Skip to content
Related Articles

Related Articles

Find the fractional (or n/k – th) node in linked list
  • Difficulty Level : Easy
  • Last Updated : 28 Nov, 2019

Given a singly linked list and a number k, write a function to find the (n/k)-th element, where n is the number of elements in the list. We need to consider ceil value in case of decimals.

Examples:

Input : list = 1->2->3->4->5->6 
        k = 2
Output : 3
Since n = 6 and k = 2, we print (6/2)-th node 
which is 3.

Input : list = 2->7->9->3->5
        k = 3
Output : 7 
Since n is 5 and k is 3, we print ceil(5/3)-th 
node which is 2nd node, i.e., 7.

  1. Take two pointers temp and fractionalNode and initialize them with null and head respectively.
  2. For every k jumps of the temp pointer, make one jump of the fractionalNode pointer.

C++




// C++ program to find fractional node in a linked list
#include <bits/stdc++.h>
  
/* Linked list node */
struct Node {
    int data;
    Node* next;
};
  
/* Function to create a new node with given data */
Node* newNode(int data)
{
    Node* new_node = new Node;
    new_node->data = data;
    new_node->next = NULL;
    return new_node;
}
  
/* Function to find fractional node in the linked list */
Node* fractionalNodes(Node* head, int k)
{
    // Corner cases
    if (k <= 0 || head == NULL)
        return NULL;
  
    Node* fractionalNode = NULL;
      
    // Traverse the given list 
    int i = 0;
    for (Node* temp = head; temp != NULL; temp = temp->next) {
  
        // For every k nodes, we move fractionalNode one
        // step ahead. 
        if (i % k == 0) {
  
            // First time we see a multiple of k
            if (fractionalNode == NULL)
                fractionalNode = head;
  
            else
                fractionalNode = fractionalNode->next;
        }
        i++;
    }
    return fractionalNode;
}
  
// A utility function to print a linked list
void printList(Node* node)
{
    while (node != NULL) {
        printf("%d ", node->data);
        node = node->next;
    }
    printf("\n");
}
  
/* Driver program to test above function */
int main(void)
{
    Node* head = newNode(1);
    head->next = newNode(2);
    head->next->next = newNode(3);
    head->next->next->next = newNode(4);
    head->next->next->next->next = newNode(5);
    int k = 2;
  
    printf("List is ");
    printList(head);
  
    Node* answer = fractionalNodes(head, k);
    printf("\nFractional node is ");
    printf("%d\n", answer->data);
  
    return 0;
}


Java




// Java program to find fractional node in
// a linked list
public class FractionalNodell
{
    /* Linked list node */
    static class Node{
        int data;
        Node next;
  
        //Constructor
        Node (int data){
            this.data = data;
        }
    }
  
    /* Function to find fractional node in the
       linked list */
    static Node fractionalNodes(Node head, int k)
    {
        // Corner cases
        if (k <= 0 || head == null)
            return null;
  
        Node fractionalNode = null;
  
        // Traverse the given list
        int i = 0;
        for (Node temp = head; temp != null;
                          temp = temp.next){
  
            // For every k nodes, we move
            // fractionalNode one step ahead.
            if (i % k == 0){
  
                // First time we see a multiple of k
                if (fractionalNode == null)
                    fractionalNode = head;
                else
                    fractionalNode = fractionalNode.next;
            }
            i++;
        }
        return fractionalNode;
    }
  
    // A utility function to print a linked list
    static void printList(Node node)
    {
        while (node != null)
        {
            System.out.print(node.data+" ");
            node = node.next;
        }
        System.out.println();
    }
  
    /* Driver program to test above function */
    public static void main(String[] args) {
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        head.next.next.next = new Node(4);
        head.next.next.next.next = new Node(5);
        int k =2;
  
        System.out.print("List is ");
        printList(head);
  
        Node answer = fractionalNodes(head, k);
        System.out.println("Fractional node is "+
                                      answer.data);
    }
}
// This code is contributed by Sumit Ghosh


Python3




# Python3 program to find fractional node
# in a linked list
import math
  
# Linked list node
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None
  
# Function to create a new node 
# with given data
def newNode(data):
    new_node = Node(data)
    new_node.data = data
    new_node.next = None
    return new_node
  
# Function to find fractional node 
# in the linked list
def fractionalNodes(head, k):
      
    # Corner cases
    if (k <= 0 or head == None):
        return None
  
    fractionalNode = None
  
    # Traverse the given list
    i = 0
    temp = head
    while (temp != None):
          
        # For every k nodes, we move 
        # fractionalNode one step ahead. 
        if (i % k == 0):
  
            # First time we see a multiple of k
            if (fractionalNode == None):
                fractionalNode = head
  
            else:
                fractionalNode = fractionalNode.next
  
        i = i + 1
        temp = temp.next
  
    return fractionalNode
  
# A utility function to print a linked list
def prList(node):
    while (node != None):
        print(node.data, end = ' ')
        node = node.next
  
# Driver Code
if __name__ == '__main__':
    head = newNode(1)
    head.next = newNode(2)
    head.next.next = newNode(3)
    head.next.next.next = newNode(4)
    head.next.next.next.next = newNode(5)
    k = 2
  
    print("List is", end = ' ')
    prList(head)
  
    answer = fractionalNodes(head, k)
    print("\nFractional node is", end = ' ')
    print(answer.data)
  
# This code is contributed by Srathore


C#




// C# program to find fractional node in
// a linked list
using System;
  
public class FractionalNodell
{
    /* Linked list node */
    public class Node
    {
        public int data;
        public Node next;
  
        //Constructor
        public Node (int data)
        {
            this.data = data;
        }
    }
  
    /* Function to find fractional node in the
    linked list */
    static Node fractionalNodes(Node head, int k)
    {
        // Corner cases
        if (k <= 0 || head == null)
            return null;
  
        Node fractionalNode = null;
  
        // Traverse the given list
        int i = 0;
        for (Node temp = head; temp != null;
                        temp = temp.next)
        {
  
            // For every k nodes, we move
            // fractionalNode one step ahead.
            if (i % k == 0)
            {
  
                // First time we see a multiple of k
                if (fractionalNode == null)
                    fractionalNode = head;
                else
                    fractionalNode = fractionalNode.next;
            }
            i++;
        }
        return fractionalNode;
    }
  
    // A utility function to print a linked list
    static void printList(Node node)
    {
        while (node != null)
        {
            Console.Write(node.data+" ");
            node = node.next;
        }
        Console.WriteLine();
    }
  
    /* Driver code */
    public static void Main(String[] args) 
    {
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        head.next.next.next = new Node(4);
        head.next.next.next.next = new Node(5);
        int k =2;
  
        Console.Write("List is ");
        printList(head);
  
        Node answer = fractionalNodes(head, k);
        Console.WriteLine("Fractional node is "+
                                    answer.data);
    }
}
  
// This code is contributed by Rajput-Ji



Output:


List is 1 2 3 4 5 
Fractional node is 3

This article is contributed by Prakriti Gupta. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up
Recommended Articles
Page :