# Delete Nth node from the end of the given linked list

• Difficulty Level : Medium
• Last Updated : 18 Nov, 2021

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. Letâ€™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``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 Arnab Kundu`

## 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 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

 ``
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 ``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
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()`
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