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Solution: Roman to Integer

seanpgallivan on February 20, 2021

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 upv...

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UsamaHafeez0 • Aug 21 '22 • Edited

There is an issue here

but we can clear that by multiplying num by any number between 2 and 4 before comparing it to ans,

as the algo will produce incorrect result when num is multiplied by 2. As Alish Satani (i tried but couldn't mention idk how to :P) commented that this algo produces wrong results on "MDCCCLXXXIV" this roman.
It returns 1664 while the correct answer is 1884. This only happens when num is multiplied by 2.
Dry Run:
Let's try a little dry run (MDCCCLXXXIV):
num = ans = 0;
Starting from right side
We read V: 5 * 2 < 0 (ans initially is 0) = false, so ans becomes 5
Next we read I: 1 * 2 < 5 = true, so we subtract 1 from 5 and ans = 4 now
Next we read X: 10 * 2 < 4 = false, we add 10 to ans, ans = 14
Again we get X: 10 * 2 < 14 = false, we add 10 to ans, ans = 24
Here is the issue we get X again (3rd consecutive): 10 * 2 < 24 "TRUE", now we remove 10 from 24 instead of adding it.
So the trouble begins when the algo encounters same roman element for 3 consecutive times in case we multiply it with 2. You can see this pattern to repeat again in case of DCCCL (3 Cs).

Reason:
Reason for this is as the roman elements can be repeated upto max 3 consecutive times so the sum of all romans on its right side are going to be less then 3 times of the the repeating roman. Like in case of XXX the sum of romans on the right is going to be less then 3 * 10 but when we use 2 as a constant then we don't account for the possibility of 3rd consecutive roman and like in case of XXX 2 * 10 is not less then XXIV (or XX.....).

FIX:
so, the article should say

but we can clear that by multiplying num by number 3 or 4 before comparing it to ans,

I haven't seen any example to fail in case of 3 or 4.
Thanks

seanpgallivan • Aug 21 '22

Thanks, and you're correct. Article has been updated.

Sai Nitesh • Jan 20 '22 • Edited

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

if(4* num<ans) ans-=num;

can u please explain

seanpgallivan • Jan 20 '22

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.

Sai Nitesh • Jan 30 '22

Thank you so much, i got it cleared :)

Jake6868 • May 2 '22

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

seanpgallivan • May 2 '22

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).

Jake6868 • May 2 '22

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.

張旭 • Jun 9 '22

you could check the dev.to/seanpgallivan/solution-inte...

Alfredo Salzillo • Feb 21 '21

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);

seanpgallivan • Feb 21 '21

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));

Rahman • Dec 11 '21

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
    }
}

t3chnicallyinclined • Sep 15 '22 • Edited

Thanks for the help, after digging into each method I found the following:

There's no need for: .into_iter() since the rev() method returns an interator
and num does not need to be mutable in this case.

So here would be a little leaner version:

use std::collections::HashMap;

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

        let map = HashMap::from([
            ("I", 1),
            ("V", 5),
            ("X", 10),
            ("L", 50),
            ("C", 100),
            ("D", 500),
            ("M", 1000),
        ]);

        s.chars().rev().for_each(|c| {
            let num = map.get(c.to_string().as_str()).unwrap();

            if 4 * num < ans {
                ans -= num;
            }
            else {
                ans += num;
            };

        });
        ans
    }
}

SadAdolf • Sep 9 '22

Never Underestimate the POWER OF IF ELSE MUHAHAHAHA

/**
 * @param {string} s
 * @return {number}
 */
var romanToInt = function(s) {
    let letters = s.split('');
    let letterreversered = [];
    let numerual = []
    for (let x = letters.length ;x >= 0 ;x--){
        if (letters[x] === 'I'){
            if (letters[x+1] == 'V' | letters[x+1] == 'X'){
                numerual.push(0)
                letterreversered.push(0)
            }
            else if (letters[x-1] == 'V' | letters[x-1] == 'X'){
                numerual.push(0)
                letterreversered.push(0)
            }
            else {
                numerual.push(1)
                letterreversered.push(1)
            }

        }
        else if (letters[x] === 'V'){
            if (letters[x-1] === 'I'){
                numerual.push(4)   
                letterreversered.push(4) 
            }else if (letters[x+1] === 'I') {
                if (letters[x+2]=== 'X' | letters[x+2]=== 'V'){
                    numerual.push(5)
                    letterreversered.push(5)
                }else {
                    numerual.push(6)
                    letterreversered.push(6)   
                }
                // numerual.push(5)
                // letterreversered.push(5)
            }else {
                numerual.push(5)
                letterreversered.push(5)
            }

        }
        else if (letters[x] === 'X'){
            if (letters[x-1] === 'I'){
                numerual.push(9)  
                letterreversered.push(9)  
            }else if (letters[x+1] === 'L' | letters[x+1] === 'C' ) {
                numerual.push(0)
                letterreversered.push(0)
            }else if (letters[x+1] === 'I') {
                if (letters[x+2]=== 'X' | letters[x+2]=== 'V'){
                    numerual.push(10)
                    letterreversered.push(10)
                }else {
                    numerual.push(11)
                    letterreversered.push(11)   
                }
            }else {
                numerual.push(10)
                letterreversered.push(10)
            }
        }
        else if (letters[x] === 'L'){
            if (letters[x-1] === 'X'){
                numerual.push(40)   
                letterreversered.push(40) 
            }
            else {
                numerual.push(50)
                letterreversered.push(50)
            }

        }
        else if (letters[x] === 'C'){
            if (letters[x-1] === 'X'){
                numerual.push(90)   
                letterreversered.push(90) 
            }else if (letters[x+1] === 'M' | letters[x+1] === 'D' ) {
                numerual.push(0)
                letterreversered.push(0)
            }
            else {
                numerual.push(100)
                letterreversered.push(100)
            }
        }
        else if (letters[x] === 'D'){
            if (letters[x-1] === 'C'){
                numerual.push(400)   
                letterreversered.push(400) 
            }
            else {
                numerual.push(500)
                letterreversered.push(500)
            }
        }
        else if (letters[x] === 'M'){
            if (letters[x-1] === 'C'){
                numerual.push(900)   
                letterreversered.push(900) 
            }
            else {
                numerual.push(1000)
                letterreversered.push(1000)
            }
        }
    }
    // label.innerText = letters
    // label.innerText = letterreversered
    const reducer = (accumulator, curr) => accumulator + curr;
    let finalsum = numerual.reduce(reducer)
    // sum.innerText = finalsum


    return finalsum
};

Andrel Karunia S • Jul 13 '21

hello there

i have not checked yet the speed, but this accepted

Javascript solution:

Idea:

number reduced if the next number bigger than current number
number added if next number is smaller than or equal the current number

Dalcio • Oct 31 '21

/**
 *
 * @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));

Satyagraha • Jul 27 '22

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...

Apoorav Misra • Jan 9 '22

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;

Rahul Gawade • Aug 11 '21 • Edited

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;
}
}

Ian Pride • Feb 20 '21

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.