Ruby index of array elements

Question

Consider I have an array of integer elements. In which I am trying to find the index where a long sequence of repetitive numbers begins.

my_array = [100, 101, 100, 102, 100, 100, 101, 100, 250, 251, 253, 260, 250, 200, 100, 100, 100, 100, 100, 100, 100, 100, 100, 120]

Bellow is the way I am trying to find the index. Can anyone suggest me more optimised and correct way of doing it?

my_array.each with_index do |e, x|
  match = e.to_s * 5
  next_10 = my_array[x + 1, 5].join()

  if match == next_10
    puts "index #{x}"
    break
  end
end

#index 14

Answer 1

my_array.index.with_index{|value,index| my_array[index,6].uniq.size==1}

This is a kind of tweak, if you mean "optimised" just in the way code looks like.If you mean optimised performance. it does not fit.

Answer 2

In first iteration, I am getting array of repeating elements sequence and then I proceed further with logic,

groups = my_array[1..-1].inject([[my_array[0]]]) { |m, n| m.last[0] == n ? m.last << n : m << [n]; m }
# => [[100], [101], [100], [102], [100, 100], [101], [100], [250], [251], [253], [260], [250], [200], [100, 100, 100, 100, 100, 100, 100, 100, 100], [120]]

groups[0,groups.index(groups.sort { |a,b| a.count <=> b.count }.last)].flatten.count
# => 14

Using regex, it can be precise and simple.