Introduction to the Lesson

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 dig for depth in data structure understanding.

Problem 1: Preceding Smaller Elements

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.

Problem 1: Actualization

Envision analyzing daily temperatures over several months. You're interested in knowing the last 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 without breaking a sweat.

Problem 1: Naive Approach

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 unforgiving 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!

Problem 1: Efficient Approach

Enter the stack — our trusty Sherpa. 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.

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