13. Roman to Integer - Explanation

Description

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

Symbol       Value
I             1
V             5
X             10
L             50
C             100
D             500
M             1000

For example, 2 is written as II in Roman numeral, just two ones added together. 12 is written as XII, which is simply X + II. The number 27 is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

I can be placed before V (5) and X (10) to make 4 and 9.
X can be placed before L (50) and C (100) to make 40 and 90.
C can be placed before D (500) and M (1000) to make 400 and 900.

You are given a roman numeral as a string s, convert it to an integer.

Example 1:

Input: s = "III"

Output: 3

Example 2:

Input: s = "XLIX"

Output: 49

Constraints:

1 <= s.length <= 15
s contains only the characters ('I', 'V', 'X', 'L', 'C', 'D', 'M').
It is guaranteed that s is a valid roman numeral in the range [1, 3999].

Topics

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Prerequisites

Before attempting this problem, you should be comfortable with:

Hash Map - Used to store and quickly lookup the integer value for each Roman numeral character
String Iteration - Traversing through a string character by character while comparing adjacent elements
Conditional Logic - Determining when to add vs subtract based on comparing current and next values

1. Hash Map

Intuition

Roman numerals normally add values from left to right. The key insight is handling subtractive notation, where a smaller value before a larger one means subtraction (like IV = 4, not 6). As we scan left to right, if the current symbol is smaller than the next one, we subtract its value; otherwise, we add it. This single rule handles both regular addition and subtractive cases elegantly.

Algorithm

Create a hash map storing each Roman numeral character and its integer value.
Initialize the result to 0.
Iterate through each character in the string:
- If the current character's value is less than the next character's value, subtract it from the result.
- Otherwise, add it to the result.
Return the final result.

class Solution:
    def romanToInt(self, s: str) -> int:
        roman = {
            "I": 1, "V": 5, "X": 10,
            "L": 50, "C": 100, "D": 500, "M": 1000
        }
        res = 0
        for i in range(len(s)):
            if i + 1 < len(s) and roman[s[i]] < roman[s[i + 1]]:
                res -= roman[s[i]]
            else:
                res += roman[s[i]]
        return res

public class Solution {
    public int romanToInt(String s) {
        Map<Character, Integer> roman = new HashMap<>();
        roman.put('I', 1); roman.put('V', 5);
        roman.put('X', 10); roman.put('L', 50);
        roman.put('C', 100); roman.put('D', 500);
        roman.put('M', 1000);

        int res = 0;
        for (int i = 0; i < s.length(); i++) {
            if (i + 1 < s.length() && roman.get(s.charAt(i)) < roman.get(s.charAt(i + 1))) {
                res -= roman.get(s.charAt(i));
            } else {
                res += roman.get(s.charAt(i));
            }
        }
        return res;
    }
}

class Solution {
    /**
     * @param {string} s
     * @return {number}
     */
    romanToInt(s) {
        const roman = {
            I: 1,
            V: 5,
            X: 10,
            L: 50,
            C: 100,
            D: 500,
            M: 1000,
        };

        let res = 0;
        for (let i = 0; i < s.length; i++) {
            if (i + 1 < s.length && roman[s[i]] < roman[s[i + 1]]) {
                res -= roman[s[i]];
            } else {
                res += roman[s[i]];
            }
        }
        return res;
    }
}

public class Solution {
    public int RomanToInt(string s) {
        Dictionary<char, int> roman = new Dictionary<char, int> {
            {'I', 1}, {'V', 5}, {'X', 10},
            {'L', 50}, {'C', 100}, {'D', 500}, {'M', 1000}
        };

        int res = 0;
        for (int i = 0; i < s.Length; i++) {
            if (i + 1 < s.Length && roman[s[i]] < roman[s[i + 1]]) {
                res -= roman[s[i]];
            } else {
                res += roman[s[i]];
            }
        }
        return res;
    }
}

class Solution {
    fun romanToInt(s: String): Int {
        val roman = mapOf(
            'I' to 1, 'V' to 5, 'X' to 10,
            'L' to 50, 'C' to 100, 'D' to 500, 'M' to 1000
        )

        var res = 0
        for (i in s.indices) {
            if (i + 1 < s.length && roman[s[i]]!! < roman[s[i + 1]]!!) {
                res -= roman[s[i]]!!
            } else {
                res += roman[s[i]]!!
            }
        }
        return res
    }
}

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(1)$ since we have $7$ characters in the hash map.

Common Pitfalls

Always Adding Instead of Subtracting

A common mistake is to always add the value of each Roman numeral without checking for subtractive notation. For example, treating IV as I + V = 6 instead of V - I = 4. The fix is to compare the current character's value with the next character's value and subtract when the current is smaller.

Off-by-One Error When Comparing Adjacent Characters

When checking if the current numeral is smaller than the next, forgetting to verify that i + 1 is within bounds causes an index out of bounds error. Always ensure the comparison i + 1 < len(s) is checked before accessing s[i + 1].