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Rotate the sub-list of a linked list from position M to N to the right by K places

  • Difficulty Level : Medium
  • Last Updated : 30 Jun, 2021

Given a linked list and two positions ‘m’ and ‘n’. The task is to rotate the sublist from position m to n, to the right by k places.
Examples: 
 

Input: list = 1->2->3->4->5->6, m = 2, n = 5, k = 2 
Output: 1->4->5->2->3->6 
Rotate the sublist 2 3 4 5 towards right 2 times 
then the modified list are: 1 4 5 2 3 6
Input: list = 20->45->32->34->22->28, m = 3, n = 6, k = 3 
Output: 20->45->34->22->28->32 
Rotate the sublist 32 34 22 28 towards right 3 times 
then the modified list are: 20 45 34 22 28 32

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Approach: For rotating the given sublist that extends from m to n element, move the list from (n-k+1)th to nth node to starting of sub-list to finish the rotation. 
If k is greater than size of sublist then we will take its modulo with size of sublist. So traverse through list using a pointer and a counter and we will save (m-1)th node and later make it point to (n-k+1)th node and hence bring (n-k+1)th node to the start(front) of sublist. 
Similarly we will save mth node and later make nth node point to it. And for keeping rest of list intact we will make (n-k)th node point to next node of n (maybe NULL). And finally we will get the k times right rotated sublist.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Definition of node of linkedlist
struct ListNode {
    int data;
    struct ListNode* next;
};
 
// This function take head pointer of list, start and
// end points of sublist that is to be rotated and the
// number k and rotate the sublist to right by k places.
void rotateSubList(ListNode* A, int m, int n, int k)
{
    int size = n - m + 1;
 
    // If k is greater than size of sublist then
    // we will take its modulo with size of sublist
    if (k > size) {
        k = k % size;
    }
 
    // If k is zero or k is equal to size or k is
    // a multiple of size of sublist then list
    // remains intact
    if (k == 0 || k == size) {
        ListNode* head = A;
        while (head != NULL) {
            cout << head->data;
            head = head->next;
        }
        return;
    }
 
    ListNode* link = NULL;  // m-th node
    if (m == 1) {
        link = A;
    }
 
    // This loop will traverse all node till
    // end node of sublist.   
    ListNode* c = A;  // Current traversed node
    int count = 0;  // Count of traversed nodes
    ListNode* end = NULL; 
    ListNode* pre = NULL; // Previous of m-th node
    while (c != NULL) {
        count++;
 
        // We will save (m-1)th node and later
        // make it point to (n-k+1)th node
        if (count == m - 1) {
            pre = c;
            link = c->next;
        }
        if (count == n - k) {
            if (m == 1) {
                end = c;
                A = c->next;
            }
            else {
                end = c;
 
                // That is how we bring (n-k+1)th
                // node to front of sublist.
                pre->next = c->next;
            }
        }
 
        // This keeps rest part of list intact.
        if (count == n) {
            ListNode* d = c->next;
            c->next = link;
            end->next = d;
            ListNode* head = A;
            while (head != NULL) {
                cout << head->data << " ";
                head = head->next;
            }
            return;
        }
        c = c->next;
    }
}
 
// Function for creating and linking new nodes
void push(struct ListNode** head, int val)
{
    struct ListNode* new_node = new ListNode;
    new_node->data = val;
    new_node->next = (*head);
    (*head) = new_node;
}
 
// Driver code
int main()
{
    struct ListNode* head = NULL;
    push(&head, 70);
    push(&head, 60);
    push(&head, 50);
    push(&head, 40);
    push(&head, 30);
    push(&head, 20);
    push(&head, 10);
    ListNode* tmp = head;
    cout << "Given List: ";
    while (tmp != NULL) {
        cout << tmp->data << " ";
        tmp = tmp->next;
    }
    cout << endl;
 
    int m = 3, n = 6, k = 2;
    cout << "After rotation of sublist: ";
    rotateSubList(head, m, n, k);
 
    return 0;
}

Java




// Java implementation of the above approach
import java.util.*;
 
class Solution
{
 
   
// Definition of node of linkedlist
static class ListNode {
    int data;
     ListNode next;
}
   
// This function take head pointer of list, start and
// end points of sublist that is to be rotated and the
// number k and rotate the sublist to right by k places.
static void rotateSubList(ListNode A, int m, int n, int k)
{
    int size = n - m + 1;
   
    // If k is greater than size of sublist then 
    // we will take its modulo with size of sublist
    if (k > size) {
        k = k % size;
    }
   
    // If k is zero or k is equal to size or k is
    // a multiple of size of sublist then list 
    // remains intact
    if (k == 0 || k == size) {
        ListNode head = A;
        while (head != null) {
            System.out.print( head.data);
            head = head.next;
        }
        return;
    }
   
    ListNode link = null// m-th node
    if (m == 1) {
        link = A;
    }
   
    // This loop will traverse all node till
    // end node of sublist.    
    ListNode c = A;  // Current traversed node
    int count = 0// Count of traversed nodes
    ListNode end = null;  
    ListNode pre = null; // Previous of m-th node
    while (c != null) {
        count++;
   
        // We will save (m-1)th node and later
        // make it point to (n-k+1)th node
        if (count == m - 1) {
            pre = c;
            link = c.next;
        }
        if (count == n - k) {
            if (m == 1) {
                end = c;
                A = c.next;
            }
            else {
                end = c;
   
                // That is how we bring (n-k+1)th
                // node to front of sublist.
                pre.next = c.next;
            }
        }
   
        // This keeps rest part of list intact.
        if (count == n) {
            ListNode d = c.next;
            c.next = link;
            end.next = d;
            ListNode head = A;
            while (head != null) {
                System.out.print( head.data+" ");
                head = head.next;
            }
            return;
        }
        c = c.next;
    }
}
   
// Function for creating and linking new nodes
static ListNode push( ListNode head, int val)
{
     ListNode new_node = new ListNode();
    new_node.data = val;
    new_node.next = (head);
    (head) = new_node;
    return head;
}
   
// Driver code
public static void main(String args[])
{
     ListNode head = null;
    head =push(head, 70);
    head =push(head, 60);
    head =push(head, 50);
    head =push(head, 40);
    head =push(head, 30);
    head =push(head, 20);
    head =push(head, 10);
    ListNode tmp = head;
    System.out.print("Given List: ");
    while (tmp != null) {
        System.out.print( tmp.data + " ");
        tmp = tmp.next;
    }
    System.out.println();
   
    int m = 3, n = 6, k = 2;
    System.out.print( "After rotation of sublist: ");
    rotateSubList(head, m, n, k);
   
}
}
 
// This code is contributed
// by Arnab Kundu

Python3




# Python3 implementation of the above approach
import math
 
# Definition of node of linkedlist
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None
 
# This function take head pointer of list,
# start and end points of sublist that is
# to be rotated and the number k and
# rotate the sublist to right by k places.
def rotateSubList(A, m, n, k):
    size = n - m + 1
 
    # If k is greater than size of sublist then
    # we will take its modulo with size of sublist
    if (k > size):
        k = k % size
     
    # If k is zero or k is equal to size or k is
    # a multiple of size of sublist then list
    # remains intact
    if (k == 0 or k == size):
        head = A
        while (head != None):
            print(head.data)
            head = head.next
         
        return
     
    link = None # m-th node
    if (m == 1) :
        link = A
     
    # This loop will traverse all node till
    # end node of sublist.
    c = A # Current traversed node
    count = 0 # Count of traversed nodes
    end = None
    pre = None # Previous of m-th node
    while (c != None) :
        count = count + 1
 
        # We will save (m-1)th node and later
        # make it point to (n-k+1)th node
        if (count == m - 1) :
            pre = c
            link = c.next
         
        if (count == n - k) :
            if (m == 1) :
                end = c
                A = c.next
             
            else :
                end = c
 
                # That is how we bring (n-k+1)th
                # node to front of sublist.
                pre.next = c.next
             
        # This keeps rest part of list intact.
        if (count == n) :
            d = c.next
            c.next = link
            end.next = d
            head = A
            while (head != None) :
                print(head.data, end = " ")
                head = head.next
             
            return
         
        c = c.next
     
# Function for creating and linking new nodes
def push(head, val):
    new_node = Node(val)
    new_node.data = val
    new_node.next = head
    head = new_node
    return head
 
# Driver code
if __name__=='__main__':
    head = None
    head = push(head, 70)
    head = push(head, 60)
    head = push(head, 50)
    head = push(head, 40)
    head = push(head, 30)
    head = push(head, 20)
    head = push(head, 10)
    tmp = head
    print("Given List: ", end = "")
    while (tmp != None) :
        print(tmp.data, end = " ")
        tmp = tmp.next
     
    print()
 
    m = 3
    n = 6
    k = 2
    print("After rotation of sublist: ", end = "")
    rotateSubList(head, m, n, k)
 
# This code is contributed by Srathore

C#




// C# implementation of the above approach
using System;
 
class GFG
{
 
// Definition of node of linkedlist
public class ListNode
{
    public int data;
    public ListNode next;
}
 
// This function take head pointer of list,
// start and end points of sublist that is
// to be rotated and the number k and rotate
// the sublist to right by k places.
public static void rotateSubList(ListNode A, int m,
                                      int n, int k)
{
    int size = n - m + 1;
 
    // If k is greater than size of sublist
    // then we will take its modulo with
    // size of sublist
    if (k > size)
    {
        k = k % size;
    }
 
    // If k is zero or k is equal to size
    // or k is a multiple of size of sublist
    // then list remains intact
    if (k == 0 || k == size)
    {
        ListNode head = A;
        while (head != null)
        {
            Console.Write(head.data);
            head = head.next;
        }
        return;
    }
 
    ListNode link = null; // m-th node
    if (m == 1)
    {
        link = A;
    }
 
    // This loop will traverse all node till
    // end node of sublist.    
    ListNode c = A; // Current traversed node
    int count = 0;  // Count of traversed nodes
    ListNode end = null;
    ListNode pre = null; // Previous of m-th node
    while (c != null)
    {
        count++;
 
        // We will save (m-1)th node and later
        // make it point to (n-k+1)th node
        if (count == m - 1)
        {
            pre = c;
            link = c.next;
        }
        if (count == n - k)
        {
            if (m == 1)
            {
                end = c;
                A = c.next;
            }
            else
            {
                end = c;
 
                // That is how we bring (n-k+1)th
                // node to front of sublist.
                pre.next = c.next;
            }
        }
 
        // This keeps rest part of list intact.
        if (count == n)
        {
            ListNode d = c.next;
            c.next = link;
            end.next = d;
            ListNode head = A;
            while (head != null)
            {
                Console.Write(head.data + " ");
                head = head.next;
            }
            return;
        }
        c = c.next;
    }
}
 
// Function for creating and linking new nodes
public static ListNode push(ListNode head, int val)
{
    ListNode new_node = new ListNode();
    new_node.data = val;
    new_node.next = (head);
    (head) = new_node;
    return head;
}
 
// Driver code
public static void Main(string[] args)
{
    ListNode head = null;
    head = push(head, 70);
    head = push(head, 60);
    head = push(head, 50);
    head = push(head, 40);
    head = push(head, 30);
    head = push(head, 20);
    head = push(head, 10);
    ListNode tmp = head;
    Console.Write("Given List: ");
    while (tmp != null)
    {
        Console.Write(tmp.data + " ");
        tmp = tmp.next;
    }
    Console.WriteLine();
 
    int m = 3, n = 6, k = 2;
    Console.Write("After rotation of sublist: ");
    rotateSubList(head, m, n, k);
}
}
 
// This code is contributed by Shrikant13

Javascript




<script>
 
      // JavaScript implementation of the above approach
      // Definition of node of linkedlist
      class ListNode {
        constructor() {
          this.data = 0;
          this.next = null;
        }
      }
 
      // This function take head pointer of list,
      // start and end points of sublist that is
      // to be rotated and the number k and rotate
      // the sublist to right by k places.
      function rotateSubList(A, m, n, k) {
        var size = n - m + 1;
 
        // If k is greater than size of sublist
        // then we will take its modulo with
        // size of sublist
        if (k > size) {
          k = k % size;
        }
 
        // If k is zero or k is equal to size
        // or k is a multiple of size of sublist
        // then list remains intact
        if (k == 0 || k == size) {
          var head = A;
          while (head != null) {
            document.write(head.data);
            head = head.next;
          }
          return;
        }
 
        var link = null; // m-th node
        if (m == 1) {
          link = A;
        }
 
        // This loop will traverse all node till
        // end node of sublist.
        var c = A; // Current traversed node
        var count = 0; // Count of traversed nodes
        var end = null;
        var pre = null; // Previous of m-th node
        while (c != null) {
          count++;
 
          // We will save (m-1)th node and later
          // make it point to (n-k+1)th node
          if (count == m - 1) {
            pre = c;
            link = c.next;
          }
          if (count == n - k) {
            if (m == 1) {
              end = c;
              A = c.next;
            } else {
              end = c;
 
              // That is how we bring (n-k+1)th
              // node to front of sublist.
              pre.next = c.next;
            }
          }
 
          // This keeps rest part of list intact.
          if (count == n) {
            var d = c.next;
            c.next = link;
            end.next = d;
            var head = A;
            while (head != null) {
              document.write(head.data + " ");
              head = head.next;
            }
            return;
          }
          c = c.next;
        }
      }
 
      // Function for creating and linking new nodes
      function push(head, val) {
        var new_node = new ListNode();
        new_node.data = val;
        new_node.next = head;
        head = new_node;
        return head;
      }
 
      // Driver code
      var head = null;
      head = push(head, 70);
      head = push(head, 60);
      head = push(head, 50);
      head = push(head, 40);
      head = push(head, 30);
      head = push(head, 20);
      head = push(head, 10);
      var tmp = head;
      document.write("Given List: ");
      while (tmp != null) {
        document.write(tmp.data + " ");
        tmp = tmp.next;
      }
      document.write("<br>");
 
      var m = 3,
        n = 6,
        k = 2;
      document.write("After rotation of sublist: ");
      rotateSubList(head, m, n, k);
       
    </script>
Output: 
Given List: 10 20 30 40 50 60 70 
After rotation of sublist: 10 20 50 60 30 40 70

 




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