Today's lesson will build upon our foundational understanding of linked lists by diving into practical implementation exercises using Scala. These problems will sharpen your coding skills and prepare you for scenarios you might encounter in technical interviews.

Picture a scenario in which you have a sequence of events stored in a linked list. Your task is to look back in time — essentially, to reverse the chronology of these events. In technical terms, this means traversing a singly linked list in reverse order while keeping its structure intact. This skill is critical, whether for reversing transaction logs or simply navigating through a playlist from end to start.

Consider a browser's back-button functionality, where the most recently visited pages must be revisited in reverse order. This operation mirrors our task of reverse traversal in a linked list, capturing the essence of a real-world application.

One may consider creating a new linked list while iterating over the original list, inserting each element at the head of the new list. Although this approach might work, it is an overcomplicated approach that results in extra processing and memory usage that we can avoid.

A more sophisticated solution would use a stack. With a stack, we ensure an orderly collection of the nodes' values as we navigate the list. Once the traversal is complete, we extract the values in reverse, thanks to the stack's Last-In-First-Out property.

Let's visualize it with a deck of cards: We pick each card from the top (the head of the linked list) and place it into a pile (the stack). When we finish, we pick up the cards from the pile now in reverse order.

Let's tackle it with Scala:

In this code, we create a mutable.Stack[Int] to store integers. We then iterate through the linked list using a small recursive helper function. For each node visited, we push its value onto the stack. After traversing the entire list, we pop the values off the stack. This reversal is possible because stacks operate on a Last In, First Out (LIFO) principle, which means the last element added to the stack will be the first one removed, thus reversing the order of the elements.

Note on Stack Choice: While we're using scala.collection.mutable.Stack for demonstration purposes due to its explicit stack semantics, modern Scala idiomatically prefers immutable collections. For LIFO operations, consider using a List (where :: prepends and pattern matching extracts the head), or scala.collection.immutable.ArrayDeque for better performance with larger datasets. We use mutable.Stack here to clearly illustrate the stack-based approach for educational purposes.

• Time Complexity: O(n), where n is the number of nodes in the linked list. We traverse the entire list once to push elements onto the stack, and then pop all elements once.
• Space Complexity: O(n), as we store all node values in the stack, requiring additional memory proportional to the list size.

Next up, we face a seemingly simple task: determining the size of a linked list. Imagine it as a conga line at a party. You'd like to know how many people are in the line without disturbing the dance. Start at the beginning and count each person until you reach the end.

In scenarios such as buffering data streams or executing batch operations, we often need to know the size of a list of tasks to allocate resources efficiently. This real-world implication highlights the utility of accurately determining the length of a linked list.

To calculate the length efficiently, we'll employ a classic traversal technique. It's a counting exercise: start at the first node (like the first person in the conga line) and increment a counter as we follow the links until we reach the end. The counter then reflects the conga line's length.

Implementing this strategy in Scala is straightforward:

In this code, we initialize an integer length to zero, which will serve as our counter. We then use a recursive helper function loop to visit each node of the linked list, incrementing length by one each time a new node is encountered. When the end of the list is reached (which we know occurs when currentNode is None), we return the total count.

• Time Complexity: O(n), where n is the number of nodes in the linked list. We must visit each node exactly once to count them all.
• Space Complexity: O(1), as we only use a constant amount of extra space (the length counter variable) regardless of the input size.

Today, we practiced critical operations on linked lists, namely reversing a list's traversal and calculating its length. Both solutions involved traversing the linked list but with different goals in mind. We efficiently solved these problems using structurally simple yet powerful techniques in Scala. Understanding the time and space complexity trade-offs of these approaches is crucial for selecting the right solution in real-world scenarios. We will transition to a series of exercises designed to reinforce your understanding of linked lists and prepare you for tackling similar problems in real-world scenarios or technical interviews.