Symmetrical

TIP102 Unit 5 Session 2 Advanced (Click for link to problem statements)

Problem Highlights

• 💡 Difficulty: Medium
• ⏰ Time to complete: 20-30 mins
• 🛠️ Topics: Linked Lists, Two-Pointer Technique, Palindrome

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?

• What happens if the linked list is empty?
  • The function should return True since an empty list is symmetric.
• What happens if the linked list contains only one element?
  • The function should return True since a single element is symmetric.

HAPPY CASE
Input: head = Node("Bitterling") -> Node("Crawfish") -> Node("Bitterling")
Output: True
Explanation: The linked list reads the same forwards and backwards.

EDGE CASE
Input: head = Node("Bitterling") -> Node("Carp") -> Node("Koi")
Output: False
Explanation: The linked list does not read the same forwards and backwards.

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.

For Linked List palindrome problems, we want to consider the following approaches:

• Use the two-pointer technique to find the middle of the list.
• Reverse the second half of the list.
• Compare the first half and the reversed second half.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Use the two-pointer technique to find the middle of the list, reverse the second half, and then compare it with the first half.

1) Use the two-pointer technique to find the middle of the linked list.
2) Reverse the second half of the linked list.
3) Compare the first half and the reversed second half node by node.
4) If all corresponding nodes match, return `True`; otherwise, return `False`.

⚠️ Common Mistakes

• Not handling the case where the list has an odd number of elements correctly.
• Forgetting to handle the case where the list is empty.

4: I-mplement

Implement the code to solve the algorithm.

class Node:
    def __init__(self, value=None, next=None):
        self.value = value
        self.next = next

def is_symmetric(head):
    if head is None or head.next is None:
        return True  # A list with 0 or 1 element is symmetric

    # Find the middle of the list
    slow = fast = head
    while fast and fast.next:
        slow = slow.next
        fast = fast.next.next
    
    # Reverse the second half of the list
    prev = None
    while slow:
        next_node = slow.next
        slow.next = prev
        prev = slow
        slow = next_node
    
    # Compare the first half and the reversed second half
    left, right = head, prev
    while right:  # Only need to compare till the end of the second half
        if left.value != right.value:
            return False
        left = left.next
        right = right.next
    
    return True

5: R-eview

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

• Verify the correctness of the solution by checking different linked lists, including those with even and odd numbers of nodes.

Example:

head = Node("Bitterling", Node("Crawfish", Node("Bitterling")))
print(is_symmetric(head))  # Expected Output: True

head = Node("Bitterling", Node("Carp", Node("Koi")))
print(is_symmetric(head))  # Expected Output: False

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 number of nodes in the linked list. We traverse the list to find the middle, reverse the second half, and compare the halves.

Space Complexity:

• O(1), since we are only using a constant amount of extra space for pointers (slow, fast, prev, left, right). This solution is efficient because it only requires one traversal of the linked list and uses minimal extra space.