<< problem 88 - Product-sum numbers Cube digit pairs - problem 90 >>

# Problem 89: Roman numerals

For a number written in Roman numerals to be considered valid there are basic rules which must be followed.
Even though the rules allow some numbers to be expressed in more than one way there is always a "best" way of writing a particular number.

For example, it would appear that there are at least six ways of writing the number sixteen:

IIIIIIIIIIIIIIII
VIIIIIIIIIII
VVIIIIII
XIIIIII
VVVI
XVI

However, according to the rules only XIIIIII and XVI are valid, and the last example is considered to be the most efficient, as it uses the least number of numerals.

The 11K text file, roman.txt (right click and 'Save Link/Target As...'), contains one thousand numbers written in valid, but not necessarily minimal, Roman numerals;
see About... Roman Numerals for the definitive rules for this problem.

Find the number of characters saved by writing each of these in their minimal form.

Note: You can assume that all the Roman numerals in the file contain no more than four consecutive identical units.

# My Algorithm

There are two functions:

• roman2number takes a valid Roman number and converts it to an integer
• number2roman converts an integer to a minimal ("optimal") Roman number
My program reads all Roman numbers, converts them to integers and back to an optimal Roman number.
The difference of the strings' lengths is what I'm looking for.

roman2number reads the Roman numbers backwards:
• if the current letter is smaller than the previous (its right neighbor) then the current letter must be subtracted else added
• my code can subtract multiple identical letters, too, e.g. 8 = IIX which is shorter than VIII
number2roman has a list of conversion rules to convert an integer to a Roman number.
Each rule is identified by a number and that rule is applied as long as the integer is larger or equal.
Whenever rules[i] applies, one or two letters action[i] are added to the result:
irules[i]action[i]
01000M
1900CM
2500D
3400CD
4100C
590XC
650L
740XL
810X
99IX
105V
114IV
121I

## Modifications by HackerRank

I have to print the optimized Roman numbers instead of finding the sum of the length differences.

## Note

The problem can be solved by simple search'n'replace, too.
However, I liked the challenge to write a basic parser for Roman numbers.

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

This live test is based on the Hackerrank problem.

Number of test cases (1-5):

Input data (separated by spaces or newlines):

This is equivalent to
echo "1 " | ./89

Output:

Note: the original problem's input 50000000 cannot be entered
because just copying results is a soft skill reserved for idiots.

(this interactive test is still under development, computations will be aborted after one second)

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, as well as the input data, too. Or just jump to my GitHub repository.

The code contains #ifdefs to switch between the original problem and the Hackerrank version.
Enable #ifdef ORIGINAL to produce the result for the original problem (default setting for most problems).

       #include <iostream>
#include <string>

// convert valid roman numbers to binary numbers
unsigned int roman2number(const std::string& roman)
{
unsigned int result = 0;

// remember the value of the previous Roman letter
unsigned int last = 0;
// true, if the current letter is subtracted (and the next identical letters)
bool subtract = false;

// walk through the whole string from the end to the beginning ...
for (auto i = roman.rbegin(); i != roman.rend(); i++)
{
unsigned int current = 0;
switch (*i)
{
case 'M': current = 1000; break;
case 'D': current =  500; break;
case 'C': current =  100; break;
case 'L': current =   50; break;
case 'X': current =   10; break;
case 'V': current =    5; break;
case 'I': current =    1; break;
}

// smaller than its right neighbor ? => we must subtract
if (current < last)
{
subtract = true;
last = current;
}
// bigger than its right neighbor ? => we must add
else if (current > last)
{
subtract = false;
last = current;
}

// note: if current == last then we keep the variables "subtract" and "last" in their current state

if (subtract)
result -= current;
else
result += current;
}

return result;
}

std::string number2roman(unsigned int number)
{
// apply these rules in the presented order:
// - as long as number >= steps[i] add roman[i] to result
const unsigned int NumRules = 13;
const unsigned int rules[NumRules] =
{ 1000,  900, 500,  400, 100,  90,   50,   40,  10,    9,   5,    4,  1  };
const char* action[NumRules] =
{  "M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I" };

// apply all rules ...
std::string result;
for (unsigned int i = 0; i < NumRules; i++)
// ... as often as needed
while (number >= rules[i])
{
// reduce integer
number -= rules[i];
result += action[i];
}

return result;
}

int main()
{
// letters saved by optimization
unsigned int saved = 0;

unsigned int tests = 1000;
//#define ORIGINAL
#ifndef ORIGINAL
std::cin >> tests;
#endif

while (tests--)
{
std::string roman;
std::cin >> roman;

// convert it to an integer and back to an optimal Roman number
auto number    = roman2number(roman);
auto optimized = number2roman(number);

// count how many character were saved
saved += roman.size() - optimized.size();

#ifndef ORIGINAL
// print Roman number
std::cout << optimized << std::endl;
#endif
}

#ifdef ORIGINAL
std::cout << saved << std::endl;
#endif
return 0;
}


This solution contains 17 empty lines, 20 comments and 8 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++11 -DORIGINAL)

See here for a comparison of all solutions.

Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.

# Changelog

March 15, 2017 submitted solution

# Hackerrank

My code solves 4 out of 4 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 20% (out of 100%).

Hackerrank describes this problem as easy.

Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is rarely an option.

# Similar problems at Project Euler

Problem 610: Roman Numerals II

Note: I'm not even close to solving all problems at Project Euler. Chances are that similar problems do exist and I just haven't looked at them.

# Heatmap

Please click on a problem's number to open my solution to that problem:

 green solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too yellow solutions score less than 100% at Hackerrank (but still solve the original problem easily) gray problems are already solved but I haven't published my solution yet blue solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much orange problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte red problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too black problems are solved but access to the solution is blocked for a few days until the next problem is published [new] the flashing problem is the one I solved most recently

I stopped working on Project Euler problems around the time they released 617.
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The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
I scored 13526 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

Look at my progress and performance pages to get more details.

 << problem 88 - Product-sum numbers Cube digit pairs - problem 90 >>
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