Solution: Roman to Integer

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Leetcode Problem #13 (Easy): Roman to Integer

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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.

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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.

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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].

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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 either 3 or 4 before comparing it to ans, since the numerals jump in value by increments of at least 5x. (Note: 2 is too small of a multiplier due to the possibility of a triple character followed by another, ie: "XXXI" where 2 * 10 < 21)

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.

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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.

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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;        
    }
};

Comments

[UsamaHafeez0]

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]

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

[Sai Nitesh]

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]

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]

Thank you so much, i got it cleared :)

[Jake6868]

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

[seanpgallivan]

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]

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]

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

[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);

[seanpgallivan]

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]

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]

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]

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]

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

[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));

[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...

[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;

[Rahul Gawade]

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;
}
}