2130. Maximum Twin Sum of a Linked List

Problem

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, 10^5].

• 1 <= Node.val <= 10^5

Solution (Java)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public int pairSum(ListNode head) {
        if (head == null) {
            return 0;
        }
        int maxSum = Integer.MIN_VALUE;
        ListNode slow = head;
        ListNode fast = head;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        if (slow.next == null) {
            return head.val + slow.val;
        }
        ListNode tail = head;
        ListNode pivot = reverse(slow);
        while (pivot != null) {
            maxSum = Math.max(maxSum, (tail.val + pivot.val));
            tail = tail.next;
            pivot = pivot.next;
        }
        return maxSum;
    }

    private ListNode reverse(ListNode head) {
        ListNode tail = head;
        ListNode prev = null;
        while (tail != null) {
            ListNode temp = tail.next;
            tail.next = prev;
            prev = tail;
            tail = temp;
        }
        return prev;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).