Palindrome Linked List
Detailed guide and Python implementation for the 'Palindrome Linked List' problem.
1. Concept Overview
The 'Palindrome Linked List' problem is a key challenge in the Linked Lists section.
This implementation focuses on easy-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Palindrome Linked List.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Palindrome Linked List carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Problem Statement
Write a function is_palindrome_list(arr) that represents a singly linked list as a Python list arr, and returns True if it is a palindrome, and False otherwise.
- •0 <= len(arr) <= 10^5
- •0 <= arr[i] <= 9
Examples
arr = [1, 2, 2, 1]
True
The list reads the same forward and backward.
arr = [1, 2, 3]
False
The list reads [1, 2, 3] forward but [3, 2, 1] backward, which is not equal.
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
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Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
def is_palindrome_list_opt(arr):
if isinstance(arr, list):
h = build_linked_list(arr)
return is_palindrome_list_opt_helper(h)
return is_palindrome_list_opt_helper(arr)
def is_palindrome_list_opt_helper(head: ListNode) -> bool:
if not head or not head.next:
return True
slow, fast = head, head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
prev, curr = None, slow
while curr:
nxt = curr.next
curr.next = prev
prev = curr
curr = nxt
left, right = head, prev
while right:
if left.val != right.val:
return False
left = left.next
right = right.next
return TrueBrute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def is_palindrome_list_brute(arr):
if isinstance(arr, list):
h = build_linked_list(arr)
return is_palindrome_list_brute_helper(h)
return is_palindrome_list_brute_helper(arr)
def is_palindrome_list_brute_helper(head: ListNode) -> bool:
vals = []
curr = head
while curr:
vals.append(curr.val)
curr = curr.next
return vals == vals[::-1]Algorithm Pattern Checklist
When dealing with Linked Lists data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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