Given a singly linked list, rotate the linked list counter-clockwise by k nodes. Where k is a given positive integer. For example, if the given linked list is 10->20->30->40->50->60 and k is 4, the list should be modified to 50->60->10->20->30->40. Assume that k is smaller than the count of nodes in linked list.

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

To rotate the linked list, we need to change next of kth node to NULL, next of last node to previous head node, and finally change head to (k+1)th node. So we need to get hold of three nodes: kth node, (k+1)th node and last node.
Traverse the list from beginning and stop at kth node. Store pointer to kth node. We can get (k+1)th node using kthNode->next. Keep traversing till end and store pointer to last node also. Finally, change pointers as stated above.

## C/C++

```// C/C++ program to rotate a linked list counter clock wise

#include<stdio.h>
#include<stdlib.h>

struct Node
{
int data;
struct Node* next;
};

// This function rotates a linked list counter-clockwise and
// than size of linked list. It doesn't modify the list if
// k is greater than or equal to size
void rotate(struct Node **head_ref, int k)
{
if (k == 0)
return;

// Let us understand the below code for example k = 4 and
// list = 10->20->30->40->50->60.

// current will either point to kth or NULL after this loop.
//  current will point to node 40 in the above example
int count = 1;
while (count < k && current != NULL)
{
current = current->next;
count++;
}

// If current is NULL, k is greater than or equal to count
// of nodes in linked list. Don't change the list in this case
if (current == NULL)
return;

// current points to kth node. Store it in a variable.
// kthNode points to node 40 in the above example
struct Node *kthNode = current;

// current will point to last node after this loop
// current will point to node 60 in the above example
while (current->next != NULL)
current = current->next;

// Change next of last node to previous head
// Next of 60 is now changed to node 10

// Change head to (k+1)th node
// head is now changed to node 50

// change next of kth node to NULL
// next of 40 is now NULL
kthNode->next = NULL;
}

/* UTILITY FUNCTIONS */
/* Function to push a node */
void push (struct Node** head_ref, int new_data)
{
/* allocate node */
struct Node* new_node =
(struct Node*) malloc(sizeof(struct Node));

/* put in the data  */
new_node->data  = new_data;

/* link the old list off the new node */

/* move the head to point to the new node */
}

/* Function to print linked list */
void printList(struct Node *node)
{
while (node != NULL)
{
printf("%d ", node->data);
node = node->next;
}
}

/* Drier program to test above function*/
int main(void)
{

// create a list 10->20->30->40->50->60
for (int i = 60; i > 0; i -= 10)

return (0);
}
```

## Java

```// Java program to rotate a linked list

{

class Node
{
int data;
Node next;
Node(int d)
{
data = d;
next = null;
}
}

// This function rotates a linked list counter-clockwise
// smaller than size of linked list. It doesn't modify
// the list if k is greater than or equal to size
void rotate(int k)
{
if (k == 0) return;

// Let us understand the below code for example k = 4
// and list = 10->20->30->40->50->60.

// current will either point to kth or NULL after this
// loop. current will point to node 40 in the above example
int count = 1;
while (count < k && current !=  null)
{
current = current.next;
count++;
}

// If current is NULL, k is greater than or equal to count
// of nodes in linked list. Don't change the list in this case
if (current == null)
return;

// current points to kth node. Store it in a variable.
// kthNode points to node 40 in the above example
Node kthNode = current;

// current will point to last node after this loop
// current will point to node 60 in the above example
while (current.next != null)
current = current.next;

// Change next of last node to previous head
// Next of 60 is now changed to node 10

// Change head to (k+1)th node
// head is now changed to node 50

// change next of kth node to null
kthNode.next = null;

}

/*  Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
void push(int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(new_data);

/* 3. Make next of new Node as head */

/* 4. Move the head to point to new Node */
}

void printList()
{
while(temp != null)
{
System.out.print(temp.data+" ");
temp = temp.next;
}
System.out.println();
}

/* Drier program to test above functions */
public static void main(String args[])
{

// create a list 10->20->30->40->50->60
for (int i = 60; i >= 10; i -= 10)
llist.push(i);

System.out.println("Given list");
llist.printList();

llist.rotate(4);

llist.printList();
}
} /* This code is contributed by Rajat Mishra */
```

## Python

```
# Python program to rotate a linked list

# Node class
class Node:

# Constructor to initialize the node object
def __init__(self, data):
self.data = data
self.next = None

def __init__(self):

# Function to insert a new node at the beginning
def push(self, new_data):
# allocate node and put the data
new_node = Node(new_data)

# Make next of new node as head

# move the head to point to the new Node

def printList(self):
while(temp):
print temp.data,
temp = temp.next

# This function rotates a linked list counter-clockwise and
# than size of linked list. It doesn't modify the list if
# k is greater than of equal to size
def rotate(self, k):
if k == 0:
return

# Let us understand the below code for example k = 4
# and list = 10->20->30->40->50->60

# current will either point to kth or NULL after
# this loop
# current will point to node 40 in the above example
count = 1
while(count <k and current is not None):
current = current.next
count += 1

# If current is None, k is greater than or equal
# to count of nodes in linked list. Don't change
# the list in this case
if current is None:
return

# current points to kth node. Store it in a variable
# kth node points to node 40 in the above example
kthNode = current

# current will point to lsat node after this loop
# current will point to node 60 in above example
while(current.next is not None):
current = current.next

# Change next of last node to previous head
# Next of 60 is now changed to node 10

# Change head to (k+1)th node
# head is not changed to node 50

# change next of kth node to NULL
# next of 40 is not NULL
kthNode.next = None

# Driver program to test above function

# Create a list 10->20->30->40->50->60
for i in range(60, 0, -10):
llist.push(i)

llist.printList()
llist.rotate(4)

llist.printList()

# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

```

Output:
```Given linked list
10  20  30  40  50  60
50  60  10  20  30  40```

Time Complexity: O(n) where n is the number of nodes in Linked List. The code traverses the linked list only once.

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