# Find nth node of the Linked list from the end

## Given a linked list, write a program for finding the nth element from the end of the linked list

### Example

Input:  6->16->15->50->1->23->49, N= 3

Output : 1

Here, the 3rd element from the end is 1

Time Complexity : O(n),

where n is the length of the linked list.

## Algorithm

This is one of the method to find the nth element from the end of a Linked List
1. Create two pointers first, second and initialize them to the head of the linked list
2. Move the second pointer N steps
3. Till second->next not equal to Null, move first and second pointer one step ahead.
4. first is the nth element from the end.

## C++ Program

``````#include <bits/stdc++.h>
using namespace std;

struct LL{
int data;
LL *next;
};

void insertAtBeginning(struct LL**head,int dataToBeInserted)
{
struct LL* curr=new LL;

curr->data=dataToBeInserted;
curr->next=NULL;

*head=curr; //if this is first node make this as head of list

else
{
curr->next=*head; //else make the current (new) node's next point to head and make this new node a the head
}

//O(1) constant time
}

struct LL* nthNodeFromEnd(struct LL**head,int n)
{
for(int i=1;i<n;i++) // increase second pointer ahead by n steps
{
if(second)
second=second->next;
else
break;
}
if(second == NULL)
{
cout<<"List has less than specified number of nodes\n";
return NULL;
}
while(second->next!=NULL) //increase by one steps now till we reach the last node and now the first will point to Nth from end
{
first = first->next;
second = second->next;
}

return first;
}

{
while(temp!=NULL)
{
if(temp->next!=NULL)
cout<<temp->data<<" ->";
else
cout<<temp->data;

temp=temp->next; //move to next node
}
//O(number of nodes)
cout<<endl;
}

int main()
{

struct LL *head = NULL; //initial list has no elements

int N = 3; //number of node whose value is to be known

struct LL * NthFromEnd =  nthNodeFromEnd(&head,N);
if(NthFromEnd == NULL)
{	}

else
cout<<N<<" th node from end is "<<NthFromEnd->data<<endl;

return 0;
}``````

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