python 3.2 - find second smallest number in a list using recursion

Question

So I need to find the second smallest number within a list of integers using recursion but I cannot for the life of me devise a way to do it. I can do it with to find smalle

Answer 1

This should work. get_smallest_two_items() returns a tuple with the smallest 2 elements out of the 3 inputted. Implementing it should be trivial. Also note that this assumes the minimum length of the list is 2.

def smallest(int_list):
   smallest_two = smallest_helper(int_list)
   return smallest_two[0] if smallest_two[0]<=smallest_two[1] else smallest_two[1]

def smallest_helper(int_list):

  if(len(int_list) == 2):
    return (int_list[0],int_list[1])
  else:
    a_b = smallest(int_list[1:])
    a = a_b[0]
    b = a_b[1]
    c = int_list[0]

    return get_smallest_two_items(a,b,c)