Convert Roman To Numbers In C#

Dictionary < char, int > romanNumbersDictionary = new() {
    {
        'I',
        1
    }, {
        'V',
        5
    }, {
        'X',
        10
    }, {
        'L',
        50
    }, {
        'C',
        100
    }, {
        'D',
        500
    }, {
        'M',
        1000
    }
};

Now comes the logic,

Here all we gotta do is either subtraction or addition.

• For example, if string is "III" then we would do 1+1+1 = 3
• If string is "IV" then we would compare for current and its next element. If the next element's numeric value is greater than the current element's value, we would subtract else we add.
  • I = 1, V= 5: Here V > I so we substract, 1 - 5 = -4
  • then in for the last element just add the respective value so V = 5 to the equation -4 + 5 = 4

That's all. That's your whole and soul logic.

Now for our example, I have created a flow chart just for you. With the flow chart you'll get a slight idea how our logic is going to work with every iteration, and then repeat the steps for all the elements in the string.

Note
While comparing the next element we would encounter an Array IndexOutofBoundsException when we reach the last element. that is because we have exhausted the array length. for ith element we don't have i+1 element. So we need to add that validation as well.

• Just as an example of "IV" when a pointer reaches V we would simply add the value of "V" and exit the loop. I know this would be confusing. That is exactly why I have created this beautiful flowchart.

Figure 1: Flowchart Roman to Nmber

Code is pretty straight forward too,

namespace RomanToNumbersAndReverse
{
    internal class RomanToNumber
    {
        public int RomanToInt(string s)
        {
            int sum = 0;
            Dictionary<char, int> romanNumbersDictionary = new()
            {
                { 'I', 1 },
                { 'V', 5 },
                { 'X', 10 },
                { 'L', 50 },
                { 'C', 100 },
                { 'D', 500 },
                { 'M', 1000 }
            };
            for (int i = 0; i < s.Length; i++)
            {
                char currentRomanChar = s[i];
                romanNumbersDictionary.TryGetValue(currentRomanChar, out int num);
                if (i + 1 < s.Length && romanNumbersDictionary[s[i + 1]] > romanNumbersDictionary[currentRomanChar])
                {
                    sum -= num;
                }
                else
                {
                    sum += num;
                }
            }
            return sum;
        }
    }
}

Listing 1: RomanToNumber.cs

Figure 2: Code snippet in listing 1 runs against all the test cases in LeetCode