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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

Solution 1

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# 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 = []
        while head:
            s.append(head.val)
            head = head.next
        n = len(s)
        return max(s[i] + s[-(i + 1)] for i in range(n >> 1))

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/**
 * 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) {
        List<Integer> s = new ArrayList<>();
        for (; head != null; head = head.next) {
            s.add(head.val);
        }
        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;
    }
}

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/**
 * Definition for singly-linked list.
 * 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:
    int pairSum(ListNode* head) {
        vector<int> s;
        for (; head != nullptr; head = head->next) s.push_back(head->val);
        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;
    }
};

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/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func pairSum(head *ListNode) int {
    var s []int
    for ; head != nil; head = head.Next {
        s = append(s, head.Val)
    }
    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
}

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