Roman To Integer | February LeetCoding Challenge 2021 | Day20
Manas Sinha
Developer | Designer

ROMAN TO INTEGER

`PROBLEM  STATEMENT:`
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
```
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]`.
Problem is simpler to solve by working the string from back to front and using a map

Explanation

For a roman number represented as AB,
• `if A>=B integer(AB) = integer(A) + integer(B)` For Example `VI`
• `if A < B integer(AB) = -1*integer(A) + integer(B)` For Example `IV`
So we traverse the string left to right and check if the next digit is smaller than the current. If yes we add the current to the result else we subtract it from the result.
`LIX = L - I + X`
`LVI = L + V + I`
`ALGORITHM: `
• Traverse the roman number string left to right.
• Check if the current character is greater than or equal to the next.
• If yes, add it to the result.
• Else subtract it from the result.
• When you reach the last character simply add it.
• Time complexity : O(n)
• Space complexity : O(1)

C++

``````class Solution {
public:
int romanToInt(string s) {
map<char,int> R;
R['I'] = 1;
R['V'] = 5;
R['X'] = 10;
R['L'] = 50;
R['C'] = 100;
R['D'] = 500;
R['M'] = 1000;

int res = 0;
for(int i=0;i<s.length();++i){

if(i == s.length()-1 || R[s[i]]>=R[s[i+1]])
res += R[s[i]];
else res -= R[s[i]];
}
return res;
}
};``````

PYTHON

``````class Solution:
def romanToInt(self, s: str) -> int:
R = {'I':1,'V':5,'X':10,'L':50,'C':100,'D':500,'M':1000}
res = 0
for i in range(len(s)):
if(i == len(s)-1 or R[s[i]] >= R[s[i+1]]):
res += R[s[i]]
else:
res -= R[s[i]]

return res
``````