This post is part of my "LeetCode for 1x Developers" series, in which I struggle through LeetCode problems. Sometimes I figure it out, other times I don't. Either way, I give each problem my best shot, and write up my thought process through the challenges
Problem description
Roman numerals are represented by seven different symbols: I
, V
, X
, L
, C
, D
and M
.
 I: 1
 V: 5
 X: 10
 L: 50
 C: 100
 D: 500
 M: 1000
For example, two is written as II
in Roman numeral, just two one's added together. Twelve is written as, XII
, which is simply X
+ II
. The number twenty seven 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 beforeV
(5) andX
(10) to make 4 and 9. 
X
can be placed beforeL
(50) andC
(100) to make 40 and 90. 
C
can be placed beforeD
(500) andM
(1000) to make 400 and 900.
Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.
Intuition
This problem feels like bruteforce by its nature. There are a lot of rules and what feel like inconsistent patterns on the surface. So to start, I really just wanted to codify all these edge cases and different values.
At the end of function, I need to return some numeric value. So I know I'll start at 0, and all of the numerals are additive, so I'll be adding to it.
But the trick is that there are these combinations of prefixed numerals that generate distinct values. So I wrote up a pretty length if/else
branch logic. It takes the input string, checks the first character against any of the possible prefixes. If that character is sometimes found in front of other characters, we check the next one to see the possible value. We set a length
variable to 2
to indicate that this particular instance is a twocharacter value.
If there is no second character, we set length
to 1
.
We add to the result
value, based on the values of each numeral.
Finally, we subtract either 1 or 2 numerals from the front of the string and repeat this process until the input has length
0.
Here's my first pass. It's a little ugly, and I kept missing values from the prompt, so I added them in poor order:
var romanToInt = function(s) {
let result = 0;
while(s.length > 0) {
let length;
if (s.charAt(0) === 'I') {
if (s.charAt(1) === 'V') {
result += 4;
length = 2;
} else if (s.charAt(1) === 'X') {
result += 9;
length = 2;
} else {
result += 1;
length = 1;
}
} else if (s.charAt(0) === 'X') {
if (s.charAt(1) === 'L') {
result += 40
length = 2;
} else if (s.charAt(1) === 'C') {
result += 90;
length = 2;
} else {
result += 10;
length = 1;
}
} else if (s.charAt(0) === 'C') {
if (s.charAt(1) === 'D') {
result += 400;
length = 2;
} else if (s.charAt(1) === 'M') {
result += 900;
length = 2;
} else {
result += 100;
length = 1;
}
} else if (s.charAt(0) === 'V') {
result += 5;
length = 1;
} else if (s.charAt(0) === 'L') {
result += 50;
length = 1;
} else if (s.charAt(0) === 'D') {
result += 500;
length = 1;
} else if (s.charAt(0) === 'M') {
result += 1000;
length = 1;
}
s = s.substr(length);
}
return result;
};
This can be much... much cleaner. I found a great solution in the discussion section that looks like this:
Solution
var romanToInt = function(s) {
var map = {
'I': 1,
'V': 5,
'X': 10,
'L', 50,
'C': 100,
'D': 500,
'M': 1000
}
var number = 0;
var index;
if (s.indexOf('CM') != 1) number = 200;
if (s.indexOf('CD') != 1) number = 200;
if (s.indexOf('XC') != 1) number = 20;
if (s.indexOf('XL') != 1) number = 20;
if (s.indexOf('IX') != 1) number = 2;
if (s.indexOf('IV') != 1) number = 2;
for (let i=0; i<s.length; i++) {
number += map[s[i]];
}
return number;
}
This solution is super clean and I like it a lot. It sets up a map object of all the numerals and their values. Then it initializes the return value at 0
.
Next, it checks for the edge cases: CM
, CD
, XC
, XL
, IX
, and IV
. If the input string contains any of these, it subtracts from the initial value.
Then it runs a for loop against the input string and adds the value from the map of each character. Since we checked for the prefixed edge cases and subtracted the appropriate values, the final result is correct, even with the edge cases.
It took me a minute to visualize the values here. So here's an example. Consider an input: "XIV".
Without the prefix checks, the for loop would return 16
. But since the string has an indexOf not equal to 1
for IV
, we subtract 2 from the initial value. This means the naive for loop returns 14
, the correct answer.
It's a neat approach. I like it more than my big long branch. It's well organized, although I do think there's a bit of counterintuitive logic happening with the initial subtraction of the prefix values.
Top comments (3)
Special casing all those checks is asking for problems though, and I'm pretty sure it's not that neat. If we assume that the input is a valid Roman numeral (which this code does do; "IIX" is not a valid Roman numeral, but your code will parse it as 10), then the rule is that "if a numeral is smaller than the next numeral, subtract its value". Using that rule gives a much neater function:
(Apologies for using Python, but I don't trust my JS enough to write it error free) (Edited to add the vals dict so the code actually runs)
Ah yeah! Good catch. I think that's a given in the LC problem but I forgot to say it. If it isn't, none of the LC cases provide invalid numerals, haha
Interesting problem, thanks for sharing.
@everyone who might read and for the sake of not reinventing the wheel, note that there's a CSS rule for list style: