# Squareroot(n)-th node in a Linked List

• Difficulty Level : Easy
• Last Updated : 21 Jun, 2022

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.

1. Initialize two counters i and j both to 1 and a pointer sqrtn to NULL to traverse til the required position is reached.
2. Start traversing the list using head node until the last node is reached.
3. 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.
4. 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++

 `// C++ program to find sqrt(n)'th node ` `// of a linked list `   `#include ` `using` `namespace` `std;`   `// Linked list node ` `class` `Node ` `{ ` `    ``public``:` `    ``int` `data; ` `    ``Node* next; ` `}; `   `// Function to get the sqrt(n)th ` `// node of a linked list ` `int` `printsqrtn(Node* head) ` `{ ` `    ``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(Node* head) ` `{ ` `    ``while` `(head != NULL) ` `    ``{ ` `        ``cout << head->data << ``" "``; ` `        ``head = head->next; ` `    ``} ` `    ``cout<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 */` `    ``Node* head = NULL; ` `    ``push(&head, 40); ` `    ``push(&head, 30); ` `    ``push(&head, 20); ` `    ``push(&head, 10); ` `    ``cout << ``"Given linked list is:"``; ` `    ``print(head); ` `    ``cout << ``"sqrt(n)th node is "` `<< printsqrtn(head); ` `    `  `    ``return` `0; ` `} `   `// This is code is contributed by rathbhupendra`

## C

 `// C program to find sqrt(n)'th node ` `// of a linked list`   `#include` `#include`   `// 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 ` `// beginning 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;` `}`

## Java

 `// Java program to find sqrt(n)'th node ` `// of a linked list `   `class` `GfG ` `{`   `// Linked list node ` `static` `class` `Node ` `{ ` `    ``int` `data; ` `    ``Node next; ` `}` `static` `Node head = ``null``;`   `// Function to get the sqrt(n)th ` `// node of a linked list ` `static` `int` `printsqrtn(Node head) ` `{ ` `    ``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; ` `} `   `static` `void` `print(Node head) ` `{ ` `    ``while` `(head != ``null``) ` `    ``{ ` `        ``System.out.print( head.data + ``" "``); ` `        ``head = head.next; ` `    ``} ` `    ``System.out.println(); ` `} `   `// function to add a new node at the ` `// beginning of the list ` `static` `void` `push( ``int` `new_data) ` `{ ` `    ``// allocate node ` `    ``Node new_node = ``new` `Node(); ` `    `  `    ``// put in the data ` `    ``new_node.data = new_data; ` `    `  `    ``// link the old list off the new node ` `    ``new_node.next = head; ` `    `  `    ``// move the head to point to the new node ` `    ``head = new_node; ` `} `   `/* Driver code*/` `public` `static` `void` `main(String[] args) ` `{ ` `    ``/* Start with the empty list */` `    ``push( ``40``); ` `    ``push( ``30``); ` `    ``push( ``20``); ` `    ``push( ``10``); ` `    ``System.out.print(``"Given linked list is:"``); ` `    ``print(head); ` `    ``System.out.print(``"sqrt(n)th node is "` `+` `                        ``printsqrtn(head)); ` `} ` `} `   `// This code is contributed by prerna saini`

## Python3

 `# Python3 program to find sqrt(n)'th node ` `# of a linked list `   `# Node class ` `class` `Node: `   `    ``# Function to initialise the node object ` `    ``def` `__init__(``self``, data): ` `        ``self``.data ``=` `data ` `        ``self``.``next` `=` `None`   `# Function to get the sqrt(n)th ` `# node of a linked list ` `def` `printsqrtn(head) :`   `    ``sqrtn ``=` `None` `    ``i ``=` `1` `    ``j ``=` `1` `    `  `    ``# Traverse the list ` `    ``while` `(head !``=` `None``) :` `    `  `        ``# check if j = sqrt(i) ` `        ``if` `(i ``=``=` `j ``*` `j) :` `        `  `            ``# for first node ` `            ``if` `(sqrtn ``=``=` `None``) :` `                ``sqrtn ``=` `head ` `            ``else``:` `                ``sqrtn ``=` `sqrtn.``next` `            `  `            ``# increment j if j = sqrt(i) ` `            ``j ``=` `j ``+` `1` `        `  `        ``i ``=` `i ``+` `1` `        `  `        ``head ``=` `head.``next` `    `  `    ``# return node's data ` `    ``return` `sqrtn.data `   `def` `print_1(head) :`   `    ``while` `(head !``=` `None``) :` `        ``print``( head.data, end ``=` `" "``) ` `        ``head ``=` `head.``next` `    ``print``(``" "``)`   `# function to add a new node at the ` `# beginning of the list ` `def` `push(head_ref, new_data) :`   `    ``# allocate node ` `    ``new_node ``=` `Node(``0``)` `    `  `    ``# 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 ` `    ``return` `head_ref`   `# Driver Code ` `if` `__name__``=``=``'__main__'``: `   `    ``# Start with the empty list ` `    ``head ``=` `None` `    ``head ``=` `push(head, ``40``) ` `    ``head ``=` `push(head, ``30``) ` `    ``head ``=` `push(head, ``20``) ` `    ``head ``=` `push(head, ``10``) ` `    ``print``(``"Given linked list is:"``) ` `    ``print_1(head) ` `    ``print``(``"sqrt(n)th node is "``, ` `              ``printsqrtn(head))` `    `  `# This code is contributed by Arnab Kundu`

## C#

 `// C# program to find sqrt(n)'th node ` `// of a linked list ` `using` `System;` `public` `class` `GfG ` `{ `   `// Linked list node ` `class` `Node ` `{ ` `    ``public` `int` `data; ` `    ``public` `Node next; ` `} ` `static` `Node head = ``null``; `   `// Function to get the sqrt(n)th ` `// node of a linked list ` `static` `int` `printsqrtn(Node head) ` `{ ` `    ``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; ` `} `   `static` `void` `print(Node head) ` `{ ` `    ``while` `(head != ``null``) ` `    ``{ ` `        ``Console.Write( head.data + ``" "``); ` `        ``head = head.next; ` `    ``} ` `    ``Console.WriteLine(); ` `} `   `// function to add a new node at the ` `// beginning of the list ` `static` `void` `push( ``int` `new_data) ` `{ ` `    ``// allocate node ` `    ``Node new_node = ``new` `Node(); ` `    `  `    ``// put in the data ` `    ``new_node.data = new_data; ` `    `  `    ``// link the old list off the new node ` `    ``new_node.next = head; ` `    `  `    ``// move the head to point to the new node ` `    ``head = new_node; ` `} `   `/* Driver code*/` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``/* Start with the empty list */` `    ``push( 40); ` `    ``push( 30); ` `    ``push( 20); ` `    ``push( 10); ` `    ``Console.Write(``"Given linked list is:"``); ` `    ``print(head); ` `    ``Console.Write(``"sqrt(n)th node is "` `+ ` `                        ``printsqrtn(head)); ` `} ` `} `   `/* This code is contributed by 29AjayKumar */`

## Javascript

 ``

Output:

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

Complexity Analysis:

Time Complexity: O(n).

Space Complexity: O(1).

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