Question

I have a list of email address which I want to distribute evenly by domain.

For example:

let the list be,

[e-mail] [e-mail] [e-mail] [e-mail] [e-mail] [e-mail] 

The output should be

[e-mail] [e-mail] [e-mail] [e-mail] [e-mail] [e-mail] 

The source list is not sorted by domain as in example, but can be sorted by domain, if that can help. What would be an efficient (single/two pass?) algorithm of doing this?

raj

Answer 1

Here is my solution to this problem. Given an input like

[['A', 13], ['B', 5], ['C', 3], ['D', 1]]  

The output is

AABABAACABAACABAACABAD 

Ruby source of the program is:

require 'pp'  def shuffle (total, num)   ret_arr = Array.new   intervel = total/num.to_f   0.upto(num-1) do |i|     val = i * intervel     ret_arr << val.floor   end   return ret_arr end  freq_table = [['A', 13], ['B', 5], ['C', 3], ['D', 1]]  pp freq_table total = 0 freq_table.collect {|i| total += i[1] } final_array = Array.new(total,0) print  final_array.to_s + '\n\n' placed = 0  freq_table.each do |i|   placements =  shuffle(total - placed, i[1])   placements.each do |j|     free_index = -1     0.upto final_array.size do |k|       free_index += 1 if (final_array[k] == 0 || final_array[k] == i[0])       if j == free_index         final_array[k] = i[0]         break       end     end   end   print 'Placing #{i[1]} #{i[0]}s over #{total - placed} positions\n'   pp placements   print  final_array.to_s + '\n\n'   placed += i[1] end 

The idea is to take the alphabet with highest frequency and distribute it first across an array whose size is the total number of all elements. Next distribute the alphabet with second highest frequency and distribute it across the free spaces and so on.

If you have questions let me know and I am happy to answer.

raj