From a coding interviewâ€™s perspective Linked list plays an important role whether it be an online test or a face-to-face interview therefore there is a question â€śDelete the last occurrence of an item from the linked listâ€ť. Please think about the problem independently and try to figure out the approach on how to delete the last occurrence of an item from the linked list.

### Problem Statement

In this problem, we are given a linked list and asked to delete the last occurrence of an item X from the linked list.

### Problem Statement Understanding

We have been given a linked list and a value X. We have to delete the last occurrence of the item X from the linked list.

Letâ€™s understand this problem with an examples.

If the linked list given to us is: headâ†’1â†’2â†’7â†’5â†’2â†’10 and X = 2.

- According to the problem statement, first, we will have to find the last occurrence of
**X = 2**in the linked list, i.e., last node towards the end of the linked list which had**node -> data == X**, and then we will have to delete it from the linked list. - Our linked list after deleting the last occurrence of
**X = 2**will be: headâ†’1â†’2â†’7â†’5â†’10.

Suppose if the linked list is: headâ†’1â†’3â†’5â†’7â†’7â†’7 and X=7.

- Then, in this case, after deleting the last occurrence of X = 7 from the linked list our linked list will be headâ†’1â†’3â†’5â†’7â†’7.

##### Some more examples:

Sample Input 1: headâ†’1â†’1â†’3â†’5â†’1â†’9, X = 1.

Sample Output 1: headâ†’1â†’1â†’3â†’5â†’9

Sample Input 2: headâ†’1â†’3â†’3â†’5â†’3â†’9â†’3â†’5â†’3â†’3, X = 3.

Sample Output 2: headâ†’1â†’3â†’3â†’5â†’3â†’9â†’3â†’5â†’3

Now I think from the above example, the problem statement is clear. So let’s see how we will approach it.

Before moving to the approach section, try to think about how you can approach this problem.

- If stuck, no problem, we will thoroughly see how we can approach the problem in the next section.

Letâ€™s move to the approach section.

### Approach to delete the last occurrence of an item from the linked list

Our approach will be simple:

- The idea is to initialize a special pointer variable and start traversing through the linked list.
- During the traversal, Whenever the current nodeâ€™s data is matching with value X, i.e.,
**node -> data == X**, move the special pointer to the current node. - When we reach the end of the linked list, the special pointer will be pointing to the last occurrence of X in the linked list.
- Now we will have to delete the node to which special pointer is pointing.
- Finally, after deleting the node to which the special pointer points, our objective will be satisfied.

Let’s move to the algorithm section.

### Algorithm to delete the last occurrence of an item from the linked list

- Initialize a pointer variable named
**special**with NULL. - Start traversing through the linked list.
- If the current node data is equal to value X
**(curr->data == X)**, move the special pointer to this current node. - Continue the above step till the end of the linked list.
- When we reach the end, At that point of time, the special pointer points to the node that is the last occurring node with value X in the linked list.
- Now we have to delete that node.
- For deletion of that node, run a loop and reach the node prior to the node containing the last occurrence, i.e., the node to which special pointer is pointing.
- Now link the node before the special pointer and node after the special pointer together, and delete the special node.
- Finally, after deleting the node, our objective of deleting the last occurrence of X from the linked list will be satisfied.

### Dry Run on delete the last occurrence of an item from the linked list

### Implementation of delete the last occurrence of an item from the linked list

#include <stdio.h> #include <stdlib.h> // A linked list Node struct Node { int data; struct Node* next; }; void deleteLast(struct Node* head, int x) { struct Node *temp = head, *ptr = NULL; while (temp) { if (temp->data == x) ptr = temp; temp = temp->next; } // If the last occurrence is the last node if (ptr != NULL && ptr->next == NULL) { temp = head; while (temp->next != ptr) temp = temp->next; temp->next = NULL; } // If it is not the last node if (ptr != NULL && ptr->next != NULL) { ptr->data = ptr->next->data; temp = ptr->next; ptr->next = ptr->next->next; free(temp); } } struct Node* newNode(int x) { struct Node* node = malloc(sizeof(struct Node*)); node->data = x; node->next = NULL; return node; } void display(struct Node* head) { struct Node* temp = head; if (head == NULL) { printf("NULL\n"); return; } while (temp != NULL) { printf("%d -> ", temp->data); temp = temp->next; } printf("NULL\n"); } int main() { struct Node* head = newNode(1); head->next = newNode(2); head->next->next = newNode(7); head->next->next->next = newNode(5); head->next->next->next->next = newNode(2); head->next->next->next->next->next = newNode(10); printf("Input Linked list: "); display(head); deleteLast(head, 2); printf("List after deletion of 4: "); display(head); return 0; }

#include <bits/stdc++.h> using namespace std; // A linked list Node class Node { public: int data; Node* next; }; void deleteLast(Node* head, int X){ // Initialize special pointer with NULL Node* special = NULL; // Start from head and find the last Occurrence // of node with value X Node* temp = head; while (temp) { // If we found the key, update special if (temp->data == X) special = temp; temp = temp->next; } // key occurs at-least once if (special != NULL) { Node* removeNode = head; while(removeNode->next != special){ removeNode = removeNode->next; } removeNode->next = special->next; special->next = NULL; delete special; } } // function to create a new node with given data Node* newNode(int data){ Node* temp = new Node; temp->data = data; temp->next = NULL; return temp; } // function to print the linked list void printList(Node* node){ while (node != NULL) { cout<<node->data<<" "; node = node->next; } } int main(){ Node* head = newNode(1); head->next = newNode(2); head->next->next = newNode(7); head->next->next->next = newNode(5); head->next->next->next->next = newNode(2); head->next->next->next->next->next = newNode(10); cout<<"Input Linked List: "; printList(head);cout<<endl; deleteLast(head, 2); cout<<"Output Linked List: "; printList(head); return 0; }

class DeleteOccurances { static class Node { int data; Node next; }; // Function to delete the last occurrence static void deleteLast(Node head, int x) { Node temp = head, ptr = null; while (temp!=null) { // If found key, update if (temp.data == x) ptr = temp; temp = temp.next; } // If the last occurrence is the last node if (ptr != null && ptr.next == null) { temp = head; while (temp.next != ptr) temp = temp.next; temp.next = null; } // If it is not the last node if (ptr != null && ptr.next != null) { ptr.data = ptr.next.data; temp = ptr.next; ptr.next = ptr.next.next; System.gc(); } } static Node newNode(int x) { Node node = new Node(); node.data = x; node.next = null; return node; } static void display(Node head) { Node temp = head; if (head == null) { System.out.print("null\n"); return; } while (temp != null) { System.out.printf("%d --> ", temp.data); temp = temp.next; } System.out.print("null\n"); } /* Driver code*/ public static void main(String[] args) { Node head = newNode(11); head.next = newNode(2); head.next.next = newNode(3); head.next.next.next = newNode(4); head.next.next.next.next = newNode(5); head.next.next.next.next.next = newNode(4); head.next.next.next.next.next.next = newNode(4); System.out.print("Created Linked list: "); display(head); deleteLast(head, 4); System.out.print("List after deletion of 4: "); display(head); } }

class Node: def __init__(self, new_data): self.data = new_data self.next = None def deleteLast(head, x): temp = head ptr = None while (temp != None): if (temp.data == x): ptr = temp temp = temp.next if (ptr != None and ptr.next == None): temp = head while (temp.next != ptr): temp = temp.next temp.next = None if (ptr != None and ptr.next != None): ptr.data = ptr.next.data temp = ptr.next ptr.next = ptr.next.next return head def newNode(x): node = Node(0) node.data = x node.next = None return node def display(head): temp = head if (head == None): print("NULL\n") return while (temp != None): print( temp.data, end = " ") temp = temp.next print("NULL") head = newNode(1) head.next = newNode(2) head.next.next = newNode(7) head.next.next.next = newNode(5) head.next.next.next.next = newNode(2) head.next.next.next.next.next = newNode(10) print("Created Linked list: ", end = '') display(head) # Pass the address of the head pointer head = deleteLast(head, 2) print("List after deletion of 4: ", end = '') display(head)

#### Output

Input Linked List: 1 2 7 5 2 10

Output Linked List: 1 2 7 5 10

**Time Complexity:** O(N), Since we traversed through the list two times, One traversal for finding the last occurred node of value X in the linked list and another traversal for the deletion process.

In this blog, we have explained the problem of â€śHow to delete the last occurrence of an item from the linked listâ€ť very effectively and we will hope this blog will help you to enhance your knowledge regarding the linked list concepts. And if you want to solve more problems on Linked List, which is curated by our expert mentors at PrepBytes, you can follow this link Linked List.

## FAQs

**How do you remove the last element in a linked list in java?****How to find the last node in a linked list?**- Take a temp variable.
- Iterate the linked list, until the condition fails.
- temp.next != null
- The temp pointer points to the last node of the linked list.
**How do you delete an item from a linked list?**- Find the previous node of the node which we have to delete.
- Change the next pointer of the previous node.
- Free the memory for the node which will be deleted.

In Java, there is a method called LinkedList.removeLast() which is used to remove the last element from the linked list.

You can find the last node in such a way:

To delete a node from a linked list: