Euler problem 1 - Generic Solution

# defines the list of integers from 'n' downto 0
def Zi (a, b)
   (a.downto b).lazy
end

# the equation numberOfOccurances(smallestMultiple + largestMultiple)/2 gives the sum of all the multiples for a given range
# so, below, multipleCounts[i](multiples[i] + lastMultiples[i])/2 => multipleCounts[i](lastAndFirstMultipleSums[i])/2
# gives the arbitrary sum of multiples of a give number in multiples for the specified range.
def find_sum_for_each_multiple(from, to, *multiples)
   myReverseArr = Zi to, from
   lastMultiples = multiples.map { |m| myReverseArr.find { |i| i%m == 0 } }
   multipleCounts = lastMultiples.zip(multiples).map { |ms| ms.reduce(:/) }
   lastAndFirstMultipleSums = lastMultiples.zip(multiples).map { |ms| ms.reduce(:+) }
   sumsOfEachMultiple = lastAndFirstMultipleSums.zip(multipleCounts).map { |xs| xs.reduce(:*) }.map { |x| x/2 }
end

def find_sum_of_multiples(from, to, *multiples)
   sumsOfEachMultiple = find_sum_for_each_multiple(from, to, multiples)

   commonMultiples = []
   (0..(multiples.length - 1)).each do |i|
      ((i+1)..(multiples.length - 1)).each do |j|
         commonMultiples << (multiples[i] * multiples[j])
      end
   end

   sumsOfCommonMultiples = find_sum_for_each_multiple(from, to, commonMultiples)

   totalSum = (sumsOfEachMultiple.inject { |sum, x| sum + x }) - (sumsOfCommonMultiples.inject { |sum, x| sum + x })
end

puts find_sum_of_multiples(0, 999, 3, 5)