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
// 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 listint 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 listvoid 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
// 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
# 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#
// 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 */ |
Javascript
<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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