Input
["MyLinkedList", "addAtHead", "addAtTail", "addAtIndex", "get", "deleteAtIndex", "get"]
[[], [1], [3], [1, 2], [1], [1], [1]]
Output
[null, null, null, null, 2, null, 3]

Explanation
MyLinkedList myLinkedList = new MyLinkedList();
myLinkedList.addAtHead(1);
myLinkedList.addAtTail(3);
myLinkedList.addAtIndex(1, 2);    // linked list becomes 1->2->3
myLinkedList.get(1);              // return 2
myLinkedList.deleteAtIndex(1);    // now the linked list is 1->3
myLinkedList.get(1);              // return 3

Solution Explanation:

This problem requires designing a linked list data structure. Two solutions are provided: one using a singly linked list and another using an array-based implementation that simulates a linked list.

Solution 1: Singly Linked List

This solution uses a singly linked list, where each node contains a value (val) and a pointer to the next node (next). A dummy node is used at the head to simplify the operations.

Data Structure:

• dummy: A dummy node at the beginning of the list. This simplifies the add and delete operations at the head because you don't have to check for an empty list.
• cnt: Keeps track of the number of elements in the linked list.

Methods:

• get(index): Iterates through the list until the indexth node is found and returns its value. Returns -1 if the index is out of bounds.
• addAtHead(val): Adds a new node with the given value to the beginning of the list.
• addAtTail(val): Adds a new node with the given value to the end of the list.
• addAtIndex(index, val): Adds a new node with the given value before the indexth node. Handles edge cases where index is 0 (add to head) or equal to cnt (add to tail). Does nothing if index is greater than cnt.
• deleteAtIndex(index): Deletes the indexth node from the list. Handles edge cases where index is 0 (delete from head) or out of bounds.

Time Complexity:

• get(index): O(index) - Linear time because it might need to traverse up to index nodes.
• addAtHead(val): O(1) - Constant time as it adds to the head.
• addAtTail(val): O(n) - Linear time because it needs to iterate to the tail.
• addAtIndex(index, val): O(index) - Linear time because it might need to traverse up to index nodes.
• deleteAtIndex(index): O(index) - Linear time because it might need to traverse up to index nodes.

Space Complexity: O(n) - Linear space due to the linked list storing n nodes.

Solution 2: Array-Based Simulation

This solution simulates a linked list using arrays. e stores the values, ne stores the index of the next node, head stores the index of the first node, idx stores the index of the next available slot in the arrays, and cnt stores the number of nodes.

Data Structures:

• e: Array to store the values of the nodes.
• ne: Array to store the index of the next node for each node.
• head: Index of the first node in the list.
• idx: Index of the next available slot in the arrays.
• cnt: Number of nodes in the list.

Methods: The methods are analogous to those in Solution 1, but they use array indices to represent the links between nodes.

Time Complexity:

• get(index): O(index)
• addAtHead(val): O(1)
• addAtTail(val): O(n) - In the worst case (adding to the end), it might have to traverse the entire array to find the tail.
• addAtIndex(index, val): O(index)
• deleteAtIndex(index): O(index)

Space Complexity: O(1010) - Constant space because the arrays are pre-allocated. While technically not strictly constant, the space is fixed and doesn't depend on the number of nodes, making it effectively constant within the constraint of at most 2000 calls.

Choosing between Solutions 1 and 2:

Solution 1 (singly linked list) is generally preferred for its dynamic memory management and scalability. Solution 2 (array simulation) is provided as an alternative approach that can be more efficient in terms of memory in some scenarios if the maximum number of nodes is known and relatively small (as in this problem's constraints). For very large linked lists where dynamic allocation becomes expensive, an array-based approach might be slightly better. But for most practical applications, Solution 1 is better because it can handle many more nodes without constraints.