Squareroot(n)-th node in a Linked List
Last Updated :
25 Oct, 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.
- Initialize two counters i and j both to 1 and a pointer sqrtn to NULL to traverse till 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++
// C++ program to find sqrt(n)'th node
// of a linked list
#include <bits/stdc++.h>
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<<endl;
}
// function to add a new node at the
// beginning of the list
void push(Node** head_ref, int new_data)
{
// allocate node
Node* new_node = new Node();
// put in the data
new_node->data = new_data;
// link the old list of 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 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
// 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 of 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 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 of 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 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 of 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# 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 of 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 */
<script>
// JavaScript program to find sqrt(n)'th node
// of a linked list
// Linked list node
class Node {
constructor(val) {
this.data = val;
this.next = null;
}
}
var head = null;
// Function to get the sqrt(n)th
// node of a linked list
function printsqrtn(head) {
var sqrtn = null;
var 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;
}
function print(head) {
while (head != null) {
document.write(head.data + " ");
head = head.next;
}
document.write("<br/>");
}
// function to add a new node at the
// beginning of the list
function push(new_data) {
// allocate node
var new_node = new Node();
// put in the data
new_node.data = new_data;
// link the old list of the new node
new_node.next = head;
// move the head to point to the new node
head = new_node;
}
/* Driver code */
/* Start with the empty list */
push(40);
push(30);
push(20);
push(10);
document.write("Given linked list is:");
print(head);
document.write("sqrt(n)th node is " + printsqrtn(head));
// This code contributed by Rajput-Ji
</script>
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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