Roman to Integer Leetcode Solution

In this post, we are going to solve the Roman to Integer Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

Table of Contents

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 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].

Now, let’s see the leetcode solution of Roman to Integer Leetcode Solution.

Roman to Integer Leetcode Solution in Python

class Solution(object):
    def romanToInt(self, s):
        """
        :type s: str
        :rtype: int
        """
        roman_dic = {
            "I": 1,
            "V": 5,
            "X": 10,
            "L": 50,
            "C": 100,
            "D": 500,
            "M": 1000,
            }

  
        skip_next_n = False
        int_sum = 0
        for idx, n in enumerate(s):
            if skip_next_n == True:
                skip_next_n = False
                continue
            try:
                if roman_dic[n] < roman_dic[s[idx + 1]]:
                    int_sum +=  roman_dic[s[idx + 1]] - roman_dic[n]
                    skip_next_n = True
                    continue
                else:
                    int_sum += roman_dic[n]
            except:
                int_sum += roman_dic[n]
                
        return int_sum

Roman to Integer Leetcode Solution in CPP

class Solution {
public:
int romanToInt(string s) {
    unordered_map<char,int> mp{
        {'I',1},
        {'V',5},
        {'X',10},
        {'L',50},
        {'C',100},
        {'D',500},
        {'M',1000},
    };
    int ans =0;
    for(int i=0;i<s.size();i++){
        if(mp[s[i]]<mp[s[i+1]])
            ans-=mp[s[i]];
        else
            ans+=mp[s[i]];
    }
    return ans;
    
}
};

Roman to Integer Leetcode Solution in Java

public int romanToInt(String s) {
         int ans = 0, num = 0;
        for (int i = s.length()-1; i >= 0; i--) {
            switch(s.charAt(i)) {
                case 'I': num = 1; break;
                case 'V': num = 5; break;
                case 'X': num = 10; break;
                case 'L': num = 50; break;
                case 'C': num = 100; break;
                case 'D': num = 500; break;
                case 'M': num = 1000; break;
            }
            if (4 * num < ans) ans -= num;
            else ans += num;
        }
        return ans;
    }

Note: This problem Roman to Integer is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.

NEXT: Longest Common Prefix Leetcode Solution