Roman to Integer easy

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 one's 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.

Example 1:

Input: s = "III" Output: 3 Explanation: III = 3.

Example 2:

Input: s = "LVIII" Output: 58 Explanation: L = 50, V= 5, III = 3.

Steps

1. Create a Dictionary: Map Roman symbols to their integer values.
2. Iterate through the Roman numeral string: Process each symbol.
3. Handle Subtraction: If the current symbol's value is less than the next symbol's value, subtract the current value from the total; otherwise, add the current value to the total.
4. Return the Total: After processing all symbols, return the accumulated integer value.

Explanation

The key to solving this problem efficiently is recognizing the subtractive nature of certain Roman numeral combinations. Instead of iterating and simply adding values, we need to look ahead to see if a subtraction is necessary. The dictionary provides a quick lookup for the integer value of each Roman symbol, and the conditional statement in the loop handles the subtraction logic.

Code

using System;
using System.Collections.Generic;

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

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

Complexity

• Time Complexity: O(n), where n is the length of the Roman numeral string. We iterate through the string once.
• Space Complexity: O(1). The dictionary's size is constant regardless of the input string's length.