Introduction

In this blog post, we will explore how to solve the LeetCode problem "13" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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. 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].

Explanation

Here's the breakdown of the solution:

• Iterate and Accumulate: Traverse the Roman numeral string from left to right.
• Subtraction Handling: If a smaller value precedes a larger value (e.g., "IV"), subtract the smaller value from the total. Otherwise, add the value to the total.

• Lookup Table: Use a dictionary to efficiently map Roman numeral characters to their corresponding integer values.

• Runtime Complexity: O(n), where n is the length of the Roman numeral string. Storage Complexity: O(1) - The dictionary uses constant space.

Code

    def romanToInt(s: str) -> int:
    roman_map = {'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000}
    result = 0
    prev_value = 0

    for i in range(len(s) - 1, -1, -1):
        curr_value = roman_map[s[i]]

        if curr_value < prev_value:
            result -= curr_value
        else:
            result += curr_value

        prev_value = curr_value

    return result