Maximum Twin Sum of a Linked List

Description

In a linked list of size n, where n is even, the ith node (0-indexed) of the linked list is known as the twin of the (n-1-i)th node, if 0 <= i <= (n / 2) - 1.

For example, if n = 4, then node 0 is the twin of node 3, and node 1 is the twin of node 2. These are the only nodes with twins for n = 4.

The twin sum is defined as the sum of a node and its twin.

Given the head of a linked list with even length, return the maximum twin sum of the linked list.

Example 1:
Input: head = [5, 4, 2, 1]
Output: 6
Explanation: Nodes 0 and 1 are twins of nodes 3 and 2 respectively. Twin sums are: 5 + 1 = 6, 4 + 2 = 6. The maximum twin sum is 6.

Example 2:
Input: head = [4, 2, 2, 3]
Output: 7
Explanation: Nodes 0 and 1 are twins of nodes 3 and 2 respectively. Twin sums are: 4 + 3 = 7, 2 + 2 = 4. The maximum twin sum is 7.

Example 3:
Input: head = [1, 100000]
Output: 100001
Explanation: Only one pair of twins (node 0 and node 1). Twin sum is 1 + 100000 = 100001.

Constraints

The number of nodes in the list is an even integer in the range [2, 10^5]
1 <= Node.val <= 10^5

Signature

def solution(head: ListNode) -> int: ...

Complexity

Time: O(n)
Space: O(1)

Hints

Pattern: Two pointers with linked list reversal
Approach: Use slow/fast pointers to find middle, reverse second half, then sum pairs from start and reversed second half
Complexity: Reverse second half, then compare first and reversed second half

Solutions

Python

from typing import Optional


class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next


def solution(head: Optional[ListNode]) -> int:
    # n = head
    slow = fast = head
    # prev = None
    while fast and fast.next:
        # print(n.val)
        # n = n.next
        # prev = slow
        slow = slow.next
        fast = fast.next.next

    # print(prev.val)
    # print(slow.val)
    curr = slow
    prev = None
    while curr:
        # print(curr.val, curr.next)
        next_node = curr.next
        curr.next = prev
        prev = curr
        curr = next_node

    # print(prev.val)
    # print(head.val)
    p1 = head
    p2 = prev
    ans = 0
    while p2:
        ans = max(ans, p1.val + p2.val)
        p1 = p1.next
        p2 = p2.next

    return ans