Maximum Twin Sum of a Linked List

Problem Statement:-

In a linked list of size n, where n is even, the i<sup>th</sup> node (0-indexed) of the linked list is known as the twin of the (n-1-i)<sup>th</sup> 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.

Link: https://leetcode.com/problems/maximum-twin-sum-of-a-linked-list/description/

Problem Explanation with examples:-

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<sup>5</sup>].

• 1 <= Node.val <= 10<sup>5</sup>

Intuition:-

• Simply store the values in an array and take two pointers at the start and end of the array.

• Keep updating the sum and return the maximum sum.

Solution:-

• Create an empty list.

• Traverse the linked list and store the values in the list.

• Take two pointers i and j at the start and end of the list.

• Take a sum variable and initialize it to 0.

• Traverse the list till i < j and keep updating the sum by the maximum of sum and arr[i]+arr[j].

• Return the maximum sum.

Code:-

JAVA Solution

class Solution {
    public int pairSum(ListNode head) {
        long maxTwinSum = 0,currentTwinSum = 0;
        ArrayList<Integer> arr = new ArrayList<>();
        ListNode temp;
        temp = head;
        while (temp.next != null)
        {
            arr.add(temp.val);
            temp = temp.next;
        }
        arr.add(temp.val);
        int i = 0,j = arr.size()-1;
        while (i<j){
            currentTwinSum = arr.get(i) + arr.get(j);
            maxTwinSum = Math.max(maxTwinSum,currentTwinSum);
            i++;
            j--;
        }

        return (int) maxTwinSum;
    }
}

Python Solution

class Solution:
    def pairSum(self, head: Optional[ListNode]) -> int:
        arr = []
        while head:
            arr.append(head.val)
            head = head.next
        n = len(arr)
        i = 0 
        j = n-1
        sm = 0
        while i < j:
            sm = max(sm,arr[i]+arr[j])
            i += 1
            j -= 1

        return sm

Complexity Analysis:-

TIME:-

The time complexity is O(n) where n is the number of nodes in the linked list and we traverse the linked list once to store the values in the list and once to find the maximum sum.

SPACE:-

The space complexity is O(n) where n is the number of nodes in the linked list and we store the values in the list.

References:-

• Linked List

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