Delete every kth node from a circular linked list until only one node is left. Also, print the intermediate lists.
Examples:
Input : n=4, k=2, list = 1->2->3->4 Output : 1->2->3->4->1 1->2->4->1 2->4->2 2->2 Input : n=9, k=4, list = 1->2->3->4->5->6->7->8->9 Output : 1->2->3->4->5->6->7->8->9->1 1->2->3->4->6->7->8->9->1 1->2->3->4->6->7->8->1 1->2->3->6->7->8->1 2->3->6->7->8->2 2->3->6->8->2 2->3->8->2 2->3->2 2->2
Algorithm:
Repeat the following steps until there is only one node left in the list.
- Case 1: The list is empty.
If the list is empty, simply return it. - Case 2: The list has only one node.
If the list has only one node left, we will print the list and return it as our goal is reached. - Case 3: The list has more than one node.
Define two pointers curr and prev and initialize the pointer curr with the head node.
Traverse the list using curr pointer by iterating it k times.
- The node to be deleted is the first node of the list.
Conditions to check this( curr == head && curr->next == head).
If yes, then move prev until it reaches the last node. After prev reaches the last node, set head = head -> next and prev -> next = head. Delete curr. - The node to be deleted is the last node in the list.
The condition to check this is (curr -> next == head).
If curr is the last node. Set prev -> next = head and delete the node curr for free(curr). - The one to be deleted is neither the first node nor the last node, then set prev -> next = temp -> next and delete curr.
Implementation:
C++
// C++ program to delete every kth Node from// circular linked list.#include <bits/stdc++.h>using namespace std;/* structure for a Node */struct Node { int data; Node* next; Node(int x) { data = x; next = NULL; }};/*Utility function to print the circular linked list*/void printList(Node* head){ if (head == NULL) return; Node* temp = head; do { cout << temp->data << "->"; temp = temp->next; } while (temp != head); cout << head->data << endl;}/*Function to delete every kth Node*/void deleteK(Node** head_ref, int k){ Node* head = *head_ref; // If list is empty, simply return. if (head == NULL) return; // take two pointers - current and previous Node *curr = head, *prev; while (true) { // Check if Node is the only Node\ // If yes, we reached the goal, therefore // return. if (curr->next == head && curr == head) break; // Print intermediate list. printList(head); // If more than one Node present in the list, // Make previous pointer point to current // Iterate current pointer k times, // i.e. current Node is to be deleted. for (int i = 0; i < k; i++) { prev = curr; curr = curr->next; } // If Node to be deleted is head if (curr == head) { prev = head; while (prev->next != head) prev = prev->next; head = curr->next; prev->next = head; *head_ref = head; free(curr); } // If Node to be deleted is last Node. else if (curr->next == head) { prev->next = head; free(curr); } else { prev->next = curr->next; free(curr); } }}/* Function to insert a Node at the end of a Circular linked list */void insertNode(Node** head_ref, int x){ // Create a new Node Node* head = *head_ref; Node* temp = new Node(x); // if the list is empty, make the new Node head // Also, it will point to itself. if (head == NULL) { temp->next = temp; *head_ref = temp; } // traverse the list to reach the last Node // and insert the Node else { Node* temp1 = head; while (temp1->next != head) temp1 = temp1->next; temp1->next = temp; temp->next = head; }}/* Driver program to test above functions */int main(){ // insert Nodes in the circular linked list struct Node* head = NULL; insertNode(&head, 1); insertNode(&head, 2); insertNode(&head, 3); insertNode(&head, 4); insertNode(&head, 5); insertNode(&head, 6); insertNode(&head, 7); insertNode(&head, 8); insertNode(&head, 9); int k = 4; // Delete every kth Node from the // circular linked list. deleteK(&head, k); return 0;} |
Java
// Java program to delete every kth Node from // circular linked list. class GFG{ /* structure for a Node */static class Node{ int data; Node next; Node(int x) { data = x; next = null; } }; /*Utility function to print the circular linked list*/static void printList(Node head) { if (head == null) return; Node temp = head; do { System.out.print( temp.data + "->"); temp = temp.next; } while (temp != head); System.out.println(head.data ); } /*Function to delete every kth Node*/static Node deleteK(Node head_ref, int k) { Node head = head_ref; // If list is empty, simply return. if (head == null) return null; // take two pointers - current and previous Node curr = head, prev=null; while (true) { // Check if Node is the only Node\ // If yes, we reached the goal, therefore // return. if (curr.next == head && curr == head) break; // Print intermediate list. printList(head); // If more than one Node present in the list, // Make previous pointer point to current // Iterate current pointer k times, // i.e. current Node is to be deleted. for (int i = 0; i < k; i++) { prev = curr; curr = curr.next; } // If Node to be deleted is head if (curr == head) { prev = head; while (prev.next != head) prev = prev.next; head = curr.next; prev.next = head; head_ref = head; } // If Node to be deleted is last Node. else if (curr.next == head) { prev.next = head; } else { prev.next = curr.next; } } return head;} /* Function to insert a Node at the end of a Circular linked list */static Node insertNode(Node head_ref, int x) { // Create a new Node Node head = head_ref; Node temp = new Node(x); // if the list is empty, make the new Node head // Also, it will point to itself. if (head == null) { temp.next = temp; head_ref = temp; return head_ref; } // traverse the list to reach the last Node // and insert the Node else { Node temp1 = head; while (temp1.next != head) temp1 = temp1.next; temp1.next = temp; temp.next = head; } return head;} /* Driver code */public static void main(String args[]){ // insert Nodes in the circular linked list Node head = null; head = insertNode(head, 1); head = insertNode(head, 2); head = insertNode(head, 3); head = insertNode(head, 4); head = insertNode(head, 5); head = insertNode(head, 6); head = insertNode(head, 7); head = insertNode(head, 8); head = insertNode(head, 9); int k = 4; // Delete every kth Node from the // circular linked list. head = deleteK(head, k); } }// This code is contributed by Arnab Kundu |
Python3
# Python3 program to delete every kth Node from# circular linked list.import math# structure for a Node class Node: def __init__(self, data): self.data = data self.next = None# Utility function to print the circular linked listdef printList(head): if (head == None): return temp = head print(temp.data, end = "->") temp = temp.next while (temp != head): print(temp.data, end = "->") temp = temp.next print(head.data)# Function to delete every kth Nodedef deleteK(head_ref, k): head = head_ref # If list is empty, simply return. if (head == None): return # take two pointers - current and previous curr = head prev = None while True: # Check if Node is the only Node\ # If yes, we reached the goal, therefore # return. if (curr.next == head and curr == head): break # Print intermediate list. printList(head) # If more than one Node present in the list, # Make previous pointer point to current # Iterate current pointer k times, # i.e. current Node is to be deleted. for i in range(k): prev = curr curr = curr.next # If Node to be deleted is head if (curr == head): prev = head while (prev.next != head): prev = prev.next head = curr.next prev.next = head head_ref = head # If Node to be deleted is last Node. elif (curr.next == head) : prev.next = head else : prev.next = curr.next# Function to insert a Node at the end of #a Circular linked list def insertNode(head_ref, x): # Create a new Node head = head_ref temp = Node(x) # if the list is empty, make the new Node head # Also, it will po to itself. if (head == None): temp.next = temp head_ref = temp return head_ref # traverse the list to reach the last Node # and insert the Node else : temp1 = head while (temp1.next != head): temp1 = temp1.next temp1.next = temp temp.next = head return head# Driver Codeif __name__=='__main__': # insert Nodes in the circular linked list head = None head = insertNode(head, 1) head = insertNode(head, 2) head = insertNode(head, 3) head = insertNode(head, 4) head = insertNode(head, 5) head = insertNode(head, 6) head = insertNode(head, 7) head = insertNode(head, 8) head = insertNode(head, 9) k = 4 # Delete every kth Node from the # circular linked list. deleteK(head, k)# This code is contributed by Srathore |
C#
// C# program to delete every kth Node from // circular linked list.using System;class GFG{ /* structure for a Node */public class Node{ public int data; public Node next; public Node(int x) { data = x; next = null; } }; /*Utility function to print the circular linked list*/static void printList(Node head) { if (head == null) return; Node temp = head; do { Console.Write( temp.data + "->"); temp = temp.next; } while (temp != head); Console.WriteLine(head.data ); } /*Function to delete every kth Node*/static Node deleteK(Node head_ref, int k) { Node head = head_ref; // If list is empty, simply return. if (head == null) return null; // take two pointers - current and previous Node curr = head, prev = null; while (true) { // Check if Node is the only Node\ // If yes, we reached the goal, therefore // return. if (curr.next == head && curr == head) break; // Print intermediate list. printList(head); // If more than one Node present in the list, // Make previous pointer point to current // Iterate current pointer k times, // i.e. current Node is to be deleted. for (int i = 0; i < k; i++) { prev = curr; curr = curr.next; } // If Node to be deleted is head if (curr == head) { prev = head; while (prev.next != head) prev = prev.next; head = curr.next; prev.next = head; head_ref = head; } // If Node to be deleted is last Node. else if (curr.next == head) { prev.next = head; } else { prev.next = curr.next; } } return head;} /* Function to insert a Node at the end of a Circular linked list */static Node insertNode(Node head_ref, int x) { // Create a new Node Node head = head_ref; Node temp = new Node(x); // if the list is empty, make the new Node head // Also, it will point to itself. if (head == null) { temp.next = temp; head_ref = temp; return head_ref; } // traverse the list to reach the last Node // and insert the Node else { Node temp1 = head; while (temp1.next != head) temp1 = temp1.next; temp1.next = temp; temp.next = head; } return head;} /* Driver code */public static void Main(String []args){ // insert Nodes in the circular linked list Node head = null; head = insertNode(head, 1); head = insertNode(head, 2); head = insertNode(head, 3); head = insertNode(head, 4); head = insertNode(head, 5); head = insertNode(head, 6); head = insertNode(head, 7); head = insertNode(head, 8); head = insertNode(head, 9); int k = 4; // Delete every kth Node from the // circular linked list. head = deleteK(head, k); } }// This code has been contributed by 29AjayKumar |
Javascript
<script>// javascript program to delete every kth Node from // circular linked list. /* structure for a Node */ class Node { constructor(val) { this.data = val; this.next = null; } } /* * Utility function to print the circular linked list */ function printList(head) { if (head == null) return; var temp = head; do { document.write(temp.data + "->"); temp = temp.next; } while (temp != head); document.write(head.data+"<br/>"); } /* Function to delete every kth Node */ function deleteK(head_ref , k) { var head = head_ref; // If list is empty, simply return. if (head == null) return null; // take two pointers - current and previous var curr = head, prev = null; while (true) { // Check if Node is the only Node\ // If yes, we reached the goal, therefore // return. if (curr.next == head && curr == head) break; // Print intermediate list. printList(head); // If more than one Node present in the list, // Make previous pointer point to current // Iterate current pointer k times, // i.e. current Node is to be deleted. for (i = 0; i < k; i++) { prev = curr; curr = curr.next; } // If Node to be deleted is head if (curr == head) { prev = head; while (prev.next != head) prev = prev.next; head = curr.next; prev.next = head; head_ref = head; } // If Node to be deleted is last Node. else if (curr.next == head) { prev.next = head; } else { prev.next = curr.next; } } return head; } /* * Function to insert a Node at the end of a Circular linked list */ function insertNode(head_ref , x) { // Create a new Node var head = head_ref; var temp = new Node(x); // if the list is empty, make the new Node head // Also, it will point to itself. if (head == null) { temp.next = temp; head_ref = temp; return head_ref; } // traverse the list to reach the last Node // and insert the Node else { var temp1 = head; while (temp1.next != head) temp1 = temp1.next; temp1.next = temp; temp.next = head; } return head; } /* Driver code */ // insert Nodes in the circular linked list var head = null; head = insertNode(head, 1); head = insertNode(head, 2); head = insertNode(head, 3); head = insertNode(head, 4); head = insertNode(head, 5); head = insertNode(head, 6); head = insertNode(head, 7); head = insertNode(head, 8); head = insertNode(head, 9); var k = 4; // Delete every kth Node from the // circular linked list. head = deleteK(head, k);// This code is contributed by todaysgaurav</script> |
Output
1->2->3->4->5->6->7->8->9->1 1->2->3->4->6->7->8->9->1 1->2->3->4->6->7->8->1 1->2->3->6->7->8->1 2->3->6->7->8->2 2->3->6->8->2 2->3->8->2 2->3->2 2->2
Complexity Analysis:
- Time Complexity: O(n*n), as we are using nested loops to traverse n*n times. for deleting and printing the linked list. Where n is the number of nodes in the linked list.
- Auxiliary Space: O(1), as we are not using any extra space.
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