Let LIST be a datatype for an implementation of linked list defined as follows:typedef struct list { int data; struct list *next; } LIST;Suppose a program has created two linked lists, L1 and L2, whose contents are given in the figure below (code for creating L1 and L2 is not provided here). L1 contains 9 nodes, and L2 contains 7 nodes. Consider the following C program segment that modifies the list L1. The number of nodes that will be there in L1 after the execution of the code segment is _________ . (Answer in integer)int find (int query, LIST *list) { while (list != NULL) { if(list->data == query) return 1; list = list->next; } return 0; } int main () { ... ... ... ptr1=L1; ptr2=L2; while (ptr1->next != NULL) { query = ptr1->next->data; if (find (query, L2)) ptr1->next = ptr1->next->next; else ptr1 = ptr1->next; } ... ... ... return 0; }

Initial List States:

• $\text{L1} = [1 \to 7 \to 12 \to 3 \to 9 \to 5 \to 11 \to 15 \to 8]$

• $\text{L2} = [1 \to 11 \to 6 \to 9 \to 15 \to 12 \to 4]$

• Elements present in $\text{L2}$ are $\{1, 4, 6, 9, 11, 12, 15\}$.

Functionality of the Code:

• The function find(query, L2) returns 1 if query is in $\text{L2}$, and 0 otherwise.

• The while (ptr1->next != NULL) loop in main() checks every node in $\text{L1}$ following ptr1 (i.e., starting from the 2nd node of $\text{L1}$):

  • If the next node's value is in $\text{L2}$, it bypasses/deletes it from $\text{L1}$ via ptr1->next = ptr1->next->next without advancing ptr1.

  • If it is not in $\text{L2}$, it advances ptr1 by ptr1 = ptr1->next.

Step-by-step Trace:

• Initially, ptr1 points to node with value $1$.

• Step 1:query = $7$ (next node). $7 \notin \text{L2} \implies$ ptr1 advances to $7$.

• Step 2:query = $12$. $12 \in \text{L2} \implies$ node $12$ is removed. ptr1 still points to $7$, and its next becomes $3$.

• Step 3:query = $3$. $3 \notin \text{L2} \implies$ ptr1 advances to $3$.

• Step 4:query = $9$. $9 \in \text{L2} \implies$ node $9$ is removed. ptr1 still points to $3$, and its next becomes $5$.

• Step 5:query = $5$. $5 \notin \text{L2} \implies$ ptr1 advances to $5$.

• Step 6:query = $11$. $11 \in \text{L2} \implies$ node $11$ is removed. ptr1 still points to $5$, and its next becomes $15$.

• Step 7:query = $15$. $15 \in \text{L2} \implies$ node $15$ is removed. ptr1 still points to $5$, and its next becomes $8$.

• Step 8:query = $8$. $8 \notin \text{L2} \implies$ ptr1 advances to $8$.

• Step 9:ptr1->next is NULL, loop terminates.

Final List $\text{L1}$:

The remaining number of nodes in $\text{L1}$ is 5.