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In this post, you will find the solution for the **Roman to Integer** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Roman to Integer– LeetCode Problem

Roman to Integer – LeetCode Problem

**Problem:**

Roman numerals are represented by seven different symbols: `I`

, `V`

, `X`

, `L`

, `C`

, `D`

and `M`

.

SymbolValueI 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:3Explanation:III = 3.

**Example 2:**

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

**Example 3:**

Input:s = "MCMXCIV"Output:1994Explanation: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]`

.

Roman to Integer– LeetCode Solutions

class Solution { public: int romanToInt(string s) { int ans = 0; vectorroman(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()]; } };

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)]; } }

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]]