How do you solve Maximum Twin Sum of a Linked List on LeetCode?

Maximum Twin Sum of a Linked List (LeetCode #2130) 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: Reversing the second half of the list allows direct comparison with the first half.

What is the brute force solution for Maximum Twin Sum of a Linked List?

The brute force approach for Maximum Twin Sum of a Linked List is Brute Force, with O(n²) time complexity and O(n) space complexity. We can calculate the twin sum for every pair of nodes by iterating through the linked list. This approach is straightforward but inefficient, as it requires checking each possible pair. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Twin Sum of a Linked List?

Maximum Twin Sum of a Linked List uses the following concepts: Linked List, Two Pointers, Stack. The recommended patterns to study are: Two Pointers, Linked List.

What are the interview tips for Maximum Twin Sum of a Linked List?

Explain your thought process clearly while coding. Consider edge cases, such as the minimum input size. Be ready to discuss the time and space complexity of your solution.

What are common mistakes when solving Maximum Twin Sum of a Linked List?

Not handling the linked list traversal correctly. Forgetting to reverse the second half before calculating sums.

#2130

Maximum Twin Sum of a Linked List

Medium

Linked ListTwo PointersStackTwo PointersLinked List

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

By reversing the second half of the linked list, we can directly compute the twin sums without needing extra space for an array. This approach is efficient and leverages the structure of the linked list.

⚙️

Algorithm

3 steps

1Step 1: Use a slow and fast pointer to find the midpoint of the linked list.
2Step 2: Reverse the second half of the linked list.
3Step 3: Calculate the twin sums by iterating through the first half and the reversed second half, keeping track of the maximum sum.

solution.py24 lines

1# Full working Python code
2class ListNode:
3    def __init__(self, val=0, next=None):
4        self.val = val
5        self.next = next
6
7def pairSum(head):
8    slow = fast = head
9    while fast and fast.next:
10        slow = slow.next
11        fast = fast.next.next
12    prev = None
13    while slow:
14        temp = slow.next
15        slow.next = prev
16        prev = slow
17        slow = temp
18    max_sum = 0
19    left, right = head, prev
20    while right:
21        max_sum = max(max_sum, left.val + right.val)
22        left = left.next
23        right = right.next
24    return max_sum

ℹ

Complexity note: The time complexity is O(n) because we traverse the list a constant number of times (finding the midpoint, reversing, and calculating sums). The space complexity is O(1) since we only use a few pointers and do not require extra space proportional to the input size.

1Reversing the second half of the list allows direct comparison with the first half.
2Using pointers efficiently reduces space complexity.

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