In this session, we will delve into the world of coding interviews by focusing on stack-based problems. We will endeavor to decode interview questions that leverage the Last-In, First-Out (LIFO) behavior of stacks to offer elegantly efficient solutions. After today, not only will you be able to handle stacks with ease, but you'll also be able to articulate and apply this knowledge when faced with interview questions that require a deep understanding of data structures.

Consider a sequence of integers like the peaks and valleys of a mountain range. Each peak has a height represented by an integer, and you're hiking from left to right, recording peaks shorter than the one you're currently on. For each peak, we want to find out the height of the nearest preceding peak that's lower than it — a classic problem where stacks excel.

Envision analyzing daily temperatures over several months. You're interested in knowing the most recent cooler day for each day you examine. This mirrors our array problem, where we're seeking the previous smaller number before each entry in the array. It’s these kinds of time-sensitive queries that stack operations handle effortlessly.

You might be tempted to approach this problem with the vigor of a brute force assault — looking behind each element to find a smaller one. However, this could mean reviewing elements multiple times and spending excessive time as you consider each element repeatedly. In a vast data set, this would be akin to retracing your steps on each day's hike to find a shorter peak — an exhausting proposition!

As we progress through the array, we push peaks onto the stack. When we encounter a peak (arr[i]), we pop entries from the stack that aren't shorter than the current one. The stack's top now reveals the nearest preceding smaller peak, which we note before adding the current peak to the stack.

Let's lace up our boots and start the ascent by iterating through the array of peak heights and interacting with our stack.

In our code, we traverse each element in the array (arr). Our conditions within the loop perform the 'pop' operation — discarding any peak that isn't lower than our current one, ensuring that only useful candidates remain. Then, we record the result — either -1 if no such peak exists or the last peak remaining on the stack. Before moving on, we add our current peak to the stack.

The naive approach for finding preceding smaller elements involves checking each element against all previous elements, resulting in a time complexity of O(n^2). This is because for each element, you may need to traverse back through all preceding elements to find a smaller one.

In contrast, the stack-based approach optimizes this process to O(n). As we iterate through the array, each element is pushed and popped from the stack at most once, ensuring that the total number of operations is linear relative to the number of elements.

Think about a real-time inventory tracking system in a warehouse where items are stacked based on the order of arrival. However, you must keep an ongoing record of the lightest item in stock for quick retrieval. This scenario highlights the need for a system that efficiently maintains a snapshot of the minimum item as stack operations proceed.

Consider tagging each item with its weight and then brute-forcing through the stack to find the minimum every time it's needed. However, this is like rummaging through the entire stock each time a request is made — an excessive and inefficient undertaking.

We can approach this minimum element problem by utilizing two stacks instead of one. The secondary stack acts as a memory, holding the minimum value attained with each element pushed to the primary stack. This way, when the current minimum leaves the stack, the next one in line is right at the top of the auxiliary stack.

It's time to manifest this brainchild into Kotlin code. Here's the skeletal structure of our MinStack:

The push method introduces the key player — our minValues list, which retains the minimum value observed so far every time we add a new entry. Meanwhile, the pop operation is like a relay race transition, handing off the title "minimum" to the next contender when the current titleholder is knocked off the podium.

Simulating the pushing of various elements onto the stack and invoking getMin would yield the correct minimum every time, thanks to our additional minValues list.

The naive approach for maintaining a minimum trace in a stack involves scanning through the stack to find the minimum each time it's needed, leading to a time complexity of O(n) for each minimum retrieval.

By utilizing two stacks, the MinStack approach allows us to maintain the minimum value in constant time O(1). The auxiliary stack keeps track of the minimum values, ensuring that each getMin operation retrieves the minimum element without additional traversal, thus achieving constant time efficiency.

Our expedition has shown us that stacks can be the clever trick up your sleeve for certain types of interview questions. We have seen how to keep track of past element states with 'Preceding Smaller Elements' and maintain instant access to the minimum element in our 'MinStack'. From trails to inventory — stacks reveal their flexibility and efficiency. Thus, your toolbox of algorithms has just received a shiny new set of tools, bolstering your confidence for what lies ahead — practice!