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seanpgallivan
seanpgallivan

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Solution: Roman to Integer

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:


(Jump to: Solution Idea || Code: JavaScript | Python | Java | C++)

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:


(Jump to: Problem Description || Code: JavaScript | Python | Java | C++)

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:


(Jump to: Problem Description || Solution Idea)

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:


(Jump to: Problem Description || Solution Idea)

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:


(Jump to: Problem Description || Solution Idea)

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:


(Jump to: Problem Description || Solution Idea)

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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Discussion (20)

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sainiteshb profile image
Sai Nitesh • Edited on

Thank you for very detailed explanation.
But i didnt understand why u used 4 in the statement

if(4* num<ans) ans-=num;
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can u please explain

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seanpgallivan profile image
seanpgallivan Author

This part of the explanation covers the 4:

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.

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sainiteshb profile image
Sai Nitesh

Thank you so much, i got it cleared :)

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jake6868 profile image
Jake6868

why is it wrong when I put the test case with roman number :"CMDM"

Thread Thread
seanpgallivan profile image
seanpgallivan Author

Because that's not a valid Roman numeral sequence. Roman numbers are written in descending value order, with the exception for 9s and 4s and their equivalents in orders of magnitude.

At the start, the "CM" would be read as 900, because "C" is 100 and "M" is 1000. But then you have a "D", which would be another 500, so "CMD" would be 1400. But you'd never write it that way instead of "MCD" which is 1000+400 (rather than 900+500).

Then you have another "M", which is another 1000, but then that "M" should go at the beginning.

So assuming that the "CM" is supposed to be 900 (rather than 1100), then this should be written as "MMCD", or 1000+1000+400 (M+M+CD).

Thread Thread
jake6868 profile image
Jake6868

what do you mean by "with the exception for 9s and 4s and their equivalents in orders of magnitude." I don't understand the "s" meaning and why "9s" and "4s". Sorry, if you have time can you explain to me.

Thread Thread
zx1986 profile image
張旭
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alfredosalzillo profile image
Alfredo Salzillo

Kind of a one-liner.
It's better to sum in reverse order.

const romanMap = {
  'I':1,
  'V':5,
  'X':10,  
  'L':50,
  'C':100,
  'D':500,
  'M':1000,
}
const roman = (v) => romanMap[v];
const romanToNumber = (s) => s
    .split('')
    .reverse()
    .map(roman)
    .reduce((
        sum,
        value,
        i,
        { [i - 1]: prev = 0 },
        ) => (sum + Math.sign(value - prev) || 1) * value), 0);
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seanpgallivan profile image
seanpgallivan Author

The one-liners are never quite as performant as the more standard code, but I do love one-line solutions!

Suggestions: You can simplify/speed up the solution a bit by condensing the .split() and .map(), while converting to a faster 16-bit typed array with Uint16Array.from(). Then, you can also simplify the .reduce() a bit as well.

const romanMap = {
  'I':1,
  'V':5,
  'X':10,  
  'L':50,
  'C':100,
  'D':500,
  'M':1000,
}
const romanToNumber = s => Uint16Array.from(s, n => romanMap[n])
    .reverse()
    .reduce((sum,value) => sum + (value * 4 < sum ? -value : value));
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mochidaz profile image
Rahman

This one's written in Rust. Used macro rules for hashmap so that i won't have to insert all the romans and their values one by one into the hashmap.

use std::collections::HashMap;

macro_rules! hashmap {
    ($( $key: expr => $val: expr ),*) => {{
         let mut map = ::std::collections::HashMap::new();
         $( map.insert($key, $val); )*
         map
    }}
}

impl Solution {
    pub fn roman_to_int(s: String) -> i32 {
        let mut ans: i32 = 0;

        let map = hashmap!["I" => 1, "V" => 5, "X" => 10, "L" => 50, "C" => 100, "D" => 500, "M" => 1000];

        s.chars().rev().into_iter().for_each(|c| {
            let mut num = map.get(c.to_string().as_str()).unwrap();
            if 4 * num < ans {
                ans -= num;
            }
            else {
                ans += num;
            };
        });
        ans
    }
}
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gaweki profile image
Andrel Karunia S

hello there

i have not checked yet the speed, but this accepted

Javascript solution:

var romanToInt = function(s) {
const roman = {'I':1,'V':5,'X':10,'L':50,'C':100,'D':500,'M':1000}
let ans = 0
for(let i = 0; i < s.length; i++){
if(roman[s[i]] < roman[s[i+1]]){
ans -= roman[s[i]]
} else {
ans += (roman[s[i]] || 0) // still accepted using zero or not
}
}
return ans
};

Idea:

  • number reduced if the next number bigger than current number
  • number added if next number is smaller than or equal the current number
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dalcio profile image
Dalcio
/**
 *
 * @param {string} s
 * @return {number}
 */

const romanToInt = function (s) {
  const roman = { I: 1, V: 5, X: 10, L: 50, C: 100, D: 500, M: 1000 };

  let num = 0;

  for (let index = 0; index < s.length - 1; index++) {
    const curr = roman[s[index]];
    const next = roman[s[index + 1]];
    if (curr >= next) {
      num += curr;
    } else {
      num -= curr;
    }
  }
  num += roman[s[s.length - 1]];

  return Math.abs(num);
};

const s = "MCMXCIV";

console.log(romanToInt(s));
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prateekool109 profile image
Satyagraha

Simple Explanation | Java code with detailed comments

Simple Explanation video - youtu.be/iOjKZ4_xQPM

Java code with detailed comments in repository - github.com/prateekgoyal511/dsa/blo...

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thefluxapex profile image
Ian Pride

Very impressive, you have me inspired to write this in Rust. I have actually considered doing this many years ago, but, of course, I never got around to it and forgot about it.

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seanpgallivan profile image
seanpgallivan Author

Awesome. Feel free to drop the code in the comments here if you do!!

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rahulgawade1000 profile image
Rahul Gawade

what do you think of mine one ?

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apoorav_misra profile image
Apoorav Misra

public static int romanToInt(String s){
int year = 0;
for(int i=0;i<s.length();i++){
char c = s.charAt(i);
if(c == 'I') {
if (i != s.length() - 1 && s.charAt(i + 1) == 'V')
year = year - 1;
else if (i != s.length() - 1 && s.charAt(i + 1) == 'X')
year = year - 1;
else
year = year + 1;
}
else if(c == 'V')
year = year + 5;
else if(c == 'X'){
if (i != s.length() - 1 && s.charAt(i + 1) == 'L')
year = year - 10;
else if (i != s.length() - 1 && s.charAt(i + 1) == 'C')
year = year - 10;
else
year = year + 10;
}
else if(c == 'L')
year = year + 50;
else if(c == 'C'){
if (i != s.length() - 1 && s.charAt(i + 1) == 'D')
year = year - 100;
else if (i != s.length() - 1 && s.charAt(i + 1) == 'M')
year = year - 100;
else
year = year + 100;
}
else if(c =='D')
year = year + 500;
else if(c == 'M')
year = year + 1000;
}
return year;

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rahulgawade1000 profile image
Rahul Gawade • Edited on

I did traverse Left to right. Its Accepted .

class Solution {
public int romanToInt(String s) {
Map map=new HashMap<>();
map.put('I',1);
map.put('V',5);
map.put('X',10);
map.put('L',50);
map.put('C',100);
map.put('D',500);
map.put('M',1000);
char [] romanChar=s.toCharArray();
int sum=0;
int i=0;
for(char ch:romanChar){
int flag=1;
if(!(i==s.length()-1) ){
if (map.get(ch) < map.get(s.charAt(i + 1))) {
sum = sum - map.get(ch) ;
flag=0;
}
}
if(flag==1) {
sum = sum + map.get(ch);
}
i++;
}
return sum;
}
}

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alish_satani profile image
Alish Satani

It's wrong for "MDCCCLXXXIV" .
ans should be 1884 and it returns 1664.

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seanpgallivan profile image
seanpgallivan Author

Just checked and all four of my codeblocks are properly returning 1884 for that input. Can you copy and paste the code you're using?