0013 - Roman to Integer (Easy)

Problem Link

https://leetcode.com/problems/roman-to-integer/

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].

Approach 1: Iterating over the string

The solution used was iterating over the string and executing a condition that meets the subtraction principles described above in the problem statement, in order to return the result corresponding to the final sum.

The condition says: if the current character is greater than the previous character then subtract the previous character value from the $result$, otherwise increment the value of the previous character to the $result$.

For example, if we consider the string $s = 'XIV'$, the first character 'X' whose value is 10 will satisfy the condition, since the $previousChar$ variable is initialized with 0, then $result$ still remains 0. The second character 'I' whose value is 1 won't satisfy the condition once 1 isn't greater than $previousChar$ which is now 10. So $result$ is incremented by 10 and $previousChar$ is updated to 1. Finally, the third character 'V' whose value is 5 will satisfy the condition because is greater than $previousChar$. So from the $result$ is subtracted 1 and $previousChar$ is updated to 5. The code finishes the loop with $result = 9$ and finally increments the value of $previousChar$ to the $result$. So our final $result$ becomes 14.

Time Complexity: $O(n)$

This solution will be $O(n)$ as the time varies proportionally to the length of the string.

Space Complexity: $O(1)$

The space complexity for this solution is $O(1)$ as we only created variables for the counters and they're not related to the input size.

Python

Written by @jessicaribeiroalves

romanNumeralsDict = {
    "I": 1,
    "V": 5,
    "X": 10,
    "L": 50,
    "C": 100,
    "D": 500,
    "M": 1000
}

class Solution(object):
    def romanToInt(self, s):
        result = 0
        previousChar = 0

        for char in s:
            if romanNumeralsDict[char] > previousChar:
                result -= previousChar
            else:
                result += previousChar

            previousChar = romanNumeralsDict[char]

        result += previousChar
        return result

JavaScript

Written by @radojicic23

/**
 * @param {string} s
 * @return {number}
 */
var romanToInt = function (s) {
  const roman = {
    I: 1,
    V: 5,
    X: 10,
    L: 50,
    C: 100,
    D: 500,
    M: 1000,
  };
  let res = 0;
  for (i = 0; i < s.length; i++) {
    if (i + 1 < s.length && roman[s[i]] < roman[s[i + 1]]) {
      res -= roman[s[i]];
    } else {
      res += roman[s[i]];
    }
  }
  return res;
};

C++

Written by @radojicic23

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