# 2130. Maximum Twin Sum of a Linked List

## 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.

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, 105].
• 1 <= Node.val <= 105

## Solutions

• /**
* 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 {
List<Integer> s = new ArrayList<>();
}
int ans = 0, n = s.size();
for (int i = 0; i < (n >> 1); ++i) {
ans = Math.max(ans, s.get(i) + s.get(n - 1 - i));
}
return ans;
}
}

• /**
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode() : val(0), next(nullptr) {}
*     ListNode(int x) : val(x), next(nullptr) {}
*     ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
vector<int> s;
int ans = 0, n = s.size();
for (int i = 0; i < (n >> 1); ++i) ans = max(ans, s[i] + s[n - i - 1]);
return ans;
}
};

• # Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
def pairSum(self, head: Optional[ListNode]) -> int:
s = []
n = len(s)
return max(s[i] + s[-(i + 1)] for i in range(n >> 1))


• /**
* type ListNode struct {
*     Val int
*     Next *ListNode
* }
*/
var s []int
}
ans, n := 0, len(s)
for i := 0; i < (n >> 1); i++ {
if ans < s[i]+s[n-i-1] {
ans = s[i] + s[n-i-1]
}
}
return ans
}

• /**
* class ListNode {
*     val: number
*     next: ListNode | null
*     constructor(val?: number, next?: ListNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.next = (next===undefined ? null : next)
*     }
* }
*/

function pairSum(head: ListNode | null): number {
const arr = [];
while (node) {
arr.push(node.val);
node = node.next;
}
const n = arr.length;
let ans = 0;
for (let i = 0; i < n >> 1; i++) {
ans = Math.max(ans, arr[i] + arr[n - 1 - i]);
}
return ans;
}


• // Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
pub fn pair_sum(head: Option<Box<ListNode>>) -> i32 {
let mut arr = Vec::new();
while node.is_some() {
let t = node.as_ref().unwrap();
arr.push(t.val);
node = &t.next;
}
let n = arr.len();
let mut ans = 0;
for i in 0..n >> 1 {
ans = ans.max(arr[i] + arr[n - 1 - i]);
}
ans
}
}