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For our second problem, we’ll continue extending our `LinkedList`

class.

Write a method which returns the node that is `n`

nodes from the tail of the linked list.

If `n`

is `0`

, we would return the tail node, if `n`

is `1`

, we would return the second to last node, and so on.

# a -> b -> c -> d -> e linked_list.n_from_last(0) # 'e' node linked_list.n_from_last(3) # 'b' node

We’ll need to be creative in how we solve this problem since we only have a reference to the head node and not the tail!

The `.size()`

method on `LinkedList`

may be helpful.

To recap:

- write a method in the
`LinkedList`

class:`.n_from_last()`

. `.n_from_last()`

takes one argument:- the number of nodes
**counting from the tail**.

- the number of nodes
- return the node instance at that location.

### Instructions

**1.**

There are multiple ways we can solve this problem. Try to answer it on your own, and use the hint if you get stuck.

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