# How to Convert a Given Roman Numeral to its Equivalent Integer Value

### **Problem statement**

Roman numerals are represented by seven symbols: `I`, `V`, `X`, `L`, `C`, `D`, and `M`.

```plaintext
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 from largest to smallest and 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

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

#### Example 2

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

#### Example 3

```plaintext
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]`.
    

### Solution: Mapping and summing the values

And to treat the subtraction cases easier, you can iterate the string `s` backward.

#### Code

```cpp
#include <iostream>
#include <unordered_map>
using namespace std;
int romanToInt(string s) {
    unordered_map<char, int> value = {
        {'I', 1},
        {'V', 5},
        {'X', 10},
        {'L', 50},
        {'C', 100},
        {'D', 500},
        {'M', 1000}
    };
    int i = s.length() - 1;
    int result = value[s[i--]];
    while (i >= 0) {
        if (value[s[i]] < value[s[i+1]]) {
            result -= value[s[i--]]; 
        } else {
            result += value[s[i--]];
        }
    }
    return result;
}
int main() {
    cout << romanToInt("III") << endl;
    cout << romanToInt("LVIII") << endl;
    cout << romanToInt("MCMXCIV") << endl;
}
```

```plaintext
Output:
3
58
1994
```

This solution efficiently converts a Roman numeral string into an integer by iterating through the string from right to left and applying the rules of Roman numerals, where subtractive combinations are subtracted and additive combinations are added to calculate the total value. The unordered map is used to look up the values of Roman numerals.

#### Complexity

* Runtime: `O(N)` where `N = s.length`.
    
* Extra space: `O(1)` (the map `value` is very small).
    

### Conclusion

The problem is about encoding/decoding between a string and an integer, you could use an [unordered\_map](https://en.cppreference.com/w/cpp/container/unordered_map) here.

This is a very interesting problem since it is about history and mathematics. You can read more about it on [Wikipedia](https://en.wikipedia.org/wiki/Roman_numerals).

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