Roman to Integer

Problem 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.

Given a roman numeral, convert it to an integer.

Examples

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].

Look-Ahead Comparison Rule

For any character at index i, we look at the value of the character at i + 1:

• If the current character’s value is smaller than the next character’s value, we subtract the current value from our total (e.g. for IV, I (1) is less than V (5), so we subtract 1).
• If the current value is greater than or equal to the next value, we add the current value to our total.

Solution: Look-Ahead Scan

import java.util.*;

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

        int total = 0;
        int n = s.length();

        for (int i = 0; i < n; i++) {
            int currentVal = values.get(s.charAt(i));
            if (i + 1 < n && currentVal < values.get(s.charAt(i + 1))) {
                total -= currentVal;
            } else {
                total += currentVal;
            }
        }

        return total;
    }
}

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

#include <string>
#include <unordered_map>

class Solution {
public:
    int romanToInt(std::string s) {
        std::unordered_map<char, int> values = {
            {'I', 1}, {'V', 5}, {'X', 10}, {'L', 50},
            {'C', 100}, {'D', 500}, {'M', 1000}
        };

        int total = 0;
        int n = s.length();

        for (int i = 0; i < n; i++) {
            int currentVal = values[s[i]];
            if (i + 1 < n && currentVal < values[s[i + 1]]) {
                total -= currentVal;
            } else {
                total += currentVal;
            }
        }

        return total;
    }
};

Complexity Analysis:

• Time Complexity: O(N) where N is the length of the string s. (Since N is bounded by 15, execution takes less than a microsecond).
• Space Complexity: O(1) auxiliary space.

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