Squareroot(n)-th node in a Linked List


The linked list is one of the most important concepts 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 have been given a linked list. Our task is to write a function that accepts the head node of the linked list as a parameter and returns the value of the node present at (floor(sqrt(n)))th position in the linked list.

Here n is the count of the number of nodes present in the linked list.

Problem Statement Understanding

Let’s try to understand the problem with help of examples by referring online coding classes.

Suppose we are given a linked list A → B → C → D → E → F and now according to the problem statement we have to find the node at (floor(sqrt(n)))th position in the linked list and return its value as output.

For the above given linked list n = 6, as there are 6 nodes in the linked list, so

  • (floor(sqrt(n))) = floor(sqrt(6)) = 2, it means that we have to return the value of 2nd node of the linked list as output.

Output: B
Explanation: We will return B in output as B is the value of (floor(sqrt(n)))th node of the above given linked list.

If the linked list is:

For the above linked list n = 9 (as there are 9 nodes in the linked list), then the value of (floor(sqrt(n))) is:

  • (floor(sqrt(n))) = (floor(sqrt(9))) = 3, so we will return the value of 3rd node of the linked list as output.

Output: C

Note: Here in this problem we are considering 1 based indexing.

Some more Examples:

Input :

Output : 3

Input :

Output : 60

Input :

Output : 14

Now I think from the above examples the problem is clear, and now we have to think how can we approach it.

Approach 1

A simple approach is:

  • First, loop through the linked list to find the number of nodes n in the linked list.
  • Then calculate the value of floor(sqrt(n)), where n is the total number of nodes in the linked list.
  • Now starting from the first node of the list traverse to this position given by floor(sqrt(n)) and return the data of the node at this position.

Time Complexity: O(n), as we are traversing the complete length of the list
Space Complexity: O(1).

This approach traverses the linked list 2 times. Let's try to think, can we do better than this? Can we find out the (floor(sqrt(n)))th node of the linked list in one single traversal?

Let’s see the approach 2 to get the answers to the above questions.

Approach 2

As we know that:

  • sqrt(1) = 1
  • sqrt(4) = 2
  • sqrt(9) = 3
  • sqrt(16) = 4 ..........

One little conclusion which we can make about floor(sqrt(n)) is that:

  • Every number in range from 1 to 3 will have floor(sqrt(n)) equal to 1.
  • Every number in range from 4 to 8 will have floor(sqrt(n)) equal to 2.
  • Similarly, every number in range from 9 to 15 will have floor(sqrt(n)) equal to 3.
  • It goes on like this for range of numbers between every consequtive perfect squares.

Note: The final conclusion is that every number num between two consequtive perfect squares (a,b) (excluding the bigger perfect square b) have (floor(sqrt(num))) = sqrt(a).

Now we will try to utilize the above conclusion to solve this problem, and it will help us to find the (floor(sqrt(n)))th node of the linked list in single traversal.

The complete algorithm for this approach is explained below.


  • Firstly, we will initialize 2 pointers x and y both to 1 and a pointer req_node to NULL to traverse till the required position in the list is reached.
  • We will start traversing the list using the head node until the last node is reached.
  • While traversing the list, we will check if the value of y is equal to sqrt(x).
    • If the value is equal,we will increment both x and y by 1 and make req_node to req_node->next.
    • Otherwise, we will increment only x.
  • Now, when we reach the last node of the list x will contain the value of n, y will contain the value of sqrt(x) and req_node will point to the node at yth position.
  • Finally, we will return the data contained in the node pointed by pointer req_node.

Dry Run

Code Implementation

using namespace std;

class Node
    int data;
    Node* next;

int printsqrtn(Node* head)
    Node* req_node = NULL;
    int x = 1, y = 1;
    while (head!=NULL)
        if (x == y*y)
            if (req_node == NULL)
                req_node = head;
                req_node = req_node->next;
    return req_node->data;
void printList(Node* head)
    while (head != NULL)
        cout << head->data << " ";
        head = head->next;
    cout<data = new_data;
    new_node->next = (*head_ref);
    (*head_ref) = new_node;

int main()
    Node* head = NULL;
    push(&head, 4);
    push(&head, 3);
    push(&head, 2);
    push(&head, 1);
    cout << "Given linked list is: ";
    cout << "sqrt(n)th node is " << printsqrtn(head);
    return 0;


Given linked list is: 1 2 3 4
sqrt(n)th node is 2

Time Complexity: O(n), as we are traversing the complete length of the list.

So, in this article, you have learnt how to get the value of the node present at (floor(sqrt(n)))th position in the Linked List. If you want to solve more questions on Linked List, which are curated by our expert mentors at PrepBytes, you can follow this link Linked List.

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