XOR linked list: Reverse last K nodes of a Linked List
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<–>11Input: 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:
- Reverse the given XOR Linked List.
- Now, reverse the first K nodes of the reversed linked list.
- After completing the above steps, reverse the linked list again to get the resultant linked list.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include<bits/stdc++.h>using namespace std;// Structure of a node of XOR Linked Liststruct 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 nodesstruct 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 positionstruct 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 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 = new 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 Listvoid 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 std::cout << 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 Kstruct 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 Listvoid 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 Codeint 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);}// This code is contributed by ajaymakvana. |
C
// C program for the above approach#include <inttypes.h>#include <stdio.h>#include <stdlib.h>// Structure of a node// of XOR Linked Liststruct 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 nodesstruct 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 positionstruct 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 Listvoid 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 Kstruct 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 Listvoid 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 Codeint 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), as we are using a loop to traverse N times, where N is the length of the linked list.
Auxiliary Space: O(1), as we are not using any extra space.
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