#### Get Node Value Hackerrank

You’re given the pointer to the head node of a linked list and a specific position. Counting backwards from the tail node of the linked list, get the value of the node at the given position. A position of 0 corresponds to the tail, 1 corresponds to the node before the tail and so on.

Input Format

You have to complete the int GetNode(Node* head, int positionFromTail) method which takes two arguments – the head of the linked list and the position of the node from the tail. positionFromTail will be at least 0 and less than the number of nodes in the list. You should NOT read any input from stdin/console.

**Constraints **

Position will be a valid element in linked list.

**Output Format **

Find the node at the given position counting backwards from the tail. Then return the data contained in this node. Do NOT print anything to stdout/console.

**Sample Input**

1 -> 3 -> 5 -> 6 -> NULL, positionFromTail = 0

1 -> 3 -> 5 -> 6 -> NULL, positionFromTail = 2

**Sample Output**

6

3

##### Solution

To get the nth node from the tail of the linked list, we can calculate the length of the entire list by traversing the list once. Let this length be l. nth node from the tail is the (l-n-1)th node from the start (everything is 0-based) Or we can use two pointers. We can increment one of these to point to the nth node from the start and then increment both of these simultaneously till the first node reaches the end of the list. The second node points to the nth node from the tail.

Pseudocode:

GetNthNode(Node head, N)

ptr1=head //First Pointer

ptr2=head //Second Pointer

count=0

//Increment ptr1 till it points to Nth node from the head

while ptr1 is not NULL and count

count++

while ptr1->next is not NULL

ptr1=ptr1->next

ptr2=ptr2->next

//ptr2 points to the Nth node from the tail

return ptr2’s value

Statistics

/* Get Nth element from the end in a linked list of integers Number of elements in the list will always be greater than N. Node is defined as struct Node { int data; struct Node *next; } */ int GetNode(Node *head,int positionFromTail) { int index = 0; Node* current = head; Node* result = head; while(current!=NULL) { current=current->next; if (index++>positionFromTail) { result=result->next; } } return result->data; }

```
/*
Insert Node at the end of a linked list
head pointer input could be NULL as well for empty list
Node is defined as
class Node {
int data;
Node next;
}
*/
int GetNode(Node head,int n) {
// This is a "method-only" submission.
// You only need to complete this method.
if(head.next==null)
return head.data;
Node temp=head;
int count=0;
while(temp!=null)
{
count++;
temp=temp.next;
}
int k=count-n-1;
temp=head;
while(k>0)
{
k--;
temp=temp.next;
}
return temp.data;
}
```

```
#Body
"""
Get Node data of the Nth Node from the end.
head could be None as well for empty list
Node is defined as
class Node(object):
def __init__(self, data=None, next_node=None):
self.data = data
self.next = next_node
return back the node data of the linked list in the below method.
"""
def GetNode(head, position):
current = head
result = head
index = 0
while current is not None:
current = current.next
if index > position:
result = result.next
index+=1
return result.data
```