Min Stack Design in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to push a value onto the stack.

DSA Typescript

push(val: number): void {
  this.stack.[1](val);
}

Drag options to blanks, or click blank then click option'

Aappend

Bpush

Cadd

Dinsert

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'append' which is not a method on arrays in TypeScript.

Using 'add' or 'insert' which do not exist on arrays.

The push method adds an element to the end of the array, which is how we add to the stack.

2fill in blank

medium

Complete the code to get the current minimum value from the min stack.

DSA Typescript

getMin(): number | null {
  if (this.minStack.length === 0) return null;
  return this.minStack[[1]];
}

Drag options to blanks, or click blank then click option'

A0

Bthis.minStack.length

Cthis.minStack.length - 1

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using index 0 which is the first element, not the last.

Using length which is out of bounds.

Using -1 which is invalid in TypeScript arrays.

The minimum value is always at the top of the minStack, which is the last element. So we access it by index this.minStack.length - 1.

3fill in blank

hard

Fix the error in the pop method to correctly update the min stack.

DSA Typescript

pop(): number | null {
  if (this.stack.length === 0) return null;
  const val = this.stack.pop();
  if (val === this.minStack[[1]]) {
    this.minStack.pop();
  }
  return val;
}

Drag options to blanks, or click blank then click option'

Athis.minStack.length - 1

B0

Cthis.minStack.length

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using index 0 which is the first element, not the last.

Using length which is out of bounds.

Using -1 which returns undefined.

We need to compare val with the top element of minStack, which is the last element at index this.minStack.length - 1.

4fill in blank

hard

Fill both blanks to correctly update the min stack when pushing a new value.

DSA Typescript

push(val: number): void {
  this.stack.push(val);
  if (this.minStack.length === 0 || val [1] this.minStack[[2]]) {
    this.minStack.push(val);
  }
}

Drag options to blanks, or click blank then click option'

A<=

B>=

Cthis.minStack.length - 1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' instead of '<=' which breaks the min stack logic.

Using index 0 instead of last element index.

We push the new value onto minStack if it is less than or equal to the current minimum. The current minimum is at the last index this.minStack.length - 1.

5fill in blank

hard

Fill all three blanks to implement the getMin method safely.

DSA Typescript

getMin(): number | null {
  if (this.minStack.length === 0) return [1];
  return this.minStack[[2]];
}

// Usage example:
const minValue = stack.getMin();
if (minValue !== [3]) {
  console.log('Minimum:', minValue);
}

Drag options to blanks, or click blank then click option'

Anull

Bthis.minStack.length - 1

Cundefined

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Returning undefined instead of null.

Accessing index 0 instead of last element.

Comparing minValue to undefined instead of null.

If the minStack is empty, getMin returns null. The minimum value is at the last index this.minStack.length - 1. When checking the returned value, we compare it to null to avoid errors.

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