13. Roman to Integer 🔢

Difficulty: Easy - Tags: Hash Table, Math, String

LeetCode Problem Link

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 from left to right, from the largest to the smallest symbol. However, there are special cases 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

Example 1:

Input:

s = "III"

Output:

3

Explanation: III = 3.

Example 2:

Input:

s = "LVIII"

Output:

58

Explanation: L = 50, V = 5, III = 3.

Example 3:

Input:

s = "MCMXCIV"

Output:

1994

Explanation: M = 1000, CM = 900, XC = 90, and IV = 4.

Constraints

• The input string 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].

Solution 💡

To solve this problem, we can iterate through the Roman numeral string from left to right, adding the value of the current symbol to the total sum. If the current symbol is smaller than the next one (indicating a subtraction case), subtract the value of the current symbol instead of adding it.

Java

class Solution {
    public int romanToInt(String s) {
        Map<Character, Integer> romanMap = new HashMap<>();
        romanMap.put('I', 1);
        romanMap.put('V', 5);
        romanMap.put('X', 10);
        romanMap.put('L', 50);
        romanMap.put('C', 100);
        romanMap.put('D', 500);
        romanMap.put('M', 1000);
        
        int total = 0;
        for (int i = 0; i < s.length(); i++) {
            int current = romanMap.get(s.charAt(i));
            if (i + 1 < s.length() && current < romanMap.get(s.charAt(i + 1))) {
                total -= current;
            } else {
                total += current;
            }
        }
        return total;
    }
}

Time Complexity ⏳

• The time complexity is O(n), where n is the length of the string s.

Space Complexity 💾

• The space complexity is O(1), as we only use a few variables and a fixed-size hash map.

You can find the full solution here.