Find modular node in a linked list
Given a singly linked list and a number k, find the last node whose n%k == 0, where n is the number of nodes in the list.
Examples:
Input : list = 1->2->3->4->5->6->7 k = 3 Output : 6 Input : list = 3->7->1->9->8 k = 2 Output : 9
1. Take a pointer modularNode and initialize it with NULL. Traverse the linked list.
2. For every i%k=0, update modularNode.
C++
// C++ program to find modular 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 modular node in the linked list */ Node* modularNode(Node* head, int k) { // Corner cases if (k <= 0 || head == NULL) return NULL; // Traverse the given list int i = 1; Node* modularNode = NULL; for (Node* temp = head; temp != NULL; temp = temp->next) { if (i % k == 0) modularNode = temp; i++; } return modularNode; } /* 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; Node* answer = modularNode(head, k); printf ( "\nModular node is " ); if (answer != NULL) printf ( "%d\n" , answer->data); else printf ( "null\n" ); return 0; } |
Java
// A Java program to find modular node in a linked list public class GFG { // A Linkedlist node static class Node{ int data; Node next; Node( int data){ this .data = data; } } // Function to find modular node in the linked list static Node modularNode(Node head, int k) { // Corner cases if (k <= 0 || head == null ) return null ; // Traverse the given list int i = 1 ; Node modularNode = null ; for (Node temp = head; temp != null ; temp = temp.next) { if (i % k == 0 ) modularNode = temp; i++; } return modularNode; } // Driver code 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 ; Node answer = modularNode(head, k); System.out.print( "Modular node is " ); if (answer != null ) System.out.println(answer.data); else System.out.println( "null" ); } } // This code is contributed by Sumit Ghosh |
Python3
# Python3 program to find modular 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 modular node # in the linked list def modularNode(head, k): # Corner cases if (k < = 0 or head = = None ): return None # Traverse the given list i = 1 modularNode = None temp = head while (temp ! = None ): if (i % k = = 0 ): modularNode = temp i = i + 1 temp = temp. next return modularNode # 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 answer = modularNode(head, k) print ( "Modular node is" , end = ' ' ) if (answer ! = None ): print (answer.data, end = ' ' ) else : print ( "None" ) # This code is contributed by Srathore |
C#
// C# program to find modular node in a linked list using System; class GFG { // A Linkedlist node public class Node { public int data; public Node next; public Node( int data) { this .data = data; } } // Function to find modular node in the linked list static Node modularNode(Node head, int k) { // Corner cases if (k <= 0 || head == null ) return null ; // Traverse the given list int i = 1; Node modularNode = null ; for (Node temp = head; temp != null ; temp = temp.next) { if (i % k == 0) modularNode = temp; i++; } return modularNode; } // 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; Node answer = modularNode(head, k); Console.Write( "Modular node is " ); if (answer != null ) Console.WriteLine(answer.data); else Console.WriteLine( "null" ); } } // This code is contributed by Rajput-JI |
Javascript
<script> // A JavaScript program to find // modular node in a linked list // A Linkedlist node class Node { constructor(val) { this .data = val; this .next = null ; } } // Function to find modular node in the linked list function modularNode(head , k) { // Corner cases if (k <= 0 || head == null ) return null ; // Traverse the given list var i = 1; var modularNode = null ; for (temp = head; temp != null ; temp = temp.next) { if (i % k == 0) modularNode = temp; i++; } return modularNode; } // Driver code to test above function var 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); var k = 2; var answer = modularNode(head, k); document.write( "Modular node is " ); if (answer != null ) document.write(answer.data); else document.write( "null" ); // This code contributed by Rajput-Ji </script> |
Output:
Modular node is 4
Complexity Analysis:
Time Complexity: O(n).
Space complexity: O(1).
This article is contributed by Prakriti Gupta. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.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.