Sort an array according to the elements of another array

I have an array of ids

a1 = [1, 2, 3, 4, 5]  

and I have another array of objects with ids in random order

a2 = [(obj_with_id_5), (obj_with_id_2), (obj_with_id_1), (obj_with_id_3), (obj_with_id_4)]  

Now I need to sort a2 according to the order of ids in a1. So a2 should now become:

[(obj_with_id_1), (id_2), (id_3), (id_4), (id_5)]  

a1 might be [3, 2, 5, 4, 1] or in any order but a2 should correspond to the order of ids in a1.

I do like this:

a1.each_with_index do |id, idx|
  found_idx = a1.find_index { |c| c.id == id }
  replace_elem = a2[found_idx]
  a2[found_idx] = a2[idx]
  a2[idx] = replace_elem
end  

But this still might run into an O(n^2) time if order of elements of a2 is exactly reverse of a1. Can someone please tell me the most efficient way of sorting a2?

I'll be surprised if anything is much faster than the obvious way:

a2.sort_by{|x| a1.index x.id}

hash_object = objects.each_with_object({}) do |obj, hash| 
  hash[obj.object_id] = obj
end

[1, 2, 3, 4, 5].map { |index| hash_object[index] }
#=> array of objects in id's order

I believe that the run time will be O(n)