Given a Linked List, write a function that accepts the head node of the linked list as a parameter and returns the value of node present at (floor(sqrt(n)))th position in the Linked List, where n is the length of the linked list or the total number of nodes in the list.

Examples:

Input : 1->2->3->4->5->NULL Output : 2 Input : 10->20->30->40->NULL Output : 20 Input : 10->20->30->40->50->60->70->80->90->NULL Output : 30

**Simple method**: The simple method is to first find the total number of nodes present in the linked list, then find the value of floor(squareroot(n)) where n is the total number of nodes. Then traverse from the first node in the list to this position and return the node at this position.

This method traverses the linked list 2 times.

**Optimized approach**: In this method, we can get the required node by traversing the linked list once only. Below is the step by step algorithm for this approach.

- Initialize two counters i and j both to 1 and a pointer sqrtn to NULL to traverse til the required position is reached.
- Start traversing the list using head node until the last node is reached.
- While traversing check if the value of j is equal to sqrt(i). If the value is equal increment both i and j and sqrtn to point sqrtn->next otherwise increment only i.
- Now, when we will reach the last node of list i will contain value of n, j will contain value of sqrt(i) and sqrtn will point to node at jth position.

// C program to find sqrt(n)'th node // of a linked list #include<stdio.h> #include<stdlib.h> // Linked list node struct Node { int data; struct Node* next; }; // Function to get the sqrt(n)th // node of a linked list int printsqrtn(struct Node* head) { struct Node* sqrtn = NULL; int i = 1, j = 1; // Traverse the list while (head!=NULL) { // check if j = sqrt(i) if (i == j*j) { // for first node if (sqrtn == NULL) sqrtn = head; else sqrtn = sqrtn->next; // increment j if j = sqrt(i) j++; } i++; head=head->next; } // return node's data return sqrtn->data; } void print(struct Node* head) { while (head != NULL) { printf("%d ", head->data); head = head->next; } printf("\n"); } // function to add a new node at the // begining of the list void push(struct Node** head_ref, int new_data) { // allocate node struct Node* new_node = (struct Node*) malloc(sizeof(struct 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; } /* Driver program to test above function*/ int main() { /* Start with the empty list */ struct Node* head = NULL; push(&head, 40); push(&head, 30); push(&head, 20); push(&head, 10); printf("Given linked list is:"); print(head); printf("sqrt(n)th node is %d ",printsqrtn(head)); return 0; }

Output:

Given linked list is:10 20 30 40 sqrt(n)th node is 20

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