2130. Maximum Twin Sum of a Linked List (M)

Amazon OA

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 <= 105

Solution:

https://leetcode.com/discuss/interview-question/1546673/Amazon-or-OA-or-LinkedListSum step 1 : count the number of nodes in the list - O(N) time step 2 : break the list from middle and reverse the second half of linked list - O(N) time and O(1) space if we ignore recursion step 3 : after step 2 above, the lists would be like 1-->4 and 2-->3 now, this is as good as traversing two lists simultaneously and computing the sums and checking for max - O(N) time

step 4 : if needed, restore the structure of the linked list back - O(N) time again

class Solution {
    public int pairSum(ListNode head) {
        
        ListNode slow = head;
        ListNode fast = head;
        while(fast != null && fast.next != null)
        {
            slow = slow.next;
            fast = fast.next.next;
        }
        
        ListNode first = head;
        ListNode second = reverse(slow);
        int result = Integer.MIN_VALUE;
        while(first != null && second != null)
        {
            result = Math.max(result, first.val + second.val);
            first = first.next;
            second = second.next;
        }
        
        return result;
    }
    
    
    //Reverse the LinkedList
    public ListNode reverse(ListNode head)
    {
        if(head == null && head.next != null)
        {
            return head;
        }
        
        ListNode prev = null; 
        ListNode current = head;
        while(current != null)
        {
            ListNode post = current.next;
            current.next = prev;
            prev = current;
            current = post;
        }
        return prev;
    }
}