Maximum Twin Sum of a Linked List

class Solution {
public:
    int maxTwinSum(ListNode* head) {
        ListNode* slow_ptr = head;
        ListNode* fast_ptr = head;

        while (fast_ptr && fast_ptr->next) {
            fast_ptr = fast_ptr->next->next;
            slow_ptr = slow_ptr->next;
        }

        ListNode *nextNode, *previous = NULL;
        while (slow_ptr) {
            nextNode = slow_ptr->next;
            slow_ptr->next = previous;
            previous = slow_ptr;
            slow_ptr = nextNode;
        }

        int maxSum = 0;
        ListNode* start_ptr = head;
        while (previous) {
            maxSum = max(maxSum, start_ptr->val + previous->val);
            previous = previous->next;
            start_ptr = start_ptr->next;
        }

        return maxSum;
    }
};

The given C++ code defines a solution for finding the maximum twin sum in a linked list. A twin sum of a linked list is calculated by summing matching elements from the start and the reversed second half. This process involves finding a middle point, reversing the second half of the list, and then comparing nodes from both halves to find the maximum sum.

• Initialize two pointers, slow_ptr and fast_ptr, from the head of the list to find the middle. Move fast_ptr two steps and slow_ptr one step until fast_ptr reaches the end.
• Reverse the second half of the linked list starting from slow_ptr. Use an extra pointer, previous, to keep track of the previous node while reversing.
• Calculate the twin sums by iterating through the first half and the reversed second half simultaneously. Keep track of the maximum sum encountered during this process.
• Return the maximum sum obtained.

This solution efficiently computes the maximum twin sum with a time complexity of O(n), where n is the number of nodes in the linked list, and it operates with a space complexity of O(1), using only a few pointers.

class Solution {
    public int maximumPairSum(ListNode head) {
        ListNode slower = head;
        ListNode faster = head;

        // Find the middle point in the list
        while (faster != null && faster.next != null) {
            faster = faster.next.next;
            slower = slower.next;
        }

        // Reverse the second part of the list
        ListNode current, last = null;
        while (slower != null) {
            current = slower.next;
            slower.next = last;
            last = slower;
            slower = current;
        }

        ListNode initializer = head;
        int maxSum = 0;
        while (last != null) {
            maxSum = Math.max(maxSum, initializer.val + last.val);
            last = last.next;
            initializer = initializer.next;
        }

        return maxSum;
    }
}

This solution tackles the problem of finding the maximum twin sum in a linked list using Java. The problem requires calculating the highest sum of pairs of values, where each pair consists of nodes symmetrically situated around the center of the list.

To handle this task, the method maximumPairSum(ListNode head) is defined, which performs the following actions:

• Firstly, initialize two pointers, slower and faster. Both start at the head of the linked list. Use these pointers to find the middle of the list. The faster pointer moves at twice the speed of the slower one, ensuring that when faster reaches the end of the list, slower is at the middle.

• Once the middle is found, the next part involves reversing the second half of the linked list starting from the slower pointer. This reversal is achieved through a classic linked list reversal process using temporary pointers.

• After reversing the second half of the list, start from the head of the list with a new pointer, initializer, and iterate through the list comparing values from the start (initializer) and the reversed second half (last). Calculate the sum for each pair and update the maximum sum found (maxSum) through this traversal.

• The loop continues until the end of the list is reached, ensuring that each twin sum has been calculated.

The method ultimately returns the maximum twin sum (maxSum), successfully determining the highest sum of node values from symmetrically opposite ends of the linked list post reversal. This approach of twin pointers followed by a second-half reversal is efficient in terms of both time and space, effectively solving the problem within linear complexity.