> For the complete documentation index, see [llms.txt](https://junnie.gitbook.io/nine-chapter/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://junnie.gitbook.io/nine-chapter/5.linkedlist/2130.-maximum-twin-sum-of-a-linked-list-m.md).

# 2130. Maximum Twin Sum of a Linked List (M)

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*.

&#x20;

**Example 1:**

![](https://assets.leetcode.com/uploads/2021/12/03/eg1drawio.png)

```
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:**

![](https://assets.leetcode.com/uploads/2021/12/03/eg2drawio.png)

```
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:**

![](https://assets.leetcode.com/uploads/2021/12/03/eg3drawio.png)

```
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.
```

&#x20;

**Constraints:**

* The number of nodes in the list is an **even** integer in the range `[2, 105]`.
* `1 <= Node.val <= 105`

### Solution:

<https://leetcode.com/discuss/interview-question/1546673/Amazon-or-OA-or-LinkedListSum>\
step 1 : count the number of nodes in the list - O(N) time\
step 2 : break the list from middle and reverse the second half of linked list - O(N) time and O(1) space if we ignore recursion\
step 3 : after step 2 above, the lists would be like 1-->4 and 2-->3\
now, this is as good as traversing two lists simultaneously and computing the sums and checking for max - O(N) time

step 4 : if needed, restore the structure of the linked list back - O(N) time again

```
class Solution {
    public int pairSum(ListNode head) {
        
        ListNode slow = head;
        ListNode fast = head;
        while(fast != null && fast.next != null)
        {
            slow = slow.next;
            fast = fast.next.next;
        }
        
        ListNode first = head;
        ListNode second = reverse(slow);
        int result = Integer.MIN_VALUE;
        while(first != null && second != null)
        {
            result = Math.max(result, first.val + second.val);
            first = first.next;
            second = second.next;
        }
        
        return result;
    }
    
    
    //Reverse the LinkedList
    public ListNode reverse(ListNode head)
    {
        if(head == null && head.next != null)
        {
            return head;
        }
        
        ListNode prev = null; 
        ListNode current = head;
        while(current != null)
        {
            ListNode post = current.next;
            current.next = prev;
            prev = current;
            current = post;
        }
        return prev;
    }
}
```