"""
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 the twins of nodes 3 and 2, respectively. All have twin sum = 6.
There are no other nodes with twins in the linked list.
Thus, the maximum twin sum of the linked list is 6.

Example 2:
Input: head = [4,2,2,3]
Output: 7
Explanation:
The nodes with twins present in this linked list are:
- Node 0 is the twin of node 3 having a twin sum of 4 + 3 = 7.
- Node 1 is the twin of node 2 having a twin sum of 2 + 2 = 4.
Thus, the maximum twin sum of the linked list is max(7, 4) = 7.

Example 3:
Input: head = [1,100000]
Output: 100001
Explanation:
There is only one node with a twin in the linked list having twin sum of 1 + 100000 = 100001.

Constraints:
The number of nodes in the list is an even integer in the range [2, 105].
1 <= Node.val <= 10^5
"""
from typing import Optional

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

class MaximumTwinSumOfaLinkedList:
    def pairSum(self, head: Optional[ListNode]) -> int:
        res = 0
        def find_mid(dummy):
            slow = fast = dummy
            while fast and fast.next:
                fast = fast.next.next
                slow = slow.next
            return slow

        def reverse(cur):
            dummy = ListNode(0)
            while cur:
                post = cur.next
                cur.next = dummy.next
                dummy.next = cur
                cur = post
            return dummy.next

        a = head
        b = reverse(find_mid(head))
        while a and b:
            res = max(res, a.val + b.val)
            a = a.next
            b = b.next
        return res
