Leetcode #1265: Print Immutable Linked List in Reverse
In this guide, we solve Leetcode #1265 Print Immutable Linked List in Reverse in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given an immutable linked list, print out all values of each node in reverse with the help of the following interface: ImmutableListNode: An interface of immutable linked list, you are given the head of the list. You need to use the following functions to access the linked list (you can't access the ImmutableListNode directly): ImmutableListNode.printValue(): Print value of the current node.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Stack, Recursion, Linked List, Two Pointers
Intuition
The constraints hint that we can reason about two ends of the data at once, which is perfect for a two-pointer scan.
Moving one pointer at a time keeps the invariant intact and avoids nested loops.
Approach
Place pointers at the left and right ends and move them based on the comparison or target condition.
This yields a clean linear pass after any required sorting.
Steps:
- Set left and right pointers.
- Move a pointer based on the condition.
- Update the best answer while scanning.
Example
Input: head = [1,2,3,4]
Output: [4,3,2,1]
Python Solution
# """
# This is the ImmutableListNode's API interface.
# You should not implement it, or speculate about its implementation.
# """
# class ImmutableListNode:
# def printValue(self) -> None: # print the value of this node.
# def getNext(self) -> 'ImmutableListNode': # return the next node.
class Solution:
def printLinkedListInReverse(self, head: 'ImmutableListNode') -> None:
if head:
self.printLinkedListInReverse(head.getNext())
head.printValue()
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.