Exercise: Calculating the array mode: Write a method "mode" which takes an Array of numbers as its input and returns an Array of the most frequent values. If there's only one most-frequent value, it returns a single-element Array. For example, mode([1,2,3,3]) # => [3] mode([4.5, 0, 0]) # => [0] mode([1.5, -1, 1, 1.5]) # => [1.5] mode([1,1,2,2]) …

array_mode.rb

def mode(array)
  counter = Hash.new(0)
  # this creates a new, empty hash with no keys, but makes all defalt values zero. it will be used to store
  # the information from the array, such that the keys will be each unique number from the array (IOW, if there
  # are two or more 4's in the array, there will just be one key that is a 4), and the value for each key will
  # be the number of times that integer appears in the array.
  array.each do |i|
    counter[i] += 1
  end
  # this interates throught the array, and for each element it creates a key for that integer (if it hasn't been
  # created already) and basically pushes a +1 to the value of that key. remember, the values all start at zero
  # because the "Hash.new" language included "(0)" at the end.
  mode_array = []
  # this creates an empty array, to be filled with the most repeated number(s) from the array.
  # this will be the array that is returned at the end of the method.
  counter.each do |k, v|
    if v == counter.values.max
      mode_array << k
    end
  end
  # this loop iterates through the counter array. for each key/value pair, it compares the value to
  # the high value in the array, and if the value it's comparing is equal to the high value, it
  # adds to the mode array the key associated with that value.
  mode_array.sort
  # this returns the mode array (sorted)
end

You would get better runtime complexity if you moved counter.values.max outside of your counter.each loop by assigning it to a variable.

max_value = counter.values.max
counter.each do |k, v|
    if v == max_value
      mode_array << k
    end
  end