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XOR linked list: Reverse last K nodes of a Linked List
  • Last Updated : 05 Mar, 2021

Given a XOR Linked List and a positive integer K, the task is to reverse the last K nodes in the given XOR linked list.

Examples:

Input: LL: 7 <–> 6 <–> 8 <–> 11 <–> 3 <–> 1, K = 3
Output: 7<–>6<–>8<–>1<–>3<–>11

Input: LL: 7 <–> 6 <–> 8 <–> 11 <–> 3 <–> 1 <–> 2 <–> 0, K = 5
Output: 7<–>6<–>8<–>0<–>2<–>1<–>3<–>11

Approach: Follow the steps below to solve the given problem:



Below is the implementation of the above approach:

C




// C program for the above approach
  
#include <inttypes.h>
#include <stdio.h>
#include <stdlib.h>
  
// Structure of a node
// of 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 calculate
// Bitwise XOR of the 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 the current node
        struct Node* curr = *head;
  
        // Stores the address
        // of the previous node
        struct Node* prev = NULL;
  
        // Initialize a new Node
        struct Node* node
            = (struct Node*)malloc(
                sizeof(struct Node));
  
        // Update address of current node
        curr->nxp = XOR(node,
                        XOR(
                            NULL, curr->nxp));
  
        // Update address of the new node
        node->nxp = XOR(NULL, curr);
  
        // Update the head node
        *head = node;
  
        // Update the 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 the current node
    struct Node* curr = *head;
  
    // Stores XOR pointer
    // in the previous Node
    struct Node* prev = NULL;
  
    // Stores XOR pointer in the
    // next node
    struct Node* next;
  
    // Traverse XOR linked list
    while (curr != NULL) {
  
        // Print the current node
        printf("%d ", curr->data);
  
        // Forward traversal
        next = XOR(prev, curr->nxp);
  
        // Update the prev pointer
        prev = curr;
  
        // Update the curr pointer
        curr = next;
    }
}
  
// Function to reverse the linked
// list in the groups of K
struct Node* reverseK(struct Node** head,
                      int K, int len)
{
    struct Node* curr = *head;
  
    // If head is NULL
    if (curr == NULL)
        return NULL;
  
    // If the size of XOR linked
    // list is less than K
    else if (len < K)
        return *head;
    else {
  
        int count = 0;
  
        // Stores the XOR pointer
        // in the previous Node
        struct Node* prev = NULL;
  
        // Stores the XOR pointer
        // in the next node
        struct Node* next;
  
        while (count < K) {
  
            // Forward traversal
            next = XOR(prev, curr->nxp);
  
            // Update the prev pointer
            prev = curr;
  
            // Update the curr pointer
            curr = next;
  
            // Count the number of
            // nodes processed
            count++;
        }
  
        // Remove the prev node
        // from the next node
        prev->nxp = XOR(NULL,
                        XOR(prev->nxp,
                            curr));
  
        // Add the head pointer with prev
        (*head)->nxp = XOR(XOR(NULL,
                               (*head)->nxp),
                           curr);
  
        // Add the prev with the head
        if (curr != NULL)
            curr->nxp = XOR(XOR(curr->nxp,
                                prev),
                            *head);
        return prev;
    }
}
  
// Function to reverse last K nodes
// of the given XOR Linked List
void reverseLL(struct Node* head,
               int N, int K)
{
  
    // Reverse the given XOR LL
    head = reverseK(&head, N, N);
  
    // Reverse the first K nodes of
    // the XOR LL
    head = reverseK(&head, K, N);
  
    // Reverse the given XOR LL
    head = reverseK(&head, N, N);
  
    // Print the final linked list
    printList(&head);
}
  
// Driver Code
int main()
{
    // Stores number of nodes
    int N = 6;
  
    // Given XOR Linked List
  
    struct Node* head = NULL;
    insert(&head, 1);
    insert(&head, 3);
    insert(&head, 11);
    insert(&head, 8);
    insert(&head, 6);
    insert(&head, 7);
  
    int K = 3;
  
    reverseLL(head, N, K);
  
    return (0);
}
Output:
7 6 8 1 3 11

Time Complexity: O(N)
Auxiliary Space: O(1)

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