LeetCode: Roman to Integer

Convert a Roman numeral to an integer.

Posted on 14 Nov 2025 LeetCode Python

Difficulty: Easy

First Approach

My first approach was to iterate through the string, checking both the current character and previous character. The “trick” is that sometimes we are adding 800 to the return value, since we’ve already added 100 for the “C” and now need to add 900 for “CM”.

Runtime

7 ms | Beats 42.17%

class Solution:
    def romanToInt(self, s: str) -> int:
        value = 0
        prev_char = None

        for idx in range(len(s)):
            char = s[idx]

            if char == "M" and prev_char == "C":
                value += 800
            elif char == "D" and prev_char == "C":
                value += 300
            elif char == "C" and prev_char == "X":
                value += 80
            elif char == "L" and prev_char == "X":
                value += 30
            elif char == "X" and prev_char == "I":
                value += 8
            elif char == "V" and prev_char == "I":
                value += 3
            elif char == "M":
                value += 1000
            elif char == "D":
                value += 500
            elif char == "C":
                value += 100
            elif char == "L":
                value += 50
            elif char == "X":
                value += 10
            elif char == "V":
                value += 5
            elif char == "I":
                value += 1

            prev_char = char

        return value

Second Approach

For my second attempt, I created a mapping of both single and double character Roman numerals to their integer values. I then iterated through the string, checking for two-character matches first, and falling back to single-character matches.

This improved the time from 7 ms to 3 ms.

Runtime

3 ms | Beats 79.99%

class Solution:
    def romanToInt(self, s: str) -> int:
        char_map = {
            "CM": 900,
            "CD": 400,
            "XC": 90,
            "XL": 40,
            "IX": 9,
            "IV": 4,
            "M": 1000,
            "D": 500,
            "C": 100,
            "L": 50,
            "X": 10,
            "V": 5,
            "I": 1,
        }

        result = 0

        idx = 0
        while idx < len(s):
            # Check for two characters
            if idx < len(s) - 1 and s[idx:idx + 2] in char_map:
                result += char_map[s[idx:idx + 2]]
                idx += 2
                continue

            # Otherwise, one character
            result += char_map[s[idx]]
            idx += 1

        return result