### Problem Statement

Given a singly linked list along with a positive integer N. Write a function to delete the N^{th} node from the end of the given linked list.

### Problem Statement Understanding

Let’s understand the given problem with the help of some examples:

If our initial linked list:

It means that here we had to remove the N = 3^{rd} node from the end which is 3, so we will remove it.

After removing the 3^{rd} node from the end of the linked list, our output resultant linked list is

If our initial linked list:

It means that here we had to remove the N=1^{st} node from the end which is 6, so we will remove it.

After removing the 1^{st} node from the end of the linked list, our output resultant linked list is:

I hope that now you have understood this problem. Now, think of how we can approach this problem?

One simple approach will be to first find the length of linked list and then using it find the Nth node from the end. Let’s see how we will do this.

If length of our linked list is P, it means that we have to find (P-N)^{th} node from the beginning of the linked list and this node will be the N^{th} node from the end of the linked list.

So to delete N^{th} node using the above approach we will have to first traverse the linked list and find the length P of the linked list. Then we will have to again traverse the linked list (P-N+1) steps to reach the N^{th} node from the end.

This will work fine, but it is taking 2 traversals of linked list to find the N^{th} node from the end. Now try to think, do we actually require the length of linked list to find the N^{th} node from the end of the linked list?

So in approach we will see how we can do it in single traversal of linked list.

### Approach

Our approach would be simple.

To remove the N^{th} node from the end, we need two things, which are:

1) the N^{th} node from the end and

2) the node just before it.

First, we need to find the N^{th} node from the end of the linked list. There are many ways to do so. We will use the most efficient one, i.e., using 2 pointers.

We start with 2 pointers from the head.

- Move one of them to the N
^{th}node. - Now, both pointers are N nodes apart. If we move them forward together, that distance will be maintained.
- Now, move both the pointers together till the forward pointer points to the last node.
- Now, the backward pointer is N nodes far from the forward node. So, it will point to the nth node from the end. That is our required node to delete, so we delete it.

Think of some cases that we have to take care of while performing this operation, also try to implement the approach yourself.

Below is the step-by-step algorithm for the above approach.

### Algorithm

- Initialize 2 pointers
*i*and*j*pointing to head node. - Move the pointer
*j*, N steps ahead. - If we reach the end before reaching the N
^{th}node, this means N is greater than the length of the linked list (N^{th}node doesn’t exist). So, we simply stop and return. - Else declare a prev pointer to keep track of the node just before the node pointed by
*i*. - Now move both the pointers,
*i*and*j*, together(along with the prev pointe) till*j*reaches the end node. - Now,
*i*will be pointing to the N^{th}node from the end. - Now we have to remove it from the linked list.
- First, we simply disconnect the i
^{th}node from the linked list by doing**prev->next = i->next**. - If the node to remove is the first node, we need to modify the head pointer as well.
- At the end, delete
*i*.

### Dry Run

### Code Implementation:

#include#include struct Node { int data; struct Node* next; }; // since we are void insert_at_begenning ( struct Node **head_pointer, int data) { // allocate memory for new node struct Node *temp_node = (struct Node*) malloc(sizeof(struct Node)); // assign the data to new node temp_node->data = data; // initialize the new node to point to NULL temp_node->next = NULL; // if this is the first pointer, then this is the head pointer if (*head_pointer == NULL) { *head_pointer = temp_node; } else { // point the next of the present pointer to the head pointer temp_node->next = *head_pointer; //then move the reference of head pointer to the current pointer *head_pointer = temp_node; } } void display_list(struct Node **head_pointer) { // take a reference to head pointer for navigation struct Node *temp = *head_pointer; while(temp != NULL) { if(temp->next != NULL) printf("%d -> ", temp->data); else printf("%d", temp->data); //navigate to next pointer temp = temp->next; } printf("\n"); } struct Node * delete_node_from_end(struct Node *head, int num) { // initialize both slow_pointer and fast_pointer pointing to head pointer struct Node *slow_pointer = head, *fast_pointer = head; // move fast pointer num steps ahead for (int i = 0; i < num; i++) fast_pointer = fast_pointer->next; // while fast_pointer is not null, increment both pointers one step at a time while (fast_pointer->next) { fast_pointer = fast_pointer->next; slow_pointer = slow_pointer->next; } // once we get the node to be deleted, get that node, // store it in a local variable. This is because, it can be deleted later struct Node *node_to_be_deleted = slow_pointer->next; // link the slow pointer point to the next element slow_pointer->next = slow_pointer->next->next; // free the memory allocated for that pointer free(node_to_be_deleted); return head; } int main() { struct Node *head = NULL; insert_at_begenning(&head,80); insert_at_begenning(&head,70); insert_at_begenning(&head,60); insert_at_begenning(&head,50); insert_at_begenning(&head,40); insert_at_begenning(&head,30); insert_at_begenning(&head,20); insert_at_begenning(&head,10); printf("The present linked list is\n"); display_list(&head); head = delete_node_from_end(head, 2); printf("The linked list after deleting 2nd element from end is is\n"); display_list(&head); return 0; }

#includeusing namespace std; struct Node { int val; Node* next; Node(int value){ val = value; next = NULL; } }; void push_front(Node** head, int new_val){ Node* new_node = new Node(new_val); new_node->next = *head; *head = new_node; } void printList(Node* head){ Node* i = head; while(i){ cout< val<<" "; i = i->next; } cout<<"\n"; } void remove_nth_node_from_end(Node** head, int n){ Node *i, *j; i = j = *head; n--; // move j n steps ahead while(n>0 && j->next!=NULL){ j = j->next; n--; } if(n!=0){ cout<<"n is either <= 0 or larger than the linked list size\n"; return; } // move i and j together till j reaches the end Node *prev = i; while(j->next!=NULL){ prev = i; i = i->next; j = j->next; } // now i will point to the nth node from end // remove it prev->next = i->next; // if i points to first node // modify the head pointer as well if(i == *head){ *head = i->next; } delete i; } int main(){ Node* head = NULL; push_front(&head, 5); push_front(&head, 4); push_front(&head, 3); push_front(&head, 2); push_front(&head, 1); printList(head); // 1 2 3 4 5 // to remove 3rd node from end remove_nth_node_from_end(&head, 3); printList(head); // 1 2 4 5 }

class RemoveEnd { Node head; class Node { int data; Node next; Node(int d) { data = d; next = null; } } // Function to delete the nth node from // the end of the given linked list void deleteNode(int key) { // First pointer will point to // the head of the linked list Node first = head; // Second pointer will point to the // Nth node from the beginning Node second = head; for (int i = 0; i < key; i++) { // If count of nodes in the given // linked list is <= N if (second.next == null) { // If count = N i.e. delete the head node if (i == key - 1) head = head.next; return; } second = second.next; } // Increment both the pointers by one until // second pointer reaches the end while (second.next != null) { first = first.next; second = second.next; } // First must be pointing to the // Nth node from the end by now // So, delete the node first is pointing to first.next = first.next.next; } public void push(int new_data) { Node new_node = new Node(new_data); new_node.next = head; head = new_node; } public void printList() { Node tnode = head; while (tnode != null) { System.out.print(tnode.data + " "); tnode = tnode.next; } } public static void main(String[] args) { RemoveEnd llist = new RemoveEnd(); llist.push(7); llist.push(1); llist.push(3); llist.push(2); System.out.println("\nCreated Linked list is:"); llist.printList(); int N = 1; llist.deleteNode(N); System.out.println("\nLinked List after Deletion is:"); llist.printList(); } }

class Node: def __init__(self, data): self.data = data self.next = None class LinkedList: def __init__(self): self.head = None def create(self, x): new_node = Node(x) if self.head is None: self.head = new_node return last = self.head while last.next: last = last.next last.next = new_node def display(self): temp = self.head while temp: print(temp.data, end = " ") temp = temp.next def removeNthFromEnd(head, k): first = head second = head count = 1 while count <= k: second = second.next count += 1 if second is None: head.value = head.next.value head.next = head.next.next return while second.next is not None: first = first.next second = second.next first.next = first.next.next val = [1, 2, 3, 4, 5] k = 3 ll = LinkedList() for i in val: ll.create(i) print("Linked list before modification:"); ll.display() removeNthFromEnd(ll.head, k) print("\nLinked list after modification:"); ll.display()

### Output

1 2 3 4 5

1 2 4 5

[forminator_quiz id=”3861″]

**Space complexity:** O(1), as we are not using any extra space here.

Here N is the total number of nodes in the given linked list.

Through this article, we learned how to remove the nth node from the end of a linked list. Problems like these are good for strengthening your concepts in LinkedList. I would highly recommend you to practice more such problems from Linked List.