In a linked list of size n, where n is even, the i^th 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.
Input: head = [5,4,2,1]Output: 6Explanation: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.
Input: head = [4,2,2,3]Output: 7Explanation: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.
Input: head = [1,100000]Output: 100001Explanation:There is only one node with a twin in the linked list having twin sum of 1 + 100000 = 100001.
from leetcode_py import ListNodeclass Solution: # Time: O(n) — find middle, reverse second half, walk pairs # Space: O(1) — in-place pointer reversal def pair_sum(self, head: ListNode[int] | None) -> int: # Slow/fast pointers: slow lands on the start of the second half slow: ListNode[int] | None = head fast: ListNode[int] | None = head while fast and fast.next: assert slow is not None slow = slow.next fast = fast.next.next # Reverse the second half in place prev: ListNode[int] | None = None current: ListNode[int] | None = slow while current: nxt = current.next current.next = prev prev = current current = nxt # Walk both halves from the ends inward best = 0 first: ListNode[int] | None = head second: ListNode[int] | None = prev while second: assert first is not None best = max(best, first.val + second.val) first = first.next second = second.next return best
