XOR Linked List – Reverse a Linked List in groups of given size

Last Updated : 18 Jan, 2024

Given a XOR linked list and an integer K, the task is to reverse every K nodes in the given XOR linked list.

Examples:

Input: XLL = 7< – > 6 < – > 8 < – > 11 < – > 3, K = 3
Output: 8 < – > 6 < – > 7 < – > 3 < – > 11
Explanation:
Reversing first K(= 3) nodes modifies the Linked List to 8 < – > 6 < – > 7 < – > 11 < – > 3.
Reversing remaining nodes of the Linked List to 8 < – > 6 < – > 7 < – > 3 < – > 11.
Therefore, the required output is 8 < – > 6 < – > 7 < – > 3 < – > 11.

Input: XLL = 7 < – > 6 < – > 8 < –> 11 < – > 3 < – > 1 < – > 2 < – > 0, K = 3
Output: 8 < – > 6 < – > 7 < – > 1 < – > 3 < – > 11 < – > 0 < – > 2

Approach: The idea is to recursively reverse every K nodes of the XOR linked list in a group and connect the first node of every group of K nodes to the last node of its previous group of nodes. The recursive function is as follows:

RevInGrp(head, K, N)
{
reverse(head, min(K, N))
if (N < K) {
return head
}
head->next = RevGInGrp(next, K, N – K)
}

Follow the steps below to solve the problem:

Reverse the first K nodes of the XOR linked list and recursively reverse the remaining nodes in a group of size K. If the count of remaining nodes is less than K, then just reverse the remaining nodes.
Finally, connect the first node of every group to the last node of its previous group.

Below is the implementation of the above approach:

C++

   // C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
#include <inttypes.h>
using namespace std;
 
// Structure of a node
// in XOR linked list
struct Node {
 
    // Stores data value
    // of a node
    int data;
 
    // Stores XOR of previous
    // pointer and next pointer
    struct Node* nxp;
};
 
// Function to find the XOR of address
// of two nodes
struct Node* XOR(struct Node* a, struct Node* b)
{
    return (struct Node*)((uintptr_t)(a) ^ (uintptr_t)(b));
}
 
// Function to insert a node with
// given value at given position
struct Node* insert(struct Node** head, int value)
{
 
    // If XOR linked list is empty
    if (*head == NULL) {
 
        // Initialize a new Node
        struct Node* node = new Node;
 
        // Stores data value in
        // the node
        node->data = value;
 
        // Stores XOR of previous
        // and next pointer
        node->nxp = XOR(NULL, NULL);
 
        // Update pointer of head node
        *head = node;
    }
 
    // If the XOR linked list
    // is not empty
    else {
 
        // Stores the address
        // of current node
        struct Node* curr = *head;
 
        // Stores the address
        // of previous node
        struct Node* prev = NULL;
 
        // Initialize a new Node
        struct Node* node
            = new Node();
 
        // Update curr node address
        curr->nxp = XOR(node, XOR(NULL, curr->nxp));
 
        // Update new node address
        node->nxp = XOR(NULL, curr);
 
        // Update head
        *head = node;
 
        // Update data value of
        // current node
        node->data = value;
    }
    return *head;
}
 
// Function to print elements of
// the XOR Linked List
void printList(struct Node** head)
{
 
    // Stores XOR pointer
    // in current node
    struct Node* curr = *head;
 
    // Stores XOR pointer of
    // in previous Node
    struct Node* prev = NULL;
 
    // Stores XOR pointer of
    // in next node
    struct Node* next;
 
    // Traverse XOR linked list
    while (curr != NULL) {
 
        // Print current node
        cout << curr->data << " ";
 
        // Forward traversal
        next = XOR(prev, curr->nxp);
 
        // Update prev
        prev = curr;
 
        // Update curr
        curr = next;
    }
}
 
// Reverse the linked list in group of K
struct Node* RevInGrp(struct Node** head, int K, int len)
{
 
    // Stores head node
    struct Node* curr = *head;
 
    // If the XOR linked
    // list is empty
    if (curr == NULL)
        return NULL;
 
    // Stores count of nodes
    // reversed in current group
    int count = 0;
 
    // Stores XOR pointer of
    // in previous Node
    struct Node* prev = NULL;
 
    // Stores XOR pointer of
    // in next node
    struct Node* next;
 
    // Reverse nodes in current group
    while (count < K && count < len) {
 
        // Forward traversal
        next = XOR(prev, curr->nxp);
 
        // Update prev
        prev = curr;
 
        // Update curr
        curr = next;
 
        // Update count
        count++;
    }
 
    // Disconnect prev node from the next node
    prev->nxp = XOR(NULL, XOR(prev->nxp, curr));
 
    // Disconnect curr from previous node
    if (curr != NULL)
        curr->nxp = XOR(XOR(curr->nxp, prev), NULL);
 
    // If the count of remaining
    // nodes is less than K
    if (len < K) {
        return prev;
    }
    else {
 
        // Update len
        len -= K;
 
        // Recursively process the next nodes
        struct Node* dummy = RevInGrp(&curr, K, len);
 
        // Connect the head pointer with the prev
        (*head)->nxp = XOR(XOR(NULL, (*head)->nxp), dummy);
 
        // Connect prev with the head
        if (dummy != NULL)
            dummy->nxp = XOR(XOR(dummy->nxp, NULL), *head);
        return prev;
    }
}
 
// Driver Code
int main()
{
 
    /* Create following XOR Linked List
    head-->7<–>6<–>8<–>11<–>3<–>1<–>2<–>0*/
    struct Node* head = NULL;
    insert(&head, 0);
    insert(&head, 2);
    insert(&head, 1);
    insert(&head, 3);
    insert(&head, 11);
    insert(&head, 8);
    insert(&head, 6);
    insert(&head, 7);
 
    // Function Call
    head = RevInGrp(&head, 3, 8);
 
    // Print the reversed list
    printList(&head);
 
    return (0);
}
 
// This code is contributed by pankajsharmagfg.
 
  

C

   // C program to implement
// the above approach
 
#include <inttypes.h>
#include <stdio.h>
#include <stdlib.h>
 
// Structure of a node
// in XOR linked list
struct Node {
 
    // Stores data value
    // of a node
    int data;
 
    // Stores XOR of previous
    // pointer and next pointer
    struct Node* nxp;
};
 
// Function to find the XOR of address
// of two nodes
struct Node* XOR(struct Node* a,
                 struct Node* b)
{
    return (struct Node*)((uintptr_t)(a)
                          ^ (uintptr_t)(b));
}
 
// Function to insert a node with
// given value at given position
struct Node* insert(struct Node** head,
                    int value)
{
 
    // If XOR linked list is empty
    if (*head == NULL) {
 
        // Initialize a new Node
        struct Node* node
            = (struct Node*)malloc(
                sizeof(struct Node));
 
        // Stores data value in
        // the node
        node->data = value;
 
        // Stores XOR of previous
        // and next pointer
        node->nxp = XOR(NULL, NULL);
 
        // Update pointer of head node
        *head = node;
    }
 
    // If the XOR linked list
    // is not empty
    else {
 
        // Stores the address
        // of current node
        struct Node* curr = *head;
 
        // Stores the address
        // of previous node
        struct Node* prev = NULL;
 
        // Initialize a new Node
        struct Node* node
            = (struct Node*)malloc(
                sizeof(struct Node));
 
        // Update curr node address
        curr->nxp = XOR(node, XOR(NULL,
                                  curr->nxp));
 
        // Update new node address
        node->nxp = XOR(NULL, curr);
 
        // Update head
        *head = node;
 
        // Update data value of
        // current node
        node->data = value;
    }
    return *head;
}
 
// Function to print elements of
// the XOR Linked List
void printList(struct Node** head)
{
 
    // Stores XOR pointer
    // in current node
    struct Node* curr = *head;
 
    // Stores XOR pointer of
    // in previous Node
    struct Node* prev = NULL;
 
    // Stores XOR pointer of
    // in next node
    struct Node* next;
 
    // Traverse XOR linked list
    while (curr != NULL) {
 
        // Print current node
        printf("%d ", curr->data);
 
        // Forward traversal
        next = XOR(prev, curr->nxp);
 
        // Update prev
        prev = curr;
 
        // Update curr
        curr = next;
    }
}
 
// Reverse the linked list in group of K
struct Node* RevInGrp(struct Node** head,
                      int K, int len)
{
 
    // Stores head node
    struct Node* curr = *head;
 
    // If the XOR linked
    // list is empty
    if (curr == NULL)
        return NULL;
 
    // Stores count of nodes
    // reversed in current group
    int count = 0;
 
    // Stores XOR pointer of
    // in previous Node
    struct Node* prev = NULL;
 
    // Stores XOR pointer of
    // in next node
    struct Node* next;
 
    // Reverse nodes in current group
    while (count < K && count < len) {
 
        // Forward traversal
        next = XOR(prev, curr->nxp);
 
        // Update prev
        prev = curr;
 
        // Update curr
        curr = next;
 
        // Update count
        count++;
    }
 
    // Disconnect prev node from the next node
    prev->nxp = XOR(NULL, XOR(prev->nxp, curr));
 
    // Disconnect curr from previous node
    if (curr != NULL)
        curr->nxp = XOR(XOR(curr->nxp, prev), NULL);
 
    // If the count of remaining
    // nodes is less than K
    if (len < K) {
        return prev;
    }
    else {
 
        // Update len
        len -= K;
 
        // Recursively process the next nodes
        struct Node* dummy
            = RevInGrp(&curr, K, len);
 
        // Connect the head pointer with the prev
        (*head)->nxp = XOR(XOR(NULL,
                               (*head)->nxp),
                           dummy);
 
        // Connect prev with the head
        if (dummy != NULL)
            dummy->nxp = XOR(XOR(dummy->nxp, NULL),
                             *head);
        return prev;
    }
}
 
// Driver Code
int main()
{
 
    /* Create following XOR Linked List
    head-->7<–>6<–>8<–>11<–>3<–>1<–>2<–>0*/
    struct Node* head = NULL;
    insert(&head, 0);
    insert(&head, 2);
    insert(&head, 1);
    insert(&head, 3);
    insert(&head, 11);
    insert(&head, 8);
    insert(&head, 6);
    insert(&head, 7);
 
    // Function Call
    head = RevInGrp(&head, 3, 8);
 
    // Print the reversed list
    printList(&head);
 
    return (0);
}
 
  

Java

   // Java program to reverse a linked list in groups of 
// given size 
class LinkedList { 
    Node head; // head of list 
 
    /* Linked list Node*/
    class Node { 
        int data; 
        Node next; 
        Node(int d) 
        { 
            data = d; 
            next = null; 
        } 
    } 
 
    Node reverse(Node head, int k) 
    { 
        if(head == null) 
        return null; 
        Node current = head; 
        Node next = null; 
        Node prev = null; 
 
        int count = 0; 
 
        /* Reverse first k nodes of linked list */
        while (count < k && current != null) { 
            next = current.next; 
            current.next = prev; 
            prev = current; 
            current = next; 
            count++; 
        } 
 
        /* next is now a pointer to (k+1)th node 
        Recursively call for the list starting from 
        current. And make rest of the list as next of 
        first node */
        if (next != null) 
            head.next = reverse(next, k); 
 
        // prev is now head of input list 
        return prev; 
    } 
 
    /* Utility functions */
 
    /* Inserts a new Node at front of the list. */
    public void push(int new_data) 
    { 
        /* 1 & 2: Allocate the Node & 
                Put in the data*/
        Node new_node = new Node(new_data); 
 
        /* 3. Make next of new Node as head */
        new_node.next = head; 
 
        /* 4. Move the head to point to new Node */
        head = new_node; 
    } 
 
    /* Function to print linked list */
    void printList() 
    { 
        Node temp = head; 
        while (temp != null) { 
            System.out.print(temp.data + " "); 
            temp = temp.next; 
        } 
        System.out.println(); 
    } 
 
    /* Driver program to test above functions */
    public static void main(String args[]) 
    { 
        LinkedList llist = new LinkedList(); 
 
        /* Constructed Linked List is 1->2->3->4->5->6-> 
        7->8->8->9->null */
        llist.push(0); 
        llist.push(2); 
        llist.push(1); 
        llist.push(3); 
        llist.push(11); 
        llist.push(8); 
        llist.push(6); 
        llist.push(7);
        llist.head = llist.reverse(llist.head, 3); 
        llist.printList(); 
    } 
} 
/* This code is contributed by Rajat Mishra */
 
  

Python3

   class LinkedList:
    def __init__(self):
        self.head = None
 
    class Node:
        def __init__(self, data):
            self.data = data
            self.next = None
 
    def reverse(self, head, k):
        if head is None:
            return None
        current = head
        next_node = None
        prev = None
        count = 0
 
        while count < k and current is not None:
            next_node = current.next
            current.next = prev
            prev = current
            current = next_node
            count += 1
 
        if next_node is not None:
            head.next = self.reverse(next_node, k)
 
        return prev
 
    def push(self, new_data):
        new_node = self.Node(new_data)
        new_node.next = self.head
        self.head = new_node
 
    def print_list(self):
        temp = self.head
        while temp is not None:
            print(temp.data, end=" ")
            temp = temp.next
        print()
 
# Driver program to test above functions
if __name__ == "__main__":
    llist = LinkedList()
 
    # Constructed Linked List is 1->2->3->4->5->6->7->8->8->9->None
    llist.push(0)
    llist.push(2)
    llist.push(1)
    llist.push(3)
    llist.push(11)
    llist.push(8)
    llist.push(6)
    llist.push(7)
    llist.head = llist.reverse(llist.head, 3)
    llist.print_list()
#This code is contributed Dibyabrata Panja
 
  

C#

   using System;
 
public class LinkedList
{
    public Node head;
 
    // Node class to represent each element in the linked list
    public class Node
    {
        public int data;
        public Node next;
 
        public Node(int data)
        {
            this.data = data;
            this.next = null;
        }
    }
 
    // Function to reverse the linked list in groups of k
    public Node Reverse(Node head, int k)
    {
        if (head == null)
            return null;
 
        Node current = head;
        Node nextNode = null;
        Node prev = null;
        int count = 0;
 
        // Reverse k nodes
        while (count < k && current != null)
        {
            nextNode = current.next;
            current.next = prev;
            prev = current;
            current = nextNode;
            count++;
        }
 
        // Recursive call for the remaining nodes
        if (nextNode != null)
            head.next = Reverse(nextNode, k);
 
        return prev;
    }
 
    // Function to push a new element to the linked list
    public void Push(int newData)
    {
        Node newNode = new Node(newData);
        newNode.next = head;
        head = newNode;
    }
 
    // Function to print the linked list
    public void PrintList()
    {
        Node temp = head;
        while (temp != null)
        {
            Console.Write(temp.data + " ");
            temp = temp.next;
        }
        Console.WriteLine();
    }
 
    // Driver program to test the functions
    public static void Main()
    {
        LinkedList llist = new LinkedList();
 
        // Constructed Linked List is 1->2->3->4->5->6->7->8->8->9->None
        llist.Push(0);
        llist.Push(2);
        llist.Push(1);
        llist.Push(3);
        llist.Push(11);
        llist.Push(8);
        llist.Push(6);
        llist.Push(7);
        llist.head = llist.Reverse(llist.head, 3);
        llist.PrintList();
    }
}
 
  

Javascript

   // Node class to represent linked list nodes
class Node {
    constructor(data) {
        this.data = data;
        this.next = null;
    }
}
 
// LinkedList class
class LinkedList {
    constructor() {
        this.head = null;
    }
 
    // Function to reverse the linked list in groups of given size
    reverse(head, k) {
        if (head === null)
            return null;
 
        let current = head;
        let next = null;
        let prev = null;
        let count = 0;
 
        // Reverse first k nodes of linked list
        while (count < k && current !== null) {
            next = current.next;
            current.next = prev;
            prev = current;
            current = next;
            count++;
        }
 
        // Recursively call for the list starting from current
        // And make rest of the list as the next of the first node
        if (next !== null)
            head.next = this.reverse(next, k);
 
        // prev is now the head of the input list
        return prev;
    }
 
    // Utility function to insert a new Node at the front of the list
    push(newData) {
        const newNode = new Node(newData);
        newNode.next = this.head;
        this.head = newNode;
    }
 
    // Function to print linked list
    printList() {
        let temp = this.head;
        while (temp !== null) {
            console.log(temp.data + " ");
            temp = temp.next;
        }
        console.log();
    }
}
 
// Driver program
const llist = new LinkedList();
 
// Constructed Linked List: 1->2->3->4->5->6->7->8->9->null
llist.push(0);
llist.push(2);
llist.push(1);
llist.push(3);
llist.push(11);
llist.push(8);
llist.push(6);
llist.push(7);
 
// Reverse the linked list in groups of 3
llist.head = llist.reverse(llist.head, 3);
 
// Print the reversed linked list
llist.printList();
 
  

Output

8 6 7 1 3 11 0 2

Time Complexity: O(N)
Auxiliary Space: O(N / K)

Reverse a Linked List in groups of given size using Deque

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Practice Tags :

XOR Linked List – Reverse a Linked List in groups of given size

C++

C

Java

Python3

C#

Javascript

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