp. I'd love to know how others solved this. Because this was just the first idea I came up with and I'm sure there are many other ways to get it done!

What are the pros and cons of your solution?

Here's mine: RomanNumeral


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dmfay profile image
Dian Fay

The last code test I took for an interview involved writing a calculator in Java:

    public Abacus() {
        numerals = new LinkedHashMap<String, Integer>();

        numerals.put("M", 1000);
        numerals.put("D", 500);
        numerals.put("C", 100);
        numerals.put("L", 50);
        numerals.put("X", 10);
        numerals.put("V", 5);
        numerals.put("I", 1);
     * Find the appropriate divisor for a value, which, in the equation
     * (val - val / divisor) will resolve to the threshold at or above which
     * a subtractive maneuver is necessary to convert to Roman numerals.
    private int getDivisor(int value) {
        while (value > 10) {
            value = value / 10;

        return value;

    public String convert(int arabic) {
        StringBuilder sb = new StringBuilder();

        // preserve a pointer to the last subtractable numeral
        Map.Entry<String, Integer> lastEntry = null;    

        for (Map.Entry<String, Integer> e : numerals.entrySet()) {
            if (lastEntry != null) {
                if (arabic >= 4 * e.getValue()) {
                    // we have a valid subtraction!

                    arabic -= (lastEntry.getValue() - e.getValue());
                } else {
                    // e isn't small enough for us to have a valid subtraction
                    // yet, keep going

            // basic additive calculation: as long as the number is bigger than
            // e, keep appending e
            while (arabic >= e.getValue()) {

                arabic -= e.getValue();

            if (arabic >= e.getValue() - (e.getValue() / getDivisor(e.getValue()))) {
                // the remaining number is close enough to e that we will need
                // to perform a subtractive maneuver with reversed numerals
                lastEntry = e;

        return sb.toString();


  • the mapping only includes each Roman digit once
  • the algorithm is aesthetically pleasing (even if Java isn't the kindest to it)


  • the algorithm is extremely complicated and the simpler mapping is absolutely not worth it
  • although it satisfied the battery of unit tests I was given, one of the interviewers had run an extra on it that demonstrated a bug. We all had fun picking it apart, but I don't recall what the exact problem is.
pickleat profile image
Andy Pickle Author

Whew man, I don’t know Java at all, so this looks really complicated to me.

dmfay profile image
Dian Fay

The way it works is by subtracting until you can add based only on the one-to-one mapping in descending order. So if you're trying to convert 1990, it goes like this:

  1. 1990, evaluate M (1000); 1990 > 1000 so we have an M, leaving 990 for the next iteration
  2. 990 is less than 1000 so MM is wrong, but it's more than 900 (1000 - (1000 / 10)), so we set the lastEntry pointer to M and proceed
  3. Now we're evaluating C (100), but lastEntry is set. 990 > 900 which is CM, giving us MCM so far and leaving 90. I think this is where the bug is but it works in this case :)
  4. 90 > (100 - (100 / 10)) so lastEntry is now C
  5. Evaluate L (50); lastEntry is set, but the calculation to test whether L can be subtracted fails, so we proceed
  6. Evaluate X (10): the subtraction test succeeds, so 90 becomes XC and the full result is MCMXC.
Thread Thread
pickleat profile image
Andy Pickle Author

Nice that’s such a cool way of thinking about it!

itsasine profile image
ItsASine (Kayla)
function solution(number){
  // todo: solve without defining middle values :(
  var translations = {
    'M': 1000,
    'CM': 900,
    'D': 500,
    'CD': 400,
    'C': 100,
    'XC': 90,
    'L': 50,
    'XL': 40,
    'X': 10,
    'IX': 9,
    'V': 5,
    'IV': 4,
    'I': 1
  var answer = '';
  var value = 0;
  var thisPower = 0;

  Object.keys(translations).forEach(function(letter, index, array) {
    thisPower = translations[letter];
    value = Math.floor(number / thisPower);

    number -= value * thisPower;
    answer += letter.repeat(value);

  return answer;

Mine works... but I every time I tried doing something clever to not have to define 900, 400, 90, 40, and 9, it wouldn't work on 4, so I gave up on that. I really liked the .repeat so I didn't have to define multiples, though.

vitalipom profile image
Vitali Pomanitski

I like your solution, looks good.