# Roman to Integer

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.

Example 1:

```Input: s = "III"
Output: 3
Explanation: III = 3.
```

Example 2:

```Input: s = "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.
```

Example 3:

```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]`.
Table of Contents

### Solution 1:

Algorithm:

• Read string backwards.
• Add roman numeral to integer variable
• Subtract from integer what was added x 2 if last_char is greater than curr_char
• Update last_char variable
• Decrement i variable to progress while loop
• Return integer
``````class Solution:
def romanToInt(self, s: str) -> int:
integer = 0
i = len(s) - 1

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

last_char = s[-1]

while i >= 0:
curr_char = s[i]
integer += symbols[curr_char]

if (symbols[last_char] > symbols[curr_char]):
integer -= 2 * symbols[curr_char]

last_char = curr_char
i -= 1

return integer``````