Doubly Linked List
Detailed guide and Python implementation for the 'Doubly Linked List' problem.
1. Concept Overview
The 'Doubly 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 Doubly Linked List.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Doubly 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 create_doubly_linked_list(arr) that takes a list of integers arr, constructs a doubly linked list, and returns a tuple of two lists: (forward_traverse, backward_traverse) to verify that both the next and prev pointers are correctly configured.
- •0 <= len(arr) <= 1000
- •-10^4 <= arr[i] <= 10^4
Examples
arr = [1, 2, 3]
([1, 2, 3], [3, 2, 1])
Constructing the doubly linked list 1 <=> 2 <=> 3 allows traversing forward to get [1, 2, 3] and backward to get [3, 2, 1].
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
class DoublyListNode:
def __init__(self, val=0, next=None, prev=None):
self.val = val
self.next = next
self.prev = prev
def create_doubly_linked_list_opt(arr: list):
if not arr: return [], []
head = DoublyListNode(arr[0])
curr = head
for val in arr[1:]:
new_node = DoublyListNode(val, prev=curr)
curr.next = new_node
curr = new_node
f, b = [], []
c = head
while c:
f.append(c.val)
c = c.next
c = curr
while c:
b.append(c.val)
c = c.prev
return f, bBrute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
class DoublyListNode:
def __init__(self, val=0, next=None, prev=None):
self.val = val
self.next = next
self.prev = prev
def create_doubly_linked_list_brute(arr: list):
if not arr: return [], []
head = DoublyListNode(arr[0])
curr = head
for x in arr[1:]:
n = DoublyListNode(x)
curr.next = n
n.prev = curr
curr = n
f, b = [], []
c = head
while c:
f.append(c.val)
c = c.next
c = curr
while c:
b.append(c.val)
c = c.prev
return f, bAlgorithm Pattern Checklist
When dealing with Linked Lists data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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