Min Stack Design in DSA Typescript - Build from Scratch

Min Stack Design

📖 Scenario: Imagine you are building a special stack for a calculator app. This stack not only stores numbers but also quickly tells you the smallest number in the stack at any time.

🎯 Goal: You will create a MinStack class that supports pushing numbers, popping the top number, and getting the smallest number in the stack efficiently.

📋 What You'll Learn

Create a MinStack class with methods push, pop, and getMin.

Use two stacks internally: one for all numbers, one for tracking minimums.

Ensure getMin returns the smallest number currently in the stack.

Handle empty stack cases gracefully.

💡 Why This Matters

🌍 Real World

Min stacks are used in calculators, stock price tracking, and anywhere you need quick access to the smallest value in a changing list.

💼 Career

Understanding min stack design is useful for software engineers working on performance-critical applications and technical interviews.

1

Create the MinStack class with two empty stacks

Create a class called MinStack with two properties: stack and minStack, both initialized as empty arrays of numbers.

DSA Typescript

class MinStack {
  stack: number[] = []
  minStack: number[] = []
  // Your code here
}

Hint

Use stack: number[] = [] and minStack: number[] = [] inside the class.

2

Add the push method to add numbers and track minimum

Add a method called push that takes a num: number. It should push num to stack. Also, push num to minStack if minStack is empty or num is less than or equal to the last element in minStack.

DSA Typescript

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

  push(num: number): void {
    // Your code here
  }
}

Hint

Check if minStack is empty or num is smaller or equal to the current minimum before pushing to minStack.

3

Add the pop method to remove numbers and update minimum

Add a method called pop that removes the top number from stack. If the removed number is equal to the last element in minStack, also remove it from minStack.

DSA Typescript

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

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

  pop(): void {
    // Your code here
  }
}

Hint

Remove the top from stack. If it equals the top of minStack, remove from minStack too.

4

Add the getMin method and print the minimum after operations

Add a method called getMin that returns the last element of minStack or null if minStack is empty. Then create a MinStack instance, push 5, 3, 7, pop once, and print the result of getMin().

DSA Typescript

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

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

  pop(): void {
    if (this.stack.length === 0) return
    const popped = this.stack.pop()!
    if (popped === this.minStack[this.minStack.length - 1]) {
      this.minStack.pop()
    }
  }

  getMin(): number | null {
    // Your code here
  }
}

// Create MinStack instance and test
// Your code here

Hint

Return the last element of minStack or null if empty. Then test by pushing 5, 3, 7, popping once, and printing getMin().

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