Leetcode #1634: Add Two Polynomials Represented as Linked Lists
In this guide, we solve Leetcode #1634 Add Two Polynomials Represented as Linked Lists 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
A polynomial linked list is a special type of linked list where every node represents a term in a polynomial expression. Each node has three attributes: coefficient: an integer representing the number multiplier of the term.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Linked List, Math, 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: poly1 = [[1,1]], poly2 = [[1,0]]
Output: [[1,1],[1,0]]
Explanation: poly1 = x. poly2 = 1. The sum is x + 1.
Python Solution
# Definition for polynomial singly-linked list.
# class PolyNode:
# def __init__(self, x=0, y=0, next=None):
# self.coefficient = x
# self.power = y
# self.next = next
class Solution:
def addPoly(self, poly1: "PolyNode", poly2: "PolyNode") -> "PolyNode":
dummy = curr = PolyNode()
while poly1 and poly2:
if poly1.power > poly2.power:
curr.next = poly1
poly1 = poly1.next
curr = curr.next
elif poly1.power < poly2.power:
curr.next = poly2
poly2 = poly2.next
curr = curr.next
else:
if c := poly1.coefficient + poly2.coefficient:
curr.next = PolyNode(c, poly1.power)
curr = curr.next
poly1 = poly1.next
poly2 = poly2.next
curr.next = poly1 or poly2
return dummy.next
Complexity
The time complexity is O(n) (after optional sort O(n log n)). The space complexity is O(1).
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.