Roman to Integer: LeetCode

Have you ever wondered how Roman numerals like XIV or MCMXCIV translate into modern-day numbers? If you’re a Dart enthusiast, you’re in for a treat! In this blog, I’ll walk you through an efficient way to convert Roman numerals to integers using Dart.

Understanding Roman Numerals

Before diving into the code, let’s take a quick refresher on Roman numeral rules:

• Symbols:

  • I = 1, V = 5, X = 10, L = 50, C = 100, D = 500, M = 1000

• Basic Rules:

  • Numerals are usually written from largest to smallest (e.g., VI = 6).

  • If a smaller numeral appears before a larger numeral, it must be subtracted (e.g., IV = 4, IX = 9).

With this in mind, let’s code a Dart function that can convert any valid Roman numeral into an integer.

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Writing the Dart Function

Let’s jump straight into the implementation:

class Solution {
  int romanToInt(String s) {
    // Mapping Roman numerals to their respective values
    Map<String, int> lettersMap = {
      'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000
    };

    int res = 0;

    // Iterating through each character of the string
    for (int i = 0; i < s.length; i++) {
      // If a smaller numeral precedes a larger one, subtract it
      if (i + 1 < s.length && lettersMap[s[i]]! < lettersMap[s[i + 1]]!) {
        res -= lettersMap[s[i]]!;
      } else {
        // Otherwise, add its value to the result
        res += lettersMap[s[i]]!;
      }
    }

    return res;
  }
}

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Breaking Down the Logic

Here’s how the function works step by step:

1. Create a Map: We define a lettersMap to store Roman numeral values.

2. Initialize a Result Variable: res = 0 will store our final number.

3. Loop Through the String: We iterate through each character of the input string.

4. Handle Special Cases:

   • If a numeral is smaller than the next one, we subtract its value.

   • Otherwise, we simply add its value.

5. Return the Result: Once the loop completes, we return res, which contains the converted integer.

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Testing Our Function

A function is only as good as its test cases! Let’s run our solution with a few examples:

tvoid main() {
  Solution solution = Solution();

  print(solution.romanToInt("III"));      // Output: 3
  print(solution.romanToInt("IV"));       // Output: 4
  print(solution.romanToInt("IX"));       // Output: 9
  print(solution.romanToInt("LVIII"));    // Output: 58
  print(solution.romanToInt("MCMXCIV"));  // Output: 1994
}

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Why This Solution Works Efficiently

• Time Complexity: O(n) → We iterate through the string once.

• Space Complexity: O(1) → We use a fixed-size map and a few variables, regardless of input size.

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Final Thoughts

Do you have other ways to tackle this problem? Let’s discuss in the comments!

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