Skip to content
Related Articles

Related Articles

Delete Nth node from the end of the given linked list

Improve Article
Save Article
Like Article
  • Difficulty Level : Medium
  • Last Updated : 04 May, 2022

Given a linked list and an integer N, the task is to delete the Nth node from the end of the given linked list.
 

Examples:  

Input: 2 -> 3 -> 1 -> 7 -> NULL, N = 1 
Output: 
The created linked list is: 
2 3 1 7 
The linked list after deletion is: 
2 3 1

Input: 1 -> 2 -> 3 -> 4 -> NULL, N = 4 
Output: 
The created linked list is: 
1 2 3 4 
The linked list after deletion is: 
2 3 4 

Intuition:

Lets  K be the total nodes in the linked list.

Observation : The Nth node from the end is (K-N+1)th node from the beginning.

So the problem simplifies down to that we have to find  (K-N+1)th node from the beginning.

  • One way of doing it is to find the length (K) of the linked list in one pass and then in the second pass move (K-N+1) step from the beginning to reach the Nth node from the end.
  • To do it in one pass. Let’s take the first pointer and move N step from the beginning. Now the first pointer is (K-N+1) steps away from the last node, which is the same number of steps the second pointer require to move from the beginning to reach the Nth node from the end.

Approach:  

  • Take two pointers; the first will point to the head of the linked list and the second will point to the Nth node from the beginning.
  • Now keep incrementing both the pointers by one at the same time until the second is pointing to the last node of the linked list.
  • After the operations from the previous step, the first pointer should point to the Nth node from the end now. So, delete the node the first pointer is pointing to.

Below is the implementation of the above approach: 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
class LinkedList {
public:
    // Linked list Node
    class Node {
    public:
        int data;
        Node* next;
        Node(int d)
        {
            data = d;
            next = NULL;
        }
    };
 
    // Head of list
    Node* head;
 
    // Function to delete the nth node from the end of the
    // given linked list
    Node* deleteNode(int key)
    {
        // We will be using this pointer for holding address
        // temporarily while we delete the node
        Node* temp;
 
        // 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) {
                    temp = head;
                    head = head->next;
                    free(temp);
                }
                return head;
            }
            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
        temp = first->next;
        first->next = first->next->next;
        free(temp);
        return head;
    }
 
    // Function to insert a new Node at front of the list
    Node* push(int new_data)
    {
        Node* new_node = new Node(new_data);
        new_node->next = head;
        head = new_node;
        return head;
    }
 
    // Function to print the linked list
    void printList()
    {
        Node* tnode = head;
        while (tnode != NULL) {
            cout << (tnode->data) << (" ");
            tnode = tnode->next;
        }
    }
};
 
// Driver code
int main()
{
    LinkedList* llist = new LinkedList();
 
    llist->head = llist->push(7);
    llist->head = llist->push(1);
    llist->head = llist->push(3);
    llist->head = llist->push(2);
 
    cout << ("Created Linked list is:\n");
    llist->printList();
 
    int N = 1;
    llist->head = llist->deleteNode(N);
 
    cout << ("\nLinked List after Deletion is:\n");
    llist->printList();
}
 
// This code is contributed by Sania Kumari Gupta

C




/* C program to merge two sorted linked lists */
#include <assert.h>
#include <stdio.h>
#include <stdlib.h>
 
/* Link list node */
typedef struct Node {
    int data;
    struct Node* next;
} Node;
 
Node* deleteNode(Node* head, int key)
{
   
    // We will be using this pointer for holding address
    // temporarily while we delete the node
    Node* temp;
 
    // 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) {
                temp = head;
                head = head->next;
                free(temp);
            }
            return head;
        }
        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
    temp = first->next;
    first->next = first->next->next;
    free(temp);
    return head;
}
 
/* Function to insert a node at the beginning of the
linked list */
void push(Node** head_ref, int new_data)
{
    /* allocate node */
    Node* new_node = (Node*)malloc(sizeof(Node));
    /* put in the data */
    new_node->data = new_data;
    /* link the old list off the new node */
    new_node->next = (*head_ref);
    /* move the head to point to the new node */
    (*head_ref) = new_node;
}
 
/* Function to print nodes in a given linked list */
void printList(struct Node* node)
{
    while (node != NULL) {
        printf("%d ", node->data);
        node = node->next;
    }
}
 
// Driver program
int main()
{
    struct Node* head = NULL;
    push(&head, 7);
    push(&head, 1);
    push(&head, 3);
    push(&head, 2);
    printf("Created Linked list is:\n");
    printList(head);
    int n = 1;
    deleteNode(head, n);
    printf("\nLinked List after Deletion is:\n");
    printList(head);
    return 0;
}
 
// This code is contributed by Sania Kumari Gupta

Java




// Java implementation of the approach
class LinkedList {
 
    // Head of list
    Node head;
 
    // Linked list Node
    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;
    }
 
    // Function to insert a new Node at front of the list
    public void push(int new_data)
    {
        Node new_node = new Node(new_data);
        new_node.next = head;
        head = new_node;
    }
 
    // Function to print the linked list
    public void printList()
    {
        Node tnode = head;
        while (tnode != null) {
            System.out.print(tnode.data + " ");
            tnode = tnode.next;
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        LinkedList llist = new LinkedList();
 
        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();
    }
}

Python3




# Python3 implementation of the approach
class Node:
    def __init__(self, new_data):
        self.data = new_data
        self.next = None
class LinkedList:
    def __init__(self):
        self.head = None
 
    # createNode and make linked list
    def push(self, new_data):
        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node
 
    def deleteNode(self, n):
        first = self.head
        second = self.head
        for i in range(n):
             
            # If count of nodes in the
            # given list is less than 'n'
            if(second.next == None):
                 
                # If index = n then
                # delete the head node
                if(i == n - 1):
                    self.head = self.head.next
                return self.head
            second = second.next
         
        while(second.next != None):
            second = second.next
            first = first.next
         
        first.next = first.next.next
     
    def printList(self):
        tmp_head = self.head
        while(tmp_head != None):
            print(tmp_head.data, end = ' ')
            tmp_head = tmp_head.next
         
# Driver Code
llist = LinkedList()
llist.push(7)
llist.push(1)
llist.push(3)
llist.push(2)
print("Created Linked list is:")
llist.printList()
llist.deleteNode(1)
print("\nLinked List after Deletion is:")
llist.printList()
 
# This code is contributed by RaviParkash

C#




// C# implementation of the approach
using System;
     
public class LinkedList
{
 
    // Head of list
    public Node head;
 
    // Linked list Node
    public class Node
    {
        public int data;
        public Node next;
        public 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;
    }
 
    // Function to insert a new Node at front of the list
    public void push(int new_data)
    {
        Node new_node = new Node(new_data);
        new_node.next = head;
        head = new_node;
    }
 
    // Function to print the linked list
    public void printList()
    {
        Node tnode = head;
        while (tnode != null)
        {
            Console.Write(tnode.data + " ");
            tnode = tnode.next;
        }
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        LinkedList llist = new LinkedList();
 
        llist.push(7);
        llist.push(1);
        llist.push(3);
        llist.push(2);
 
        Console.WriteLine("\nCreated Linked list is:");
        llist.printList();
 
        int N = 1;
        llist.deleteNode(N);
 
        Console.WriteLine("\nLinked List after Deletion is:");
        llist.printList();
    }
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
// javascript implementation of the approach
 
    // Head of list
    var head;
 
    // Linked list Node
    class Node {
        constructor(val) {
            this.data = val;
            this.next = null;
        }
    }
 
    // Function to delete the nth node from
    // the end of the given linked list
    function deleteNode(key) {
 
        // First pointer will point to
        // the head of the linked list
        var first = head;
 
        // Second pointer will point to the
        // Nth node from the beginning
        var second = head;
        for (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;
    }
 
    // Function to insert a new Node at front of the list
     function push(new_data) {
        var new_node = new Node(new_data);
        new_node.next = head;
        head = new_node;
    }
 
    // Function to print the linked list
    function printList() {
        var tnode = head;
        while (tnode != null) {
            document.write(tnode.data + " ");
            tnode = tnode.next;
        }
    }
 
    // Driver code
     
        push(7);
        push(1);
        push(3);
        push(2);
 
        document.write("\nCreated Linked list is:<br/>");
        printList();
 
        var N = 1;
        deleteNode(N);
 
        document.write("<br/>Linked List after Deletion is:<br/>");
        printList();
 
// This code is contributed by todaysgaurav
</script>
Output
Created Linked list is:
2 3 1 7 
Linked List after Deletion is:
2 3 1 

we already covered the iterative version above,
Now let us see its recursive approach as well,
 

Recursive Approach :
1) Create a dummy node and create a link from dummy node to head node. i.e, dummy->next = head 
2) Then we will use the recursion stack to keep track of elements that are being pushed in recursion calls.
3) While popping the elements from recursion stack, we will decrement the N(position of target node from the end of linked list) i.e, N = N-1.
4) When we reach (N==0) that means we have reached at the target node,
5) But here is the catch, to delete the target node we require its previous node,
6) So we will now stop when (N==-1) i.e, we reached the previous node.
7) Now it is very simple to delete the node by using previousNode->next = previousNode->next->next.
 

C++




// C++ implementation of the approach
// Code is contributed by Paras Saini
#include <bits/stdc++.h>
using namespace std;
class LinkedList {
public:
    int val;
    LinkedList* next;
 
    LinkedList()
    {
        this->next = NULL;
        this->val = 0;
    }
    LinkedList(int val)
    {
        this->next = NULL;
        this->val = val;
    }
 
    LinkedList* addNode(int val)
    {
        if (this == NULL) {
            return new LinkedList(val);
        }
        else {
            LinkedList* ptr = this;
            while (ptr->next) {
                ptr = ptr->next;
            }
            ptr->next = new LinkedList(val);
            return this;
        }
    }
 
    void removeNthNodeFromEndHelper(LinkedList* head,
                                    int& n)
    {
        if (!head)
            return;
        // Adding the elements in the recursion
        // stack
        removeNthNodeFromEndHelper(head->next, n);
        // Popping the elements from recursion stack
        n -= 1;
        // If we reach the previous of target node
        if (n == -1){
            LinkedList* temp = head->next;
            head->next = head->next->next;
            free (temp);
        }
    }
 
    LinkedList* removeNthNodeFromEnd(int n)
    {
        // return NULL if we have NULL head or only
        // one node.
        if (!this or !this->next)
            return NULL;
 
        // Create a dummy node and point its next to
        // head.
        LinkedList* dummy = new LinkedList();
        dummy->next = this;
 
        // Call function to remove Nth node from end
        removeNthNodeFromEndHelper(dummy, n);
 
        // Return new head i.e, dummy->next
        return dummy->next;
    }
 
    void printLinkedList()
    {
        if (!this) {
            cout << "Empty List\n";
            return;
        }
        LinkedList* ptr = this;
        while (ptr) {
            cout << ptr->val << " ";
            ptr = ptr->next;
        }
        cout << endl;
    }
};
 
class TestCase {
private:
    void printOutput(LinkedList* head)
    {
        // Output:
        if (!head)
            cout << "Empty Linked List\n";
        else
            head->printLinkedList();
    }
    void testCase1()
    {
        LinkedList* head = new LinkedList(1);
        head = head->addNode(2);
        head = head->addNode(3);
        head = head->addNode(4);
        head = head->addNode(5);
        head->printLinkedList(); // Print: 1 2 3 4 5
        head = head->removeNthNodeFromEnd(2);
        printOutput(head); // Output: 1 2 3 5
    }
 
    void testCase2()
    {
        // Important Edge Case, where linkedList [1]
        // and n=1,
        LinkedList* head = new LinkedList(1);
        head->printLinkedList(); // Print: 1
        head = head->removeNthNodeFromEnd(2);
        printOutput(head); // Output: Empty Linked List
    }
 
    void testCase3()
    {
        LinkedList* head = new LinkedList(1);
        head = head->addNode(2);
        head->printLinkedList(); // Print: 1 2
        head = head->removeNthNodeFromEnd(1);
        printOutput(head); // Output: 1
    }
 
public:
    void executeTestCases()
    {
        testCase1();
        testCase2();
        testCase3();
    }
};
 
int main()
{
    TestCase testCase;
    testCase.executeTestCases();
    return 0;
}
Output
1 2 3 4 5 
1 2 3 5 
1 
Empty Linked List
1 2 
1 

Two Pointer Approach – Slow and Fast Pointers

This problem can be solved by using two pointer approach as below:

  • Take two pointers – fast and slow. And initialize their values as head node
  • Iterate fast pointer till the value of n.
  • Now, start iteration of fast pointer till the None value of the linked list. Also, iterate slow pointer.
  • Hence, once the fast pointer will reach to the end the slow pointer will reach the node which you want to delete.
  • Replace the next node of the slow pointer with the next to next node of the slow pointer.

Python




class Node:
    def __init__(self, data):
        self.data = data
        self.next = None
 
 
class LinkedList:
    def __init__(self):
        self.head = None
 
    def push(self, data):
        new_node = Node(data)
        new_node.next = self.head
        self.head = new_node
 
    def display(self):
        temp = self.head
        while temp != None:
            print(temp.data)
            temp = temp.next
 
    def deleteNthNodeFromEnd(self, head, n):
        fast = head
        slow = head
 
        for _ in range(n):
            fast = fast.next
 
        if not fast:
            return head.next
 
        while fast.next:
            fast = fast.next
            slow = slow.next
 
        slow.next = slow.next.next
        return head
 
 
if __name__ == '__main__':
    l = LinkedList()
    l.push(5)
    l.push(4)
    l.push(3)
    l.push(2)
    l.push(1)
    print('***** Linked List Before deletion *****')
    l.display()
 
    print('************** Delete nth Node from the End *****')
    l.deleteNthNodeFromEnd(l.head, 2)
 
    print('*********** Linked List after Deletion *****')
    l.display()

Javascript




<script>
 
class Node{
    constructor(data){
        this.data = data
        this.next = null
    }
}
 
 
class LinkedList{
    constructor(){
        this.head = null
    }
 
    push(data){
        let new_node = new Node(data)
        new_node.next =this.head
        this.head = new_node
    }
 
    display(){
        let temp =this.head
        while(temp != null){
            document.write(temp.data,"</br>")
            temp = temp.next
        }
    }
 
    deleteNthNodeFromEnd(head, n){
        let fast = head
        let slow = head
 
        for(let i=0;i<n;i++){
            fast = fast.next
        }
 
        if(!fast)
            return head.next
 
        while(fast.next){
            fast = fast.next
            slow = slow.next
        }
 
        slow.next = slow.next.next
        return head
    }
}
 
 
// driver code
 
let l = new LinkedList()
l.push(5)
l.push(4)
l.push(3)
l.push(2)
l.push(1)
document.write('***** Linked List Before deletion *****',"</br>")
l.display()
 
document.write('************** Delete nth Node from the End *****',"</br>")
l.deleteNthNodeFromEnd(l.head, 2)
 
document.write('*********** Linked List after Deletion *****',"</br>")
l.display()
 
// This code is contributed by shinjanpatra
 
</script>
Output
***** Linked List Before deletion *****
1
2
3
4
5
************** Delete nth Node from the End *****
*********** Linked List after Deletion *****
1
2
3
5

Time complexity: O(n)
 


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!