2130. Maximum Twin Sum of a Linked List

Description of Problem

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 <= 10^5

Solution 1 - using extra space

Tags: LinkedList

Explanation

Calculate the length of the linkedlist.
Loop over the first-half list: store values into the stack.
Loop over the second-half list:
- Remove values from the top of the stack.
- Calculate the sum of the value from the stack and current value.
- Calculate maximum sum.

Code

/**
 * 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) {
        int length = 0;
        ListNode * curr = head;
        while(curr != nullptr){
            length++;
            curr = curr->next;
        }

        int ans = INT_MIN;
        int halfLength = length / 2;
        vector<int> v;
        curr = head;
        while(curr != nullptr){
            if (length <= halfLength){
                ans = max(ans, curr->val + v.back());
                v.pop_back();
                
            }else{
                v.push_back(curr->val);   
            }
            length--;
            curr = curr->next;
        }

        return ans;
        
    }
};

Complexity

n is length of the linkedlist

Time Complexity

\(T(n) = \Theta(n) \)

Auxiliary Space

\(S(n) = \Theta(n) \)

Solution 2 - Cut, Reverse and Combine

Tags: LinkedList Two Pointers

Code

/**
 * 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) {
        ListNode * slow = head;
        ListNode * fast = head;

        // Cut the list
        while(fast->next != nullptr && fast->next->next != nullptr){
            slow = slow->next;
            fast = fast->next->next;
        }
        fast = slow->next;
        slow->next = nullptr;

        // Reverse the second-half list
        ListNode * prev = nullptr;
        ListNode * curr = fast;
        ListNode * node;
        while( (node = curr) != nullptr ){
            curr = node->next;
            node->next = prev;
            prev = node;
        }

        // Combine two lists
        ListNode * h1 = head;
        ListNode * h2 = prev;
        int ans = INT_MIN;
        while (h1 != nullptr && h2 != nullptr){
            ans = max(ans, h1->val + h2->val);
            h1 = h1->next;
            h2 = h2->next;
        }

        return ans;
    }
};

Complexity

n is length of the linkedlist

Time Complexity

\(T(n) = \Theta(n) \)

Auxiliary Space

\(S(n) = O(1) \)

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