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Rotate Linked List block wise

Last Updated : 15 Jun, 2022
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Given a Linked List of length n and block length k rotate in a circular manner towards right/left each block by a number d. If d is positive rotate towards right else rotate towards left.

Examples: 

Input: 1->2->3->4->5->6->7->8->9->NULL, 
        k = 3 
        d = 1
Output: 3->1->2->6->4->5->9->7->8->NULL
Explanation: Here blocks of size 3 are
rotated towards right(as d is positive) 
by 1.
 
Input: 1->2->3->4->5->6->7->8->9->10->
               11->12->13->14->15->NULL, 
        k = 4 
        d = -1
Output: 2->3->4->1->6->7->8->5->10->11
             ->12->9->14->15->13->NULL
Explanation: Here, at the end of linked 
list, remaining nodes are less than k, i.e.
only three nodes are left while k is 4. 
Rotate those 3 nodes also by d.

Prerequisite: Rotate a linked list
The idea is if the absolute value of d is greater than the value of k, then rotate the link list by d % k times. If d is 0, no need to rotate the linked list at all. 

C++
// C++ program to rotate a linked list block wise 
#include<bits/stdc++.h>
using namespace std;

/* Link list node */
class Node 
{ 
    public:
    int data; 
    Node* next; 
}; 

// Recursive function to rotate one block 
Node* rotateHelper(Node* blockHead, 
                        Node* blockTail, 
                        int d, Node** tail, 
                        int k) 
{ 
    if (d == 0) 
        return blockHead; 

    // Rotate Clockwise 
    if (d > 0) 
    { 
        Node* temp = blockHead; 
        for (int i = 1; temp->next->next && 
                        i < k - 1; i++) 
            temp = temp->next; 
        blockTail->next = blockHead; 
        *tail = temp; 
        return rotateHelper(blockTail, temp, 
                            d - 1, tail, k); 
    } 

    // Rotate anti-Clockwise 
    if (d < 0) 
    { 
        blockTail->next = blockHead; 
        *tail = blockHead; 
        return rotateHelper(blockHead->next, 
                blockHead, d + 1, tail, k); 
    } 
} 

// Function to rotate the linked list block wise 
Node* rotateByBlocks(Node* head, 
                                int k, int d) 
{ 
    // If length is 0 or 1 return head 
    if (!head || !head->next) 
        return head; 

    // if degree of rotation is 0, return head 
    if (d == 0) 
        return head; 

    Node* temp = head, *tail = NULL; 

    // Traverse upto last element of this block 
    int i; 
    for (i = 1; temp->next && i < k; i++) 
        temp = temp->next; 

    // storing the first node of next block 
    Node* nextBlock = temp->next; 

    // If nodes of this block are less than k. 
    // Rotate this block also 
    if (i < k) 
        head = rotateHelper(head, temp, d % k, 
                                    &tail, i); 
    else
        head = rotateHelper(head, temp, d % k, 
                                    &tail, k); 

    // Append the new head of next block to 
    // the tail of this block 
    tail->next = rotateByBlocks(nextBlock, k, 
                                    d % k); 

    // return head of updated Linked List 
    return head; 
} 

/* UTILITY FUNCTIONS */
/* Function to push a node */
void push(Node** head_ref, int new_data) 
{ 
    Node* new_node = new Node; 
    new_node->data = new_data; 
    new_node->next = (*head_ref); 
    (*head_ref) = new_node; 
} 

/* Function to print linked list */
void printList(Node* node) 
{ 
    while (node != NULL) 
    { 
        cout << node->data << " "; 
        node = node->next; 
    } 
} 

/* Driver code*/
int main() 
{ 
    /* Start with the empty list */
    Node* head = NULL; 

    // create a list 1->2->3->4->5-> 
    // 6->7->8->9->NULL 
    for (int i = 9; i > 0; i -= 1) 
        push(&head, i); 

    cout<<"Given linked list \n"; 
    printList(head); 

    // k is block size and d is number of 
    // rotations in every block. 
    int k = 3, d = 2; 
    head = rotateByBlocks(head, k, d); 

    cout << "\nRotated by blocks Linked list \n"; 
    printList(head); 

    return (0); 
} 

// This is code is contributed by rathbhupendra
C Java Python3 C# JavaScript

Output:

Given linked list 
1 2 3 4 5 6 7 8 9 
Rotated by blocks Linked list 
2 3 1 5 6 4 8 9 7

Time complexity: O(n) since using a single loop to traverse a linked list to rotate

Auxiliary Space: O(n) for call stack


 


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