Losing head/tail while rewiring
Wrong move: Pointer updates overwrite references before they are saved.
Usually fails on: List becomes disconnected mid-operation.
Fix: Store next pointers first and use a dummy head for safer joins.
Move from brute-force thinking to an efficient approach using linked list strategy.
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.
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:
[2, 105].1 <= Node.val <= 105Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Linked List · Two Pointers · Stack
[5,4,2,1]
[4,2,2,3]
[1,100000]
reverse-linked-list)palindrome-linked-list)middle-of-the-linked-list)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2130: Maximum Twin Sum of a Linked List
/**
* 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;
}
}
// Accepted solution for LeetCode #2130: Maximum Twin Sum of a Linked List
/**
* 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
}
# Accepted solution for LeetCode #2130: Maximum Twin Sum of a Linked List
# 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))
// Accepted solution for LeetCode #2130: Maximum Twin Sum of a Linked List
// 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();
let mut node = &head;
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
}
}
// Accepted solution for LeetCode #2130: Maximum Twin Sum of a Linked List
/**
* Definition for singly-linked list.
* 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 = [];
let node = head;
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;
}
Use this to step through a reusable interview workflow for this problem.
Copy all n nodes into an array (O(n) time and space), then use array indexing for random access. Operations like reversal or middle-finding become trivial with indices, but the O(n) extra space defeats the purpose of using a linked list.
Most linked list operations traverse the list once (O(n)) and re-wire pointers in-place (O(1) extra space). The brute force often copies nodes to an array to enable random access, costing O(n) space. In-place pointer manipulation eliminates that.
Review these before coding to avoid predictable interview regressions.
Wrong move: Pointer updates overwrite references before they are saved.
Usually fails on: List becomes disconnected mid-operation.
Fix: Store next pointers first and use a dummy head for safer joins.
Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.
Usually fails on: A valid pair can be skipped when only one side should move.
Fix: Move exactly one pointer per decision branch based on invariant.
Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.
Usually fails on: Indices point to blocked elements and outputs shift.
Fix: Pop while invariant is violated before pushing current element.