Circular Linked List Rotate

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Unit 6 Session 2 (Click for link to problem statements)

Problem Highlights

• 💡 Difficulty: Medium
• ⏰ Time to complete: 20 mins
• 🛠️ Topics: Linked Lists, Rotations

1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• Established a set (2-3) of test cases to verify their own solution later.
• Established a set (1-2) of edge cases to verify their solution handles complexities.
• Have fully understood the problem and have no clarifying questions.
• Have you verified any Time/Space Constraints for this problem?

• Q: What happens if k is greater than the length of the list?
  • A: Since we'll temporarily make the list circular, k should be taken modulo the length of the list to find the effective number of rotations.

HAPPY CASE
Input: 1 -> 2 -> 3 -> 4 -> 5, k = 2
Output: 4 -> 5 -> 1 -> 2 -> 3
Explanation: The list is rotated to the right by 2 places.

EDGE CASE
Input: 1 -> 2 -> 3 -> 4 -> 5, k = 5 (or 0)
Output: 1 -> 2 -> 3 -> 4 -> 5
Explanation: Rotating by the list's length results in the same list.

2: M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

This problem involves rotating a list, which is typically tackled by linking the list into a circle, then adjusting the links based on k.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Connect the list into a circular form, then adjust the head and tail based on the rotation count k.

1) Measure the length of the list and identify the last node.
2) Link the last node to the head to form a circular list.
3) Calculate the effective rotation using k modulo the length of the list.
4) Find the new tail (length - k % length steps from the head).
5) Make the node following the new tail the new head and break the circle there.

⚠️ Common Mistakes

• Not adjusting for k values greater than the list length.
• Not handling edge cases where the list does not change (k is 0 or a multiple of the list length).

4: I-mplement

Implement the code to solve the algorithm.

def rotate_right(head, k):
    if not head or not head.next or k == 0:
        return head

    last = head
    length = 1
    while last.next:
        last = last.next
        length += 1

    last.next = head

    steps_to_new_tail = length - k % length
    new_tail = head
    for _ in range(steps_to_new_tail - 1):
        new_tail = new_tail.next

    new_head = new_tail.next
    new_tail.next = None

    return new_head

5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

• Test the rotation with values of k that are less than, equal to, and greater than the list length to ensure proper rotation.

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

• Time Complexity: O(n) where n is the length of the list, as the list is traversed completely to find the length and to adjust the tail.
• Space Complexity: O(1) because no additional storage is used; the list nodes are just re-linked.