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XOR Linked List – Reverse a Linked List in groups of given size
  • Last Updated : 17 Jan, 2021

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

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// 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);
}

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Output:

8 6 7 1 3 11 0 2

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

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