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