Hello once again, champion of code! In this session, we will delve into the world of coding interviews by focusing on stack-based problems. We endeavor to decode interview questions that leverage the Last-In, First-Out (LIFO) magic 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 depth in data structure understanding.

Imagine a sequence of integers representing the highs and lows of a mountain range. Each integer denotes the height of a peak, and you're moving from left to right, tracking peaks that are shorter than the current one you're standing on. For every peak, the task is to identify the height of the nearest preceding peak that is shorter — a scenario perfectly suited for stacks.

Alternatively, think about monitoring daily temperatures over several months. You're interested in determining the last day when the temperature was cooler for each day you check. This is analogous to finding the previous smaller number for each entry in the array. Stacks excel at handling these types of sequence queries efficiently.

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 multiple times and spending excessive amounts of time as you consider each element repeatedly. In a vast dataset, this would be akin to retracing your steps on each day's hike to find a shorter peak — an exhausting proposition!

Enter the stack — our trusty guide. As we progress through the array, we push peaks onto the stack. When we encounter a peak, 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.

Ruby
Copy to clipboard
1def find_preceding_smaller_elements(arr)
2  result = []
3  stack = []
4
5  arr.each do |current|
6    # Pop elements from stack while they are greater than or equal to current
7    while !stack.empty? && stack.last >= current
8      stack.pop
9    end
10    # If stack is empty, no smaller element, otherwise append top element
11    result << (stack.empty? ? -1 : stack.last)
12    stack.push(current)
13  end
14
15  result
16end
17
18arr = [3, 7, 1, 5, 4, 3]
19result = find_preceding_smaller_elements(arr)
20puts result.join(" ")  # Output: -1 3 -1 1 1 1

In our code, we trek through each element in the array. Our conditions within the loop perform the 'pop' work — discarding any peak that isn't lower than our current one, ensuring that only useful candidates remain. Then, we notate 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.

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.

The stroke of genius here is using not one but two stacks. 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, ready to be the new minimum.

Let's translate our idea into Ruby code, creating an efficient stack with minimum trace functionality:

Ruby
Copy to clipboard
1class MinStack
2  def initialize
3    @stack = []
4    @min_values = []
5  end
6
7  # The push method is where most of our logic resides.
8  def push(x)
9    @min_values.push(x) if @min_values.empty? || x <= @min_values.last
10    @stack.push(x)
11  end
12
13  # Each pop requires careful coordination between our two stacks.
14  def pop
15    if !@stack.empty? && @stack.last == @min_values.last
16      @min_values.pop
17    end
18    @stack.pop unless @stack.empty?
19  end
20
21  # The top method reveals the peak of our stack cable car.
22  def top
23    @stack.empty? ? -1 : @stack.last
24  end
25
26  # GetMin serves as our on-demand minimum value provider.
27  def get_min
28    @min_values.empty? ? -1 : @min_values.last
29  end
30end
31
32# Demonstration
33min_stack = MinStack.new
34min_stack.push(5)
35min_stack.push(3)
36puts min_stack.get_min # Output: 3
37min_stack.pop
38puts min_stack.get_min # Output: 5

The push method introduces the key player — our @min_values stack, 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 get_min yields the correct minimum every time, thanks to our additional stack, @min_values.

Our expedition through stack-land today 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!