Leetcode #2807: Insert Greatest Common Divisors in Linked List
In this guide, we solve Leetcode #2807 Insert Greatest Common Divisors in Linked List 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
Given the head of a linked list head, in which each node contains an integer value. Between every pair of adjacent nodes, insert a new node with a value equal to the greatest common divisor of them.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Linked List, Math, Number Theory
Intuition
There is a mathematical invariant or formula that directly leads to the result.
Using math avoids unnecessary loops and reduces complexity.
Approach
Derive the formula or update rule, then compute the answer directly.
Handle edge cases like overflow or zero carefully.
Steps:
- Identify the math relationship.
- Compute the result with a loop or formula.
- Handle edge cases.
Example
Input: head = [18,6,10,3]
Output: [18,6,6,2,10,1,3]
Explanation: The 1st diagram denotes the initial linked list and the 2nd diagram denotes the linked list after inserting the new nodes (nodes in blue are the inserted nodes).
- We insert the greatest common divisor of 18 and 6 = 6 between the 1st and the 2nd nodes.
- We insert the greatest common divisor of 6 and 10 = 2 between the 2nd and the 3rd nodes.
- We insert the greatest common divisor of 10 and 3 = 1 between the 3rd and the 4th nodes.
There are no more adjacent nodes, so we return the linked list.
Python Solution
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def insertGreatestCommonDivisors(
self, head: Optional[ListNode]
) -> Optional[ListNode]:
pre, cur = head, head.next
while cur:
x = gcd(pre.val, cur.val)
pre.next = ListNode(x, cur)
pre, cur = cur, cur.next
return head
Complexity
The time complexity is , where is the length of the linked list, and is the maximum value of the nodes in the linked list. 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.