Solution: Roman to Integer

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Leetcode Problem #13 (Easy): Roman to Integer

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This content originally appeared on DEV Community and was authored by seanpgallivan

This is part of a series of Leetcode solution explanations (index). If you liked this solution or found it useful, please like this post and/or upvote my solution post on Leetcode's forums.

Leetcode Problem #13 (Easy): Roman to Integer

Description:

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.

Examples:

Example 1:
Input: s = "III"
Output:* 3
Example 2:
Input: s = "IV"
Output: 4
Example 3:
Input: s = "IX"
Output: 9
Example 4:
Input: s = "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.
Example 5:
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].

Idea:

The only really tricky thing about counting in roman numerals is when a numeral is used as a subtractive value rather than an additive value. In "IV" for example, the value of "I", 1, is subtracted from the value of "V", 5. Otherwise, you're simply just adding the values of all the numerals.

The one thing we should realize about the subtractive numerals is that they're identifiable because they appear before a larger number. This means that the easier way to iterate through roman numerals is from right to left, to aid in the identifying process.

So then the easy thing to do here would be to iterate backwards through S, look up the value for each letter, and then add it to our answer (ans). If we come across a letter value that's smaller than the largest one seen so far, it should be subtracted rather than added.

The standard approach would be to use a separate variable to keep track of the highest value seen, but there's an easier trick here. Since numbers generally increase in a roman numeral notation from right to left, any subtractive number must also be smaller than our current ans.

So we can avoid the need for an extra variable here. We do run into the case of repeated numerals causing an issue (ie, "III"), but we can clear that by multiplying num by any number between 2 and 4 before comparing it to ans, since the numerals jump in value by increments of at least 5x.

Once we know how to properly identify a subtractive numeral, it's a simple matter to just iterate backwards through S to find and return the ans.

Implementation:

Javascript and Python both operate with objects / disctionaries quite quickly, so we'll use a lookup table for roman numeral values.

Java and C++ don't deal with objects as well, so we'll use a switch case to function much the same way.

Javascript Code:

const roman = {'I':1,'V':5,'X':10,'L':50,'C':100,'D':500,'M':1000}

var romanToInt = function(S) {
    let ans = 0
    for (let i = S.length-1; ~i; i--) {
        let num = roman[S.charAt(i)]
        if (4 * num < ans) ans -= num
        else ans += num
    }
    return ans
};
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Python Code:

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

class Solution:
    def romanToInt(self, S: str) -> int:
        ans = 0
        for i in range(len(S)-1,-1,-1):
            num = roman[S[i]]
            if 4 * num < ans: ans -= num
            else: ans += num
        return ans
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Java Code:

class Solution {
    public int romanToInt(String S) {
        int ans = 0, num = 0;
        for (int i = S.length()-1; i >= 0; i--) {
            switch(S.charAt(i)) {
                case 'I': num = 1; break;
                case 'V': num = 5; break;
                case 'X': num = 10; break;
                case 'L': num = 50; break;
                case 'C': num = 100; break;
                case 'D': num = 500; break;
                case 'M': num = 1000; break;
            }
            if (4 * num < ans) ans -= num;
            else ans += num;
        }
        return ans;
    }
}
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C++ Code:

class Solution {
public:
    int romanToInt(string S) {
        int ans = 0, num = 0;
        for (int i = S.size()-1; ~i; i--) {
            switch(S[i]) {
                case 'I': num = 1; break;
                case 'V': num = 5; break;
                case 'X': num = 10; break;
                case 'L': num = 50; break;
                case 'C': num = 100; break;
                case 'D': num = 500; break;
                case 'M': num = 1000; break;
            }
            if (4 * num < ans) ans -= num;
            else ans += num;
        }
        return ans;        
    }
};
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This content originally appeared on DEV Community and was authored by seanpgallivan


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