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

# Challenge Notebook¶

## Constraints¶

• 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

## Algorithm¶

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.

## Code¶

In [ ]:
%run ../linked_list/linked_list.py
%load ../linked_list/linked_list.py

In [ ]:
class MyLinkedList(LinkedList):

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


## Unit Test¶

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

In [ ]:
# %load test_kth_to_last_elem.py
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')
linked_list.insert_to_front(1)
linked_list.insert_to_front(3)
linked_list.insert_to_front(5)
linked_list.insert_to_front(7)
self.assertEqual(linked_list.kth_to_last_elem(2), 3)

print('Success: test_kth_to_last_elem')

def main():
test = Test()
test.test_kth_to_last_elem()

if __name__ == '__main__':
main()


## Solution Notebook¶

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