# Question: Implement an algorithm to find the kth to last element of a singly linked list.
from linked_list import Node
def delete_middle_node(n):
# Description: Update data of current node with data of next node and delete the next node. Assumption here is that input node is neither the first nor the last node of the singly linked list.
# Time: O(1)
# Space: O(1)
n.data = n.next.data
n.next = n.next.next
if __name__ == '__main__':
head = Node(1)
head.append_to_tail(5)
head.append_to_tail(9)
head.append_to_tail(12)
head.print()
n = head
while n.data != 9:
n = n.next
print(n.data)
delete_middle_node(n)
head.print()
# Description: Hint #126 - Can you do it iteratively? Imagine if you had two pointers pointing to adjacent nodes and they were moving at the same speed through the linked list. When one hits the end of the linked list, where will the other be?
# Time: O(N)
# Space: O(1)
p1 = head
p2 = head
for i in range(k):
if not p2.next:
return
p2 = p2.next
while p2.next:
p1 = p1.next
p2 = p2.next
return p1
if __name__ == '__main__':
n = None
for i in range(10):
if not n:
n = Node(i)
else:
n.append_to_tail(i)
n.print()
k = 3
print(get_kth_to_last_element(n, k).data)
