# Program to find the K-th last node of a linked list in Python

Suppose we have a singly linked list, we have to check find the value of the kth last node (0-indexed). We have to solve this in single pass.

So, if the input is like node = [5,4,6,3,4,7], k = 2, then the output will be 3, as The second last (index 3) node has the value of 3.

To solve this, we will follow these steps −

• klast := node

• last := node

• for i in range 0 to k, do

• last := next of last

• while next of last is not null, do

• last := next of last

• klast := next of klast

• return value of klast

Let us see the following implementation to get better understanding −

## Example

Live Demo

class ListNode:
def __init__(self, data, next = None):
self.val = data
self.next = next

def make_list(elements):
for element in elements[1:]:
while ptr.next:
ptr = ptr.next
ptr.next = ListNode(element)

class Solution:
def solve(self, node, k):
klast = node
last = node
for i in range(k):
last = last.next
while last.next:
last = last.next
klast = klast.next
return klast.val

ob = Solution()
l1 = make_list([5,4,6,3,4,7])
print(ob.solve(l1, 2))

## Input

[5,4,6,3,4,7], 2

## Output

3