Python3 Program to Count rotations in sorted and rotated linked list
Last Updated :
29 Dec, 2022
Given a linked list of n nodes which is first sorted, then rotated by k elements. Find the value of k.
The idea is to traverse singly linked list to check condition whether current node value is greater than value of next node. If the given condition is true, then break the loop. Otherwise increase the counter variable and increase the node by node->next. Below is the implementation of this approach.
Python3
class Node:
def __init__( self , data):
self .data = data
self . next = None
def countRotation(head):
count = 0
min = head.data
while (head ! = None ):
if ( min > head.data):
break
count + = 1
head = head. next
return count
def push(head, data):
newNode = Node(data)
newNode.data = data
newNode. next = (head)
(head) = newNode
return head
def printList(node):
while (node ! = None ):
print (node.data, end = ' ' )
node = node. next
if __name__ = = '__main__' :
head = None
head = push(head, 12 )
head = push(head, 11 )
head = push(head, 8 )
head = push(head, 5 )
head = push(head, 18 )
head = push(head, 15 )
printList(head);
print ()
print ( "Linked list rotated elements: " ,
end = '')
print (countRotation(head))
|
Output
15 18 5 8 11 12
Linked list rotated elements: 2
Time Complexity: O(n), where n is the number of elements present in the linked list. This is because we are traversing the whole linked list in order to find the count.
Auxiliary Space: O(1), as we are not using any extra space.
Please refer complete article on Count rotations in sorted and rotated linked list for more details!
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