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13. Roman to Integer

How to decode Roman numbers

Nhut Nguyen
·Apr 24, 2023·

Problem statement

Roman numerals are represented by seven 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 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

``````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]`.

Solution: Mapping and summing the values

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

Code

``````#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;
}
``````
``````Output:
3
58
1994
``````

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 here.

This is a very interesting problem since it is about history and mathematics. You can read more about it on Wikipedia.

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