This notebook was prepared by Donne Martin. Source and license info is on GitHub.

Challenge Notebook

Problem: Find the kth to last element of a linked list.


  • Can we assume this is a non-circular, singly linked list?
    • Yes
  • Can we assume k is a valid integer?
    • Yes
  • If k = 0, does this return the last element?
    • Yes
  • What happens if k is greater than or equal to the length of the linked list?
    • Return None
  • Can you use additional data structures?
    • No
  • Can we assume we already have a linked list class that can be used for this problem?
    • Yes

Test Cases

  • Empty list -> None
  • k is >= the length of the linked list -> None
  • One element, k = 0 -> element
  • General case with many elements, k < length of linked list


Refer to the Solution Notebook. If you are stuck and need a hint, the solution notebook's algorithm discussion might be a good place to start.


In [ ]:
%run ../linked_list/
%load ../linked_list/
In [ ]:
class MyLinkedList(LinkedList):

    def kth_to_last_elem(self, k):
        # TODO: Implement me

Unit Test

The following unit test is expected to fail until you solve the challenge.

In [ ]:
# %load
import unittest

class Test(unittest.TestCase):

    def test_kth_to_last_elem(self):
        print('Test: Empty list')
        linked_list = MyLinkedList(None)
        self.assertEqual(linked_list.kth_to_last_elem(0), None)

        print('Test: k >= len(list)')
        self.assertEqual(linked_list.kth_to_last_elem(100), None)

        print('Test: One element, k = 0')
        head = Node(2)
        linked_list = MyLinkedList(head)
        self.assertEqual(linked_list.kth_to_last_elem(0), 2)

        print('Test: General case')
        self.assertEqual(linked_list.kth_to_last_elem(2), 3)

        print('Success: test_kth_to_last_elem')

def main():
    test = Test()

if __name__ == '__main__':

Solution Notebook

Review the Solution Notebook for a discussion on algorithms and code solutions.