Roman Numerals Decoder: Ruby - ruby

Question

Alright I've been working on this coding challenge for quite some time and I guess it's officially time for me to raise the flag. Help!
My task is to create a function that takes a Roman numeral as its argument and returns its value as a numeric decimal integer.
So far I've successfully created a hash mapping the numbers to its numeric values. I've also created an empty array roman_no to pass the key/value pair through.
What I am struggling with is writing the expression. Below is the full code:
def solution(roman)
# take a value of a roman numeral
roman_numeral =
{
1000 => "M",
900 => "CM",
500 => "D",
400 => "CD",
100 => "C",
90 => "XC",
50 => "L",
40 => "XL",
10 => "X",
9 => "IX",
5 => "V",
4 => "IV",
1 => "I"
}
roman_no = Array.new
roman_numeral.each do | key, value |
while
"#{roman}" >= "#{key}"
+= roman_no
"#{roman}" -= "#{key}"
end
return roman_no
solution('XXI')
How can I write an argument that will take the value from roman_numeral and return its number counter part?
for example:
solution('XXI') # should return 21

Answer 1

def solution(roman)
mapping = {
"M"=>1000,
"D"=>500,
"C"=>100,
"L"=>50,
"X"=>10,
"V"=>5,
"I"=>1
}
# split string into characters
roman.chars.map do |l|
mapping[l] # replace character with integer value
end
.compact # removes potential nils caused by invalid chars
# Splits array into chunks so that we can handle numerals such as IIX
.chunk_while do |i,j|
i <= j #
end
# each chunk will be an array like [10, 10, 100] or [1, 1, 1, 1]
.map do |chunk|
if chunk.first < chunk.last
chunk.reverse.inject(:-) # handles numerals such as IIX with subtraction
else
chunk.sum # chunk is just a list of numerals such as III
end
end
.sum # sums everything up
end