# Delete the last occurrence of an item from the linked list

### Introduction

The linked list is the most crucial concept and data structures to learn while preparing for interviews. Having a good grasp of Linked Lists can be a huge plus point in a coding interview.

### 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.

Let’s understand this problem with an examples.

If the linked list given to us is: head→1→2→7→5→2→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: head→1→2→7→5→10.

• Then, in this case, after deleting the last occurrence of X = 7 from the linked list our linked list will be head→1→3→5→7→7.
##### Some more examples:

Sample Input 1: head→1→1→3→5→1→9, X = 1.

Sample Input 2: head→1→3→3→5→3→9→3→5→3→3, X = 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.

Let’s move to the approach section.

### Approach

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 node’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

• 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   ### Code Implementation

```#include
using namespace std;

class Node {
public:
int data;
Node* next;
};

// Initialize special pointer with NULL
Node* special = NULL;

// Start from head and find the last Occurrence
// of node with value X
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) {
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<data<<" ";
node = node->next;
}
}

int main(){