2130 Maximum Twin Sum of a Linked List

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:

(5) -> [4] -> [2] -> (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:

(4) -> [2] -> [2] -> (3)
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:

(1) -> (100000)
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 <= 105

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# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next


"""
Using queue
"""
from collections import deque
class Solution:
    def pairSum(self, head: Optional[ListNode]) -> int:
        q = deque()
        while head:
            q.append(head.val)
            head = head.next
        max_sum = -1
        while q:
            max_sum = max(max_sum, q.popleft() + q.pop())
        return max_sum


"""
Reversing second half of the list
"""
class Solution:
    def pairSum(self, head: Optional[ListNode]) -> int:
        slow, fast = head, head
        maxVal = 0

        # Get middle of linked list
        while fast and fast.next:
            fast = fast.next.next
            slow = slow.next

        # Reverse second part of linked list
        curr, prev = slow, None

        while curr:       
            curr.next, prev, curr = prev, curr, curr.next   

        # Get max sum of pairs
        while prev:
            maxVal = max(maxVal, head.val + prev.val)
            prev = prev.next
            head = head.next

        return maxVal