# Roman to Integer | Leetcode solution

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 ones 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]`.

Code(Python):-

class Solution(object):
def romanToInt(self, s):
sum=0
i=0
while(i<len(s)):
if s[i]=='I':
if i+1<len(s) and s[i+1]=='V' :
sum+=4
i+=1
elif i+1<len(s) and s[i+1]=='X':
sum+=9
i+=1
else:
sum+=1
elif s[i]=='V':
sum+=5
elif s[i]=='X':
if i+1<len(s) and s[i+1]=='L' :
sum+=40
i+=1
elif i+1<len(s) and s[i+1]=='C':
sum+=90
i+=1
else:
sum+=10

elif s[i]=='L':
sum+=50
elif s[i]=='C':
if i+1<len(s) and s[i+1]=='D' :
sum+=400
i+=1
elif i+1<len(s) and s[i+1]=='M':
sum+=900
i+=1
else:
sum+=100

elif s[i]=='D':
sum+=500
elif s[i]=='M':
sum+=1000
i+=1
return sum

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