[Leetcode] 12. Roman to Integer

Table of Contents

Description

Roman numerals are represented by seven different symbols: I, V, X, L, C, D, and M.

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.

Example 3:

Input: s = “MCMXCIV”
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90, IV = 4.

Constraints:

• $1 \leq s.length \leq 15$
• 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].

Approach

We can create a map of Roman numeral values and iterate through the input string. For each character, we check if the current and next characters form a special case (e.g., “IV” or “IX”). If they do, we add the value of the special case to the total and skip the next character. Otherwise, we add the value of the current character to the total.

Solution

/**
 * @param {string} s
 * @return {number}
 */
var romanToInt = function (s) {
  // Map of Roman numeral values, including special cases for subtraction.
  const VALUE_MAP = {
    I: 1,
    V: 5,
    X: 10,
    L: 50,
    C: 100,
    D: 500,
    M: 1000,
    IV: 4,
    IX: 9,
    XL: 40,
    XC: 90,
    CD: 400,
    CM: 900,
  };
  let total = 0;

  for (let i = 0; i < s.length; i++) {
    // Check if the current and next character form a special case (e.g., "IV" or "IX").
    if (VALUE_MAP[s[i] + s[i + 1]]) {
      // Add the value of the special case to total and skip the next character.
      total += VALUE_MAP[s[i] + s[++i]];
    } else {
      // Otherwise, add the value of the current character to total.
      total += VALUE_MAP[s[i]];
    }
  }

  return total;
};

Complexity Analysis

Time Complexity

The time complexity is $O(n)$, where $n$ is the length of the input string s. We iterate through the string once.

Space Complexity

The space complexity is $O(1)$ since we use a fixed-size map to store the Roman numeral values.