commaifier.rb

def modulo_3?(number)
  # (number % 3).zero? 
  number.modulo(3).zero? # i guess this is more idiomatic Ruby, though I like the previous better
end

def commify(number)
  reversed = number.to_s.reverse
  number_size = number.to_s.size
  
  result = ""

  number_size.times do |i|
    result << reversed[i]
    result << "," if modulo_3?(i + 1)
  end

  result.reverse
end

# > commify(25164150)
# => "25,164,150"
# > commify(1234567890)
# => "1,234,567,890"

my_solution_too_large.rb

# goal: get a number like 25164150
# and transform it to: 25,164,150

# =begin
# 1) number.to_s.split("").length
# 2) start from end, count multiples of 3
# 3) insert comma
# 4) turn array back into a string
# .join(",")
# x.unshift(",")unless (x % 3 != 0)
# =end

# number = 11 223 344 556
kazu = [1,1,2,2,3,3,4,4,5,5,6]
# goal = 11,223,344,556

bloated = kazu.reverse.join.split("") # yields array of many strings
threes = 0.upto(kazu.length).find_all { |index| (index % 3).zero? }[1..-1]
# threes --> [3,6,9] # these are the comma-relevent indices

print "the solution is:  \n"
puts bloated[threes.last..-1].reverse.join + "," + threes.map { |n| (n-3)..(n-1)}.map {|r| bloated[r]}.map(&:join).join(',').reverse 

# STABBY LAMBDA TIME!!!!!!!!!!!
commify = ->(n) {
  bloated = n.to_s.reverse.split("")
  threes = 0.upto(n.to_s.split("").length).find_all { |index| (index % 3).zero? }[1..-1]
  bloated[threes.last..-1].reverse.join + "," + threes.map { |n| (n-3)..(n-1)}.map {|r| bloated[r]}.map(&:join).join(',').reverse 
}

puts "\n\nBut the COOL lambda solution is: #{commify[11223344556]}" + " !!!!!"

# 1 + 2
# 1.+(2)