Reduce a hash with arrays as values to a maximum number of array entries.

reducer.rb

#
# run like this:
# 
# > git clone https://gist.github.com/8350277.git reducer
# > cd reducer
# > gem install rspec -v 2.14.1
# > rspec reducer.rb
#
require "rspec"

class Reducer
  # Given a +hash+ of the form {user1: [1], user2: [2, 3]}, reduce the sum of all array entries
  # to +limit+, reducing the largest array first. Example: if limit is 2, the above hash
  # will be reduced to {user1: [1], user2: [2]}, since the sum of all array entries is 2.
  # 
  def self.reduce_to_max_items(hash, limit)
    while(hash.values.flatten.size > limit)
      # find the hash key which holds the most values. example:
      # if hash = {user1: [1], user2: [2, 3]}, then key is :user2, since it's array has the most items of all keys (2)
      key_with_most_values = hash.max { |(k1, v1), (k2, v2)| v1.size <=> v2.size }.first
      # remove the first value of that hash entry
      hash[key_with_most_values].shift
    end
    hash.reject { |k, v| v.empty? }
  end
end

describe "Reducer#reduce_to_max_items" do

  def reduce(hash, limit)
    Reducer.reduce_to_max_items(hash, limit)
  end

  context "Less or equal than 6 entries" do
    it "doesn't change anything when the entries belong to 1 user" do
      hash = { user1: [1,2,3,4,5,6] }
      reduce(hash, 6).should == hash
    end
    it "doesn't change anything when the entries belong to 2 users" do
      hash = { user1: [1,2,3], user2: [4,5,6] }
      reduce(hash, 6).should == hash
    end
  end

  context "More than 6 entries" do
    it "reduces to 6 when the entries belong to 1 user" do
      hash = { user1: [1,2,3,4,5,6,7] }
      reduce(hash, 6).should == { user1: [2,3,4,5,6,7] }
    end
    it "reduces to 6 when the entries belong to 2 users" do
      hash = { user1: [1,2,3], user2: [4,5,6,7] }
      reduce(hash, 6).should == { user1: [1,2,3], user2: [5,6,7] }
    end
    it "removes keys which would have empty arrays" do
      hash = { user1: [1], user2: [2], user3: [3] }
      reduce(hash, 2).should == { user2: [2], user3: [3] }
    end
  end
end