Roman to Integer (LeetCode)

Problem Statement

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

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

• 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].

Solution

Approach

Intuition

The key intuition is that in Roman numerals, when a smaller value appears before a larger value, it represents subtraction, while when a smaller value appears after or equal to a larger value, it represents addition.

Explanation

The unordered map m is created and initialized with mappings between Roman numeral characters and their corresponding integer values. For example, 'I' is mapped to 1, 'V' to 5, 'X' to 10, and so on.

The variable ans is initialized to 0. This variable will accumulate the final integer value of the Roman numeral string.

The for loop iterates over each character in the input string s.

For example, for the string "IX":

• When i is 0, the current character s[i] is 'I'. Since there is a next character ('X'), and the value of 'I' (1) is less than the value of 'X' (10), the condition m[s[i]] < m[s[i+1]] is true. In this case, we subtract the value of the current character from ans.
• When i is 1, the current character s[i] is 'X'. This is the last character in the string, so there is no next character to compare. In this case, we add the value of the current character to ans.

For the string "XI":

• When i is 0, the current character s[i] is 'X'. Since there is a next character ('I'), and the value of 'X' (10) is greater than the value of 'I' (1), the condition m[s[i]] < m[s[i+1]] is false. In this case, we add the value of the current character to ans.
• When i is 1, the current character s[i] is 'I'. This is the last character in the string, so there is no next character to compare. In this case, we add the value of the current character to ans.

After the for loop, the accumulated value in ans represents the integer conversion of the Roman numeral string, and it is returned as the result.

Code

C++ Implementation

class Solution {
public:
    int romanToInt(string s) {
        unordered_map<char, int> m = {
            {'I', 1},
            {'V', 5},
            {'X', 10},
            {'L', 50},
            {'C', 100},
            {'D', 500},
            {'M', 1000}
        };
        
        int ans = 0;
        for (int i = 0; i < s.length(); i++) {
            if (i < s.length() - 1 && m[s[i]] < m[s[i+1]]) {
                ans -= m[s[i]];
            } else {
                ans += m[s[i]];
            }
        }
        return ans;
    }
};

Java Implementation

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

Python Implementation

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

Complexity Analysis

• Time Complexity: $O(n)$ - The algorithm iterates through each character in the input string once, where n is the length of the string.
• Space Complexity: $O(1)$ - The amount of extra space used is constant, determined by the fixed size of the m map.

Conclusion

The provided solutions efficiently convert a Roman numeral to an integer by iterating through the characters and using a map to handle the conversion based on the rules of the Roman numeral system. This approach ensures that the conversion adheres to the rules while maintaining linear time complexity and constant space complexity.