# Roman to Integer – Leetcode Solution

In this post, we are going to solve the 13. Roman to Integer problem of Leetcode. This problem 13. Roman to Integer is a Leetcode easy level problem. Let’s see code, 13. Roman to Integer.

## Problem

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]`.

Now, let’s see the code of 13. Roman to Integer – Leetcode Solution.

# Roman to Integer – Leetcode Solution

### 13. Roman to Integer – Solution in Java

```class Solution {
public static int romanToInt(String s) {
if (s == null || s.length() == 0)
return -1;
HashMap<Character, Integer> map = new HashMap<Character, Integer>();
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);
int len = s.length(), result = map.get(s.charAt(len - 1));
for (int i = len - 2; i >= 0; i--) {
if (map.get(s.charAt(i)) >= map.get(s.charAt(i + 1)))
result += map.get(s.charAt(i));
else
result -= map.get(s.charAt(i));
}
return result;
}
}```

### 13. Roman to Integer – Solution in C++

```class Solution {
public:
int romanToInt(string s)
{
unordered_map<char, int> T = { { 'I' , 1 },
{ 'V' , 5 },
{ 'X' , 10 },
{ 'L' , 50 },
{ 'C' , 100 },
{ 'D' , 500 },
{ 'M' , 1000 } };

int sum = T[s.back()];
for (int i = s.length() - 2; i >= 0; --i)
{
if (T[s[i]] < T[s[i + 1]])
{
sum -= T[s[i]];
}
else
{
sum += T[s[i]];
}
}

return sum;
}
};```

### 13. Roman to Integer– Solution in Python

```class Solution:
def romanToInt(self, s: str) -> int:
translations = {
"I": 1,
"V": 5,
"X": 10,
"L": 50,
"C": 100,
"D": 500,
"M": 1000
}
number = 0
s = s.replace("IV", "IIII").replace("IX", "VIIII")
s = s.replace("XL", "XXXX").replace("XC", "LXXXX")
s = s.replace("CD", "CCCC").replace("CM", "DCCCC")
for char in s:
number += translations[char]
return number```

Note: This problem 13. Roman to Integer is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.