# 13. Roman to Integer LeetCode Solution

In this guide we will provide 13. Roman to Integer LeetCode Solution with best time and space complexity. The solution to Roman to Integer problem is provided in various programming languages like C++, Java and python. This will be helpful for you if you are preparing for placements, hackathon, interviews or practice purposes. The solutions provided here are very easy to follow and with detailed explanations.

## Table of Contents

- Problem Statement
- Roman to Integer solution in C++
- Roman to Integer soution in Java
- Roman to Integer solution Python
- Additional Resources

## Problem Statement of Roman to Integer

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].

### Complexity Analysis

**Time Complexity:**O(n)**Space Complexity:**O(128) = O(1)

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

```
class Solution {
public:
int romanToInt(string s) {
int ans = 0;
vector<int> roman(128);
roman['I'] = 1;
roman['V'] = 5;
roman['X'] = 10;
roman['L'] = 50;
roman['C'] = 100;
roman['D'] = 500;
roman['M'] = 1000;
for (int i = 0; i + 1 < s.length(); ++i)
if (roman[s[i]] < roman[s[i + 1]])
ans -= roman[s[i]];
else
ans += roman[s[i]];
return ans + roman[s.back()];
}
};
```

/* code provided by PROGIEZ */

## 13. Roman to Integer LeetCode Solution in Java

```
class Solution {
public int romanToInt(String s) {
int ans = 0;
int[] roman = new int[128];
roman['I'] = 1;
roman['V'] = 5;
roman['X'] = 10;
roman['L'] = 50;
roman['C'] = 100;
roman['D'] = 500;
roman['M'] = 1000;
for (int i = 0; i + 1 < s.length(); ++i)
if (roman[s.charAt(i)] < roman[s.charAt(i + 1)])
ans -= roman[s.charAt(i)];
else
ans += roman[s.charAt(i)];
return ans + roman[s.charAt(s.length() - 1)];
}
}
```

// code provided by PROGIEZ

## 13. Roman to Integer LeetCode Solution in Python

```
class Solution:
def romanToInt(self, s: str) -> int:
ans = 0
roman = {'I': 1, 'V': 5, 'X': 10, 'L': 50,
'C': 100, 'D': 500, 'M': 1000}
for a, b in zip(s, s[1:]):
if roman[a] < roman[b]:
ans -= roman[a]
else:
ans += roman[a]
return ans + roman[s[-1]]
```

#code by PROGIEZ

## Additional Resources

- Explore all Leetcode problems solutions at Progiez here
- Explore all problems on Leetcode website here

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