An Interest In:
Web News this Week
- April 1, 2024
- March 31, 2024
- March 30, 2024
- March 29, 2024
- March 28, 2024
- March 27, 2024
- March 26, 2024
Solution: Roman to Integer
This is part of a series of Leetcode solution explanations (index). If you liked this solution or found it useful, please like this post and/or upvote my solution post on Leetcode's forums.
Leetcode Problem #13 (Easy): Roman to Integer
Description:
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 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 beforeV
(5
) andX
(10
) to make4
and9
. -
X
can be placed beforeL
(50
) andC
(100
) to make40
and90
. -
C
can be placed beforeD
(500
) andM
(1000
) to make400
and900
.
Given a roman numeral, convert it to an integer.
Examples:
Example 1: | |
---|---|
Input: | s = "III" |
Output:* | 3 |
Example 2: | |
---|---|
Input: | s = "IV" |
Output: | 4 |
Example 3: | |
---|---|
Input: | s = "IX" |
Output: | 9 |
Example 4: | |
---|---|
Input: | s = "LVIII" |
Output: | 58 |
Explanation: | L = 50, V= 5, III = 3. |
Example 5: | |
---|---|
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]
.
Idea:
The only really tricky thing about counting in roman numerals is when a numeral is used as a subtractive value rather than an additive value. In "IV" for example, the value of "I", 1, is subtracted from the value of "V", 5. Otherwise, you're simply just adding the values of all the numerals.
The one thing we should realize about the subtractive numerals is that they're identifiable because they appear before a larger number. This means that the easier way to iterate through roman numerals is from right to left, to aid in the identifying process.
So then the easy thing to do here would be to iterate backwards through S, look up the value for each letter, and then add it to our answer (ans). If we come across a letter value that's smaller than the largest one seen so far, it should be subtracted rather than added.
The standard approach would be to use a separate variable to keep track of the highest value seen, but there's an easier trick here. Since numbers generally increase in a roman numeral notation from right to left, any subtractive number must also be smaller than our current ans.
So we can avoid the need for an extra variable here. We do run into the case of repeated numerals causing an issue (ie, "III"), but we can clear that by multiplying num by any number between 2 and 4 before comparing it to ans, since the numerals jump in value by increments of at least 5x.
Once we know how to properly identify a subtractive numeral, it's a simple matter to just iterate backwards through S to find and return the ans.
Implementation:
Javascript and Python both operate with objects / disctionaries quite quickly, so we'll use a lookup table for roman numeral values.
Java and C++ don't deal with objects as well, so we'll use a switch case to function much the same way.
Javascript Code:
const roman = {'I':1,'V':5,'X':10,'L':50,'C':100,'D':500,'M':1000}var romanToInt = function(S) { let ans = 0 for (let i = S.length-1; ~i; i--) { let num = roman[S.charAt(i)] if (4 * num < ans) ans -= num else ans += num } return ans};
Python Code:
roman = {'I':1,'V':5,'X':10,'L':50,'C':100,'D':500,'M':1000}class Solution: def romanToInt(self, S: str) -> int: ans = 0 for i in range(len(S)-1,-1,-1): num = roman[S[i]] if 4 * num < ans: ans -= num else: ans += num return ans
Java Code:
class Solution { 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; }}
C++ Code:
class Solution {public: int romanToInt(string S) { int ans = 0, num = 0; for (int i = S.size()-1; ~i; i--) { switch(S[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; }};
Original Link: https://dev.to/seanpgallivan/solution-roman-to-integer-567f
Dev To
An online community for sharing and discovering great ideas, having debates, and making friendsMore About this Source Visit Dev To