2130. Maximum Twin Sum of a Linked List

1. 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.

2. Example

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.

3. Constraints

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

4. Solutions

Two Pointers

n is the number of nodes in head
Time complexity: O(n)
Space complexity: O(1)

// Note: this solution is for illustration only
// in practice, I would not modify the input list

class Solution {
public:
    int pairSum(ListNode *head) {
        ListNode *slow = head, *fast = head->next;
        while (fast->next != nullptr) {
            slow = slow->next;
            fast = fast->next->next;
        }

        ListNode *prev_node = nullptr, *current_node = slow->next;
        while (current_node != nullptr) {
            ListNode *temp = current_node->next;
            current_node->next = prev_node;

            prev_node = current_node;
            current_node = temp;
        }

        int max_sum = 0;
        while (prev_node != nullptr) {
            max_sum = max(head->val + prev_node->val, max_sum);
            head = head->next;
            prev_node = prev_node->next;
        }

        return max_sum;
    }
};