Reverse Sublist of a Linked List

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

Problem Highlights

• 💡 Difficulty: Medium
• ⏰ Time to complete: 30 mins
• 🛠️ Topics: Linked Lists, Reverse Operations

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 m and n are the same?
  • A: No change is made to the list since the range to reverse is only one node.

HAPPY CASE
Input: 1 -> 2 -> 3 -> 4 -> 5, m = 2, n = 4
Output: 1 -> 4 -> 3 -> 2 -> 5
Explanation: Nodes between position 2 and 4 are reversed, leading to the sequence 4 -> 3 -> 2.

EDGE CASE
Input: 1 -> 2 -> 3, m = 1, n = 1
Output: 1 -> 2 -> 3
Explanation: As m and n are the same, no nodes are reversed, and the list remains unchanged.

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 is a variation of the classic linked list reversal but with a constraint to only reverse a subsection of the list.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Use a temporary node to simplify edge cases and use pointers to reverse the specified section.

1) Create a temp node to simplify operations at the head.
2) Use pointers to navigate to the start of the section to reverse.
3) Reverse the nodes between m and n using pointer manipulation.
4) Connect the reversed section back into the list.

⚠️ Common Mistakes

• Failing to reconnect the reversed section properly with the rest of the list.
• Not handling the case where the reverse starts from the head node.

4: I-mplement

Implement the code to solve the algorithm.

def reverse_between(head, m, n):
    if not head or m == n:
        return head

    # Create a temporary head node to simplify edge cases where m is 1
    temp_head = Node(0)
    temp_head.next = head
    prev = temp_head

    # Step 1: Reach the node just before position m
    for i in range(m - 1):
        prev = prev.next
    
    # `prev` now is the node just before m, and `start` will be the first node to reverse
    start = prev.next
    then = start.next

    # Step 2: Reverse from m to n
    for _ in range(n - m):
        start.next = then.next  # Remove `then` from the list
        then.next = prev.next  # Insert `then` at the beginning of the reversed section
        prev.next = then  # Move `prev` to point to `then` as the new start of the reversed section
        then = start.next  # Move `then` to the next node to be reversed

    return temp_head.next

5: R-eview

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

• Run through the code with a typical example to ensure that the section is reversed and properly linked.
• Check with the edge case where m and n are the same to confirm no unnecessary changes are made.

6: E-valuate

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

• Time Complexity: O(N) since we might need to traverse the entire list in the worst case.
• Space Complexity: O(1) as we are using a constant amount of space regardless of the input size.