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
# Program for count number of rotations in# sorted linked list.# Linked list node class Node: def __init__(self, data): self.data = data self.next = None# Function that count number of# rotation in singly linked list.def countRotation(head): # Declare count variable and assign it 1. count = 0 # Declare a min variable and assign to # data of head node. min = head.data # Check that while head not equal to None. while (head != None): # If min value is greater then head->data # then it breaks the while loop and # return the value of count. if (min > head.data): break count += 1 # head assign the next value of head. head = head.next return count# Function to push element in linked list.def push(head, data): # Allocate dynamic memory for newNode. newNode = Node(data) # Assign the data into newNode. newNode.data = data # newNode->next assign the address of # head node. newNode.next = (head) # newNode become the headNode. (head) = newNode return head# Display linked list.def printList(node): while (node != None): print(node.data, end = ' ') node = node.next # Driver codeif __name__=='__main__': # Create a node and initialize with None head = None # push() insert node in linked list. # 15 -> 18 -> 5 -> 8 -> 11 -> 12 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 = '') # Function call countRotation() print(countRotation(head))# This code is contributed by rutvik_56 |
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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