Abstract / Summary

Convert a roman numeral string into its corresponding integer value.

Note

The full problem description, including helpful examples, can be found on LeetCode.

This problem is ranked as

and has an acceptance rate of about 67% (9+ million submissions).

Typical patterns to solve this problem:

• Hash Table
• Math
• String

Every single line of code was written by me, 100% from scratch, between August 2023 and August 2024.
However, two years later I set up this website and used various AI tools to generate most of the walkthrough / explanations below.

See this dedicated page for more details.

Overview

The presented implementation converts a Roman-numeral string into its integer value. It processes the input sequentially, maintaining a running total and a reference to the previously examined character. By inspecting the current character together with the previous one, the algorithm correctly accounts for the subtractive forms defined by Roman-numeral conventions.

Core Logic

The conversion is performed inside a single for-range loop that iterates over each character of the input string. Two local variables are declared before the loop: prev, initialized to zero, stores the character processed in the prior iteration, and result, initialized to zero, accumulates the numeric value.

char prev = 0;
int result = 0;
for (auto c : s)
{
    // switch handling omitted here
    prev = c;
}
return result;

Within the loop a switch statement selects the action based on the current character c. For each Roman symbol the code either adds its base value directly or, when a subtractive pattern is detected, adjusts the addition to reflect the subtraction.

Handling Subtractive Notation

Roman numerals employ a subtractive rule where a smaller symbol placed before a larger one reduces the total (e.g., “IV” equals 4). The implementation detects these patterns by comparing c with prev. When a subtractive pair is identified, the code adds the larger value and simultaneously subtracts twice the smaller value that was previously added. This technique eliminates the need for a separate subtraction step.

case 'V': if (prev == 'I') result += 5-2*1; else result += 5; break;
case 'X': if (prev == 'I') result += 10-2*1; else result += 10; break;
case 'L': if (prev == 'X') result += 50-2*10; else result += 50; break;
case 'C': if (prev == 'X') result += 100-2*10; else result += 100; break;
case 'D': if (prev == 'C') result += 500-2*100; else result += 500; break;
case 'M': if (prev == 'C') result += 1000-2*100; else result += 1000; break;

For a pair such as “IV”, the first iteration adds 1 for ‘I’. When ‘V’ is processed, prev equals ‘I’, triggering the expression 5-2*1, which effectively adds 3, yielding a total of 4. The same principle applies to all other subtractive combinations (IX, XL, XC, CD, CM).

Edge Cases Covered

The algorithm naturally handles repeated symbols because each occurrence contributes its base value independently; for example, “III” results in three successive additions of 1. It also correctly processes a subtractive pair that appears at the start of the string, as the first character sets prev and the second character’s case logic adjusts the total accordingly. Mixed sequences that combine additive and subtractive patterns (e.g., “MCMXCIV”) are resolved step by step, with each character’s contribution determined by the immediate predecessor.

Performance Characteristics

The conversion runs in a single linear pass over the input, performing a constant amount of work per character. The use of a switch statement enables the compiler to generate an efficient jump table or branch sequence, minimizing dispatch overhead. No auxiliary containers or recursive calls are introduced, keeping the memory footprint minimal. Arithmetic adjustments for subtractive cases are performed inline, avoiding extra conditional branches or look-up structures. Consequently, the routine exhibits fast execution even for the longest valid Roman-numeral strings.

Imagine you have a big treasure chest and you are counting special gold coins that have secret letters written on them instead of regular numbers. This code acts like a helper who looks at each coin in your pile one by one to figure out how much they are worth.

The helper starts with an empty bag and looks at the first letter. If the letter is an I, it means you have one coin, so the helper adds it to the bag. Other letters like V, X, L, C, D, or M represent bigger piles of coins.

Sometimes, the letters are tricky. If you see a small letter before a big letter, it is like a little game where you have to subtract the value of the small letter instead of adding it. The helper keeps track of the letter it just saw so it knows if it needs to do this special trick. For example, if it sees an I followed by a V, it realizes you are holding a four, so it does the math to make sure the count stays exactly right. It does this work over and over until every single coin in the pile has been counted and added together to find the final total.

Several LLMs ranked my code quality on a scale from 1 (very poor) to 10 (outstanding):

Time: linear
As a rule of thumb, the maximum input size for one second execution time is

• factorial: up to 10 elements
• exponential: up to 20 elements
• polynomial: a few thousand elements
• loglinear: up to 100,000 elements
• linear: a few million elements
• logarithmic: a billion elements
• constant: practically unlimited

I wrote a basic crawler to find other solutions to that problem.
Obviously, I did that after I solved the problem myself.

In general, AlgoMonster and WalkCCC solutions are high-quality and always worth a look.
I haven't checked most of the other solutions, but they might be helpful as well.

And of course, the LeetCode forum is always a good place (or even the best) to look for alternative approaches.

(click on a column header to sort the table)

	author
author	solutions
www	AlgoMonster	C++	Python	Java										TS
	WalkCCC	C++	Python	Java
	leetcode.ca	C++	Python	Java	C#	C	Rust	JS			Ruby	Go	PHP	TS
	LeetcodeHelp	C++	Python	Java
	MemoryLimitExceeded	C++	Python
	Progiez	C++	Python	Java
github	AdelDaniel								Dart
	aylei						Rust
	begeekmyfriend					C
	BigEggStudy				C#
	black-shadows	C++
	bujosa						Rust
	chandansgowda		Python
	charles-wangkai			Java
	chenyanzhe	C++
	DimitrisJim					C
	everthis							JS
	fartem										Ruby
	FelixFan1122				C#
	fengvyi	C++
	fpdjsns	C++
	gaosq0604		Python
	gouthampradhan			Java
	halfrost											Go
	haoel	C++
	JoshCrozier							JS
	keineahnung2345	C++
	Krazune	C++
	lennylxx					C
	linouk23	C++
	lkwq007		Python
	MacJim	C++
	martianoff									Scala
	pruthvikp		Python
	qiyuangong			Java
	ruoru						Rust
	Shamik07								Dart
	shreymehul				C#
	sonugiri1043	C++
	suxrobGM				C#
	trung-hn		Python
	varunu28			Java
	vli02					C
	vthwang			Java
	XiaotaoGuo	C++
	YuriSpiridonov		Python
AlgoMonster	C++, Python, Java, TypeScript
WalkCCC	C++, Python, Java
leetcode.ca	C, C++, C#, Go, Java, JavaScript, PHP, Python, Ruby, Rust, TypeScript
LeetcodeHelp	C++, Java, Python
MemoryLimitExceeded	Python, C++
Progiez	C++, Java, Python
AdelDaniel	Dart
aylei	Rust
begeekmyfriend	C
BigEggStudy	C#
black-shadows	C++, C++
bujosa	Rust
chandansgowda	Python
charles-wangkai	Java
chenyanzhe	C++
DimitrisJim	C
everthis	JavaScript
fartem	Ruby
FelixFan1122	C#
fengvyi	C++
fpdjsns	C++
gaosq0604	Python
gouthampradhan	Java
halfrost	Go
haoel	C++
JoshCrozier	JavaScript
keineahnung2345	C++
Krazune	C++
lennylxx	C
linouk23	C++
lkwq007	Python
MacJim	C++
martianoff	Scala
pruthvikp	Python
qiyuangong	Java
ruoru	Rust
Shamik07	Dart
shreymehul	C#
sonugiri1043	C++
suxrobGM	C#
trung-hn	Python
varunu28	Java
vli02	C
vthwang	Java
XiaotaoGuo	C++
YuriSpiridonov	Python

There are a few extra pages giving you some insights into my setup:

• Tech Stack - my development environment and tools
• AI - how AI helped me create this website (NOT solving the problems)
• Progress - progression over time
• Quality - code quality