Min Stack Design in DSA Typescript - Time & Space Complexity

Time Complexity: Min Stack Design

O(1)

Understanding Time Complexity

We want to understand how fast the Min Stack operations run as the number of elements grows.

How does the time to push, pop, or get the minimum change when the stack gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following Min Stack implementation.

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

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

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

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

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

This code keeps track of the minimum value at each step using a second stack.

Identify Repeating Operations

Look at the main repeated actions when using the stack.

• Primary operation: push and pop on two stacks (main and min stack)
• How many times: once per element added or removed

How Execution Grows With Input

Each push or pop does a fixed number of steps regardless of stack size.

Input Size (n)	Approx. Operations per push/pop
10	2 to 3 operations
100	2 to 3 operations
1000	2 to 3 operations

Pattern observation: The number of steps per operation stays about the same no matter how big the stack is.

Final Time Complexity

Time Complexity: O(1)

This means each push, pop, top, or getMin operation takes the same small amount of time no matter how many elements are in the stack.

Common Mistake

[X] Wrong: "Getting the minimum value takes longer as the stack grows because we have to check all elements."

[OK] Correct: The min stack keeps track of minimums at each step, so getMin just looks at the top of minStack, which is always fast.

Interview Connect

Understanding how to keep track of extra information efficiently is a key skill. This design shows how to keep minimums handy without slowing down the main operations.

Self-Check

What if we stored minimum values only when they change instead of every time? How would that affect the time complexity?

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