Why Min Stack Design in DSA Typescript?

The Big Idea

What if you could always know the smallest item in your stack instantly, without searching?

The Scenario

Imagine you have a stack of books, and you want to quickly find the smallest book without taking all books out one by one.

Doing this manually means checking each book every time you want the smallest one.

The Problem

Manually searching for the smallest book each time is slow and tiring.

It wastes time and can cause mistakes if you lose track of which book is smallest.

The Solution

A Min Stack keeps track of the smallest book as you add or remove books.

This way, you can always know the smallest book instantly without searching.

Before vs After

✗ Before

function getMin(stack: number[]): number {
  let min = stack[0];
  for (let i = 1; i < stack.length; i++) {
    if (stack[i] < min) min = stack[i];
  }
  return min;
}

✓ After

class MinStack {
  private stack: number[] = [];
  private minStack: number[] = [];

  push(val: number) {
    this.stack.push(val);
    if (!this.minStack.length || val <= this.minStack[this.minStack.length - 1]) {
      this.minStack.push(val);
    }
  }

  pop(): number | undefined {
    const val = this.stack.pop();
    if (val !== undefined && val === this.minStack[this.minStack.length - 1]) {
      this.minStack.pop();
    }
    return val;
  }

  getMin(): number | undefined {
    return this.minStack[this.minStack.length - 1];
  }
}

What It Enables

You can instantly get the smallest value in a stack at any time, making your programs faster and simpler.

Real Life Example

In a game, you might want to quickly find the lowest score among recent players without checking all scores every time.

Key Takeaways

Manual search for minimum is slow and error-prone.

Min Stack keeps track of minimum values as you add or remove items.

This allows quick access to the smallest element anytime.

Practice

1. What is the main purpose of a Min Stack data structure?

easy

A. To retrieve the minimum element in constant time

B. To sort elements automatically

C. To store elements in ascending order

D. To remove duplicates from the stack

Solution

1. Step 1: Understand the Min Stack functionality

   A Min Stack is designed to keep track of the minimum element efficiently while supporting normal stack operations.

2. Step 2: Identify the key feature

   The key feature is retrieving the minimum element in constant time, which normal stacks cannot do efficiently.

3. Final Answer:

   To retrieve the minimum element in constant time -> Option A

4. Quick Check:

   Min Stack = constant time minimum retrieval [OK]

Hint: Min Stack always gives min element fast [OK]

Common Mistakes:

• Thinking it sorts elements automatically
• Confusing with normal stack behavior
• Assuming it removes duplicates

2. Which of the following is the correct TypeScript method signature to push an element x onto a Min Stack class?

easy

A. push(x: number): number

B. push(x: string): number

C. push(x: number): void

D. push(): number

Solution

1. Step 1: Understand push method requirements

   The push method should accept a number and return nothing (void) because it adds an element to the stack.

2. Step 2: Match method signature

   push(x: number): void matches: it takes a number and returns void, which is correct for push operation.

3. Final Answer:

   push(x: number): void -> Option C

4. Quick Check:

   Push method signature = push(x: number): void [OK]

Hint: Push takes number, returns void in stack [OK]

Common Mistakes:

• Using wrong parameter type like string
• Returning number instead of void
• Missing parameter in method

3. Consider the following TypeScript code for a Min Stack:

const minStack = new MinStack();
minStack.push(3);
minStack.push(5);
minStack.push(2);
minStack.push(1);
minStack.pop();
const min = minStack.getMin();
console.log(min);

What will be printed?

medium

A. 2

B. 1

C. 3

D. 5

Solution

1. Step 1: Trace push and pop operations

   Push 3, 5, 2, 1 onto stack. Stack is [3,5,2,1]. Then pop removes 1. Stack is now [3,5,2].

2. Step 2: Find minimum after pop

   Minimum among [3,5,2] is 2, so getMin() returns 2.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Min after pop = 2 [OK]

Hint: Pop removes top; min updates accordingly [OK]

Common Mistakes:

• Forgetting pop removes last pushed element
• Assuming min stays 1 after pop
• Confusing stack order

4. The following TypeScript Min Stack implementation has a bug. What is the error?

class MinStack {
  stack: number[] = [];
  minStack: number[] = [];

  push(x: number): void {
    this.stack.push(x);
    if (this.minStack.length === 0 || x <= this.minStack[this.minStack.length - 1]) {
      this.minStack.push(x);
    }
  }

  pop(): void {
    const val = this.stack.pop();
    if (val === this.minStack[this.minStack.length - 1]) {
      this.minStack.pop();
    }
  }

  getMin(): number {
    return this.minStack[this.minStack.length - 1];
  }
}

medium

A. push() pushes to minStack even if x > current min

B. pop() does not handle empty stack case

C. getMin() returns wrong element index

D. stack array is not initialized

Solution

1. Step 1: Analyze pop() method

   pop() calls this.stack.pop() which can return undefined if stack is empty, then compares with minStack top without check.

2. Step 2: Identify missing empty check

   There is no check if stack is empty before pop, so val can be undefined causing incorrect comparison or error.

3. Final Answer:

   pop() does not handle empty stack case -> Option B

4. Quick Check:

   pop() missing empty check = bug [OK]

Hint: Always check stack empty before pop [OK]

Common Mistakes:

• Ignoring empty stack before pop
• Assuming push always pushes to minStack
• Misreading getMin index

5. You want to design a Min Stack that supports push, pop, top, and getMin all in constant time. Which approach below correctly achieves this?

hard

A. Use one stack and scan all elements to find min on getMin

B. Use a queue to store elements and find min by sorting

C. Use a linked list and traverse to find min on getMin

D. Use two stacks: one for all elements, one for current mins

Solution

1. Step 1: Understand time complexity requirements

   All operations must be constant time, so scanning or traversing is not allowed.

2. Step 2: Identify data structure approach

   Using two stacks, one for elements and one for tracking minimums, allows constant time push, pop, and getMin.

3. Final Answer:

   Use two stacks: one for all elements, one for current mins -> Option D

4. Quick Check:

   Two stacks = constant time min tracking [OK]

Hint: Two stacks track elements and mins fast [OK]

Common Mistakes:

• Using scanning which is not constant time
• Using queue which is not stack
• Using linked list traversal for min