How do you solve Insert Greatest Common Divisors in Linked List on LeetCode?

Insert Greatest Common Divisors in Linked List (LeetCode #2807) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding GCD is crucial for this problem.

What is the brute force solution for Insert Greatest Common Divisors in Linked List?

The brute force approach for Insert Greatest Common Divisors in Linked List is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves iterating through the linked list and calculating the GCD for each pair of adjacent nodes. For every pair, we create a new node with the GCD value and link it appropriately. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Insert Greatest Common Divisors in Linked List?

Insert Greatest Common Divisors in Linked List uses the following concepts: Linked List, Math, Number Theory. The recommended patterns to study are: Linked List, Mathematical Operations.

What are the interview tips for Insert Greatest Common Divisors in Linked List?

Explain your thought process clearly while coding. Test your code with edge cases after writing. Be prepared to discuss time and space complexities.

What are common mistakes when solving Insert Greatest Common Divisors in Linked List?

Forgetting to update the current pointer after insertion. Not handling edge cases like single-node lists.

#2807

Insert Greatest Common Divisors in Linked List

Medium

Linked ListMathNumber TheoryLinked ListMathematical Operations

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal approach is similar to the brute force method but focuses on efficiently inserting nodes while traversing the list. We maintain a single pointer and adjust the links as we go, ensuring we only traverse the list once.

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Algorithm

5 steps

1Step 1: Initialize a pointer to the head of the linked list.
2Step 2: While the pointer has a next node, calculate the GCD of the current node and the next node.
3Step 3: Create a new node with the GCD value and insert it between the current node and the next node.
4Step 4: Move the pointer to the next node (skipping the newly inserted node).
5Step 5: Repeat until the end of the list is reached.

solution.py14 lines

1import math
2class ListNode:
3    def __init__(self, val=0, next=None):
4        self.val = val
5        self.next = next
6
7def insertGCD(head):
8    current = head
9    while current and current.next:
10        gcd_value = math.gcd(current.val, current.next.val)
11        new_node = ListNode(gcd_value, current.next)
12        current.next = new_node
13        current = new_node.next
14    return head

ℹ

Complexity note: The time complexity is O(n) because we only traverse the linked list once. The space complexity is O(1) as we are modifying the list in place without using any extra space.

1Understanding GCD is crucial for this problem.
2Linked list manipulation requires careful handling of pointers.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.