Roman Numeral Converter

Code Feedback
1

Hi there!

I just completed the Roman Numeral Converter project and I feel that it’s extremely repetitive with the if statements and/or there maybe a better way to doing this.

I originally used a ton of while loops to complete the project, however, I went back at it and tried to use recursion instead of 13 while loops, but I still ended up with repetitive If statements.

function convertToRoman(num) {

  let romanNumeralStr = [];

 
  function numeralParse (numIndex) {

    //Roman Numeral M
    if (numIndex >= 1000) {
      romanNumeralStr.push("M");//Pushes M into romanNum array.

      numIndex -= 1000;//Reduces our numIndex by 1000.

      /*Pushes numIndex back into this function and checks
      if numIndex is still greater than or equal to 1000.
      If it is not, it checks for 900 then 500 then 400 etc */
      return numeralParse(numIndex);
    }

    //Roman Numeral CM
    if (numIndex >= 900) {
      romanNumeralStr.push("CM");
      numIndex -= 900;
      return numeralParse(numIndex);
    }

    //Roman Numeral D
    if (numIndex >= 500) {
      romanNumeralStr.push("D");
      numIndex -= 500;
      return numeralParse(numIndex);
    }

    //Roman Numeral CD
    if (numIndex >= 400) {
      romanNumeralStr.push("CD");
      numIndex -= 400;
      return numeralParse(numIndex);
    }

    //Roman Numeral C
    if (numIndex >= 100) {
      romanNumeralStr.push("C");
      numIndex -= 100;
      return numeralParse(numIndex);
    }

    //Roman Numeral XC
    if (numIndex >= 90) {
      romanNumeralStr.push("XC");
      numIndex -= 90;
      return numeralParse(numIndex);
    }

    //Roman Numeral L
    if (numIndex >= 50) {
      romanNumeralStr.push("L");
      numIndex -= 50;
      return numeralParse(numIndex);
    }

    //Roman Numeral XL
    if (numIndex >= 40) {
      romanNumeralStr.push("XL");
      numIndex -= 40;
      return numeralParse(numIndex);
    }

    //Roman Numeral X
    if (numIndex >= 10) {
      romanNumeralStr.push("X");
      numIndex -= 10;
      return numeralParse(numIndex);
    }

    //Roman Numeral IX
    if (numIndex >= 9) {
      romanNumeralStr.push("IX");
      numIndex -= 9;
      return numeralParse(numIndex);
    }

    //Roman Numeral V
    if (numIndex >= 5) {
      romanNumeralStr.push("V");
      numIndex -= 5;
      return numeralParse(numIndex);
    }

    //Roman Numeral IV
    if (numIndex >= 4){
      romanNumeralStr.push("IV");
      numIndex -= 4;
      return numeralParse(numIndex);
    }

    //Roman Numeral I
    if (numIndex >= 1) {
      romanNumeralStr.push("I");
      numIndex -= 1;
      return numeralParse(numIndex);
    }

    return numIndex;
  }

 
 numeralParse(num);
 romanNumeralStr = romanNumeralStr.join("");

 console.log(romanNumeralStr);
 return romanNumeralStr;
}

convertToRoman(1004);

My apologies if you have a hard time reading my code and thank you in advance! :slight_smile:

2

Your solution works, even if it may not be the most elegant. To be fair, it’s actually easy to read though, as it’s a very straightforward approach.
(I spoilered it for those who may not have completed this project yet).

I’d think about using a switch statement perhaps, instead of all of those if statements.

One possible approach, to avoid going around in circles so much:

  1. Any number passed to the function could be split into its constituent digits, which represent thousands, hundreds, tens and units.

  2. If you have an array of Roman numerals M, C, X, I (ignoring for now the in-between numbers), this could correspond directly to the digits in your number.

  3. A switch statement (inside an iterative loop) could check each digit in turn and take the corresponding number of Roman numerals, depending on the integer value of the digit. If it is more than three then, of course, you’d need to factor in the in-between numerals (e.g. D, L etc). There are nifty ways of incorporating this without making your code any more verbose.

  4. Once you’ve finished iterating the digits, you should have successfully compiled the Roman numeral equivalent of the original number.

3

I didn’t think of actually splitting the digits and handling them according to their position. I’ll take the time to reformat it and make it better. Thank you!

4

Bear in mind, it’s just a suggestion… there are many different approaches, each with their own merits/demerits. My suggested approach is probably far from optimal, it’s just a different way of looking at the problem. This is the joy of problem-solving!

5

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