Lesson 3

Mastering Stacks in Interviews: Navigating Through Common Challenges

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.

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 with Stacks

With stacks, we take on the role of a librarian who cleverly keeps track of how the heights of books change as she adds new ones to the shelf. Thus, the librarian can instantly recall the last book, shorter than the current one, by looking at the top of the stack.

Problem 1: Building The Solution Step-by-Step

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

JavaScript
1function findPrecedingSmallerElements(arr) { 2 let stack = []; 3 let result = new Array(arr.length).fill(-1); 4 5 for (let i = 0; i < arr.length; i++) { 6 while (stack.length > 0 && stack[stack.length - 1].value >= arr[i]) { 7 stack.pop(); 8 } 9 if (stack.length > 0) { 10 result[i] = stack[stack.length - 1].value; 11 } 12 stack.push({ value: arr[i], index: i }); 13 } 14 return result; 15} 16 17 18console.log(findPrecedingSmallerElements([1, 3, 2, 5, 4, 7])); 19// Output: [ -1, 1, 1, 2, 2, 4 ] 20console.log(findPrecedingSmallerElements([5, 4, 3, 2, 1])); 21// Output: [ -1, -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.

Problem 2: Stack with a Minimum Trace

Our second challenge is a high-stakes game where you must keep track of your score and always be aware of your lowest score without slowing down the game. Enter the MinStack, a special stack designed to efficiently keep track of the minimal score.

Problem 2: Naive Approach

A naive solution would involve, at the end of each turn, traversing through all the scores recorded to that point to find the minimum score — similar to going through every high score ever recorded for each new score. Time-wise, this would mean when 100 scores are recorded, up to 100 checks would be needed each turn, leading to O(n^2) operations as the game progresses.

Problem 2: Efficient Approach Explanation

We envision a second stack, which shadows our main stack. As the main stack grows, this auxiliary stack captures and retains only the minimal elements seen up to each point. Thus, the minimum score is always at the top of our 'min stack'.

Problem 2: Building The Solution Step-by-Step

It's time to manifest this brainchild into JavaScript code. Here's the skeletal structure of our MinStack, waiting to be imbued with functionality:

JavaScript
1class MinStack { 2 constructor(){ 3 this.stack = []; 4 this.minStack = []; 5 } 6 7 push(val) { 8 this.stack.push(val); 9 if (this.minStack.length === 0 || val <= this.getMin()) { 10 this.minStack.push(val); 11 } 12 } 13 14 pop() { 15 const item = this.stack.pop(); 16 if (item === this.getMin()) { 17 this.minStack.pop(); 18 } 19 return item; 20 } 21 22 top() { 23 return this.stack[this.stack.length - 1]; 24 } 25 26 getMin() { 27 return this.minStack[this.minStack.length - 1]; 28 } 29}

The push method introduces the key player — our minStack stack, which retains the minimum value observed 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.

Problem 2: Utilizing Class

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

JavaScript
1let ms = new MinStack(); 2ms.push(5); 3console.log(ms.getMin()); // 5 4ms.push(7); 5console.log(ms.getMin()); // 5 6ms.push(3); 7console.log(ms.getMin()); // 3 8ms.push(1); 9console.log(ms.getMin()); // 1
Lesson Summary

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!

Enjoy this lesson? Now it's time to practice with Cosmo!

Practice is how you turn knowledge into actual skills.