Naive Approach: Brute Force

One straightforward approach is to iterate through the linked list, store the values in an array, and then calculate the twin sums by comparing elements from the start and end of the array. Finally, find the maximum twin sum.

Algorithm:

1. Traverse the linked list and store the value of each node in an array.
2. Initialize max_twin_sum to 0.
3. Iterate from i = 0 to n / 2 - 1:
   • Calculate twin_sum = array[i] + array[n - 1 - i].
   • Update max_twin_sum = max(max_twin_sum, twin_sum).
4. Return max_twin_sum.

Code (Python):

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def pairSum_brute_force(head: ListNode) -> int:
    arr = []
    curr = head
    while curr:
        arr.append(curr.val)
        curr = curr.next
    
    n = len(arr)
    max_sum = 0
    for i in range(n // 2):
        max_sum = max(max_sum, arr[i] + arr[n - 1 - i])
    
    return max_sum

Time Complexity: O(n) for traversing the linked list and O(n/2) which simplifies to O(n) for calculating twin sums, resulting in O(n) overall.

Space Complexity: O(n) for storing the linked list values in an array.

Optimal Approach: Using a Stack and Two Pointers

A more efficient approach involves using a stack to store the values of the first half of the linked list. Then, we can iterate through the second half of the list and calculate the twin sums by pairing each node with the value popped from the stack. This avoids the need for an additional array.

Algorithm:

1. Use slow and fast pointers to find the middle of the linked list. The slow pointer will point to the middle.
2. Push the first half of the linked list into a stack as you traverse it using the slow pointer.
3. Move slow pointer to the second half and pop from stack to find twin.
4. Return the maximum of the twin sums.

Code (Python):

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def pairSum(head: ListNode) -> int:
    slow, fast = head, head
    stack = []
    
    # Traverse the first half of linked list and push into stack
    while fast and fast.next:
        stack.append(slow.val)
        slow = slow.next
        fast = fast.next.next
    
    max_sum = 0
    # Traverse the second half and compare with stack
    while slow:
        max_sum = max(max_sum, slow.val + stack.pop())
        slow = slow.next
        
    return max_sum

Time Complexity: O(n) for traversing the linked list.

Space Complexity: O(n/2) which simplifies to O(n) for the stack, as it stores roughly half the elements of the linked list.

Edge Cases:

• Empty List: The problem statement specifies that the list will have at least 2 nodes, so an empty list isn't an edge case to worry about.
• List with only two nodes: Both algorithms will function correctly.

Summary:

The optimal solution provides the same O(n) time complexity, but reduces the space complexity from O(n) (for the brute force method) to O(n/2) by utilizing stack which is still O(n). This makes it more space-efficient, particularly for large linked lists.