Min Stack Design in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to push an element onto the stack.

DSA C++

void push(int x) {
    stack.push_back([1]);
    if (minStack.empty() || x <= minStack.back()) {
        minStack.push_back(x);
    }
}

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

Ax

BminStack.back()

Cstack.back()

Dx + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using minStack.back() instead of x causes wrong values in the main stack.

Using stack.back() causes pushing the last element again, not the new one.

We push the element x onto the main stack. The code pushes x to stack, so the blank must be 'x'.

2fill in blank

medium

Complete the code to pop the top element from the stack and update minStack accordingly.

DSA C++

void pop() {
    if (stack.back() == [1]) {
        minStack.pop_back();
    }
    stack.pop_back();
}

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

Astack.back()

Bstack.front()

CminStack.back()

DminStack.front()

Attempts:

3 left

💡 Hint

Common Mistakes

Using minStack.back() in the comparison causes incorrect behavior.

Using stack.front() or minStack.front() is wrong because we need the top element.

We compare the top of the main stack with the top of minStack. If equal, pop from minStack. So the blank is stack.back().

3fill in blank

hard

Fix the error in the getMin function to return the current minimum element.

DSA C++

int getMin() {
    return [1];
}

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

Astack.back()

BminStack.back()

CminStack.front()

Dstack.front()

Attempts:

3 left

💡 Hint

Common Mistakes

Returning stack.back() returns the last pushed element, not the minimum.

Using front() returns the bottom element, not the top.

The current minimum is at the top of minStack, so we return minStack.back().

4fill in blank

hard

Fill both blanks to correctly initialize the MinStack constructor.

DSA C++

MinStack() {
    stack = std::vector<int>[1];
    minStack = std::vector<int>[2];
}

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

A()

B{}

C{0}

D{-1}

Attempts:

3 left

💡 Hint

Common Mistakes

Using () is invalid syntax for vector initialization here.

Initializing with {0} or {-1} adds unwanted elements.

We initialize both vectors as empty using {}.

5fill in blank

hard

Fill all three blanks to complete the MinStack class definition with push, pop, and getMin methods.

DSA C++

class MinStack {
private:
    std::vector<int> stack;
    std::vector<int> minStack;
public:
    void push(int x) {
        stack.push_back([1]);
        if (minStack.empty() || x <= minStack.back()) {
            minStack.push_back(x);
        }
    }

    void pop() {
        if (stack.back() == [2]) {
            minStack.pop_back();
        }
        stack.pop_back();
    }

    int getMin() {
        return [3];
    }
};

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

Ax

Bstack.back()

CminStack.back()

DminStack.front()

Attempts:

3 left

💡 Hint

Common Mistakes

Using minStack.back() in push instead of x.

Comparing minStack.back() in pop instead of stack.back().

Returning stack.back() in getMin instead of minStack.back().

push uses x, pop compares stack.back(), getMin returns minStack.back().

Practice

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

easy

A. To store elements in a queue fashion

B. To retrieve the minimum element in constant time

C. To sort elements in ascending order

D. To reverse the order of elements

Solution

1. Step 1: Understand the Min Stack concept

   A Min Stack is designed to keep track of the minimum element efficiently.

2. Step 2: Identify the key feature

   It allows retrieving the minimum element in constant time, unlike a normal stack.

3. Final Answer:

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

4. Quick Check:

   Min Stack purpose = retrieve minimum element fast [OK]

Hint: Min Stack always tracks minimum for O(1) retrieval [OK]

Common Mistakes:

• Confusing Min Stack with sorting
• Thinking it behaves like a queue
• Assuming it reverses elements

2. Which of the following is the correct way to declare a stack of integers in C++?

easy

A. stack<int> s[];

B. stack s<int>;

C. stack<int> s;

D. stack<int> s()

Solution

1. Step 1: Recall C++ stack declaration syntax

   The correct syntax uses template with angle brackets and variable name after.

2. Step 2: Check each option

   stack<int> s; matches correct syntax: stack<int> s;. Others have syntax errors.

3. Final Answer:

   stack<int> s; -> Option C

4. Quick Check:

   C++ stack declaration = stack<int> s; [OK]

Hint: Use template syntax: stack<Type> variable; [OK]

Common Mistakes:

• Placing variable name inside angle brackets
• Using parentheses instead of semicolon
• Declaring as array without initialization

3. Consider the following C++ code snippet implementing a Min Stack push operation:

void push(int x) {
  s.push(x);
  if (minStack.empty() || x <= minStack.top()) {
    minStack.push(x);
  }
}

What will be the content of minStack after pushing the sequence [3, 5, 2, 2, 4]?

medium

A. [3, 5, 2, 2, 4]

B. [2, 2]

C. [3, 2]

D. [3, 2, 2]

Solution

1. Step 1: Trace push operations on minStack

   Push 3: minStack empty, push 3 -> [3] Push 5: 5 > top(3), no push -> [3] Push 2: 2 <= 3, push 2 -> [3, 2] Push 2: 2 <= 2, push 2 -> [3, 2, 2] Push 4: 4 > 2, no push -> [3, 2, 2]

2. Step 2: Confirm final minStack content

   minStack contains [3, 2, 2] after all pushes.

3. Final Answer:

   [3, 2, 2] -> Option D

4. Quick Check:

   minStack after pushes = [3, 2, 2] [OK]

Hint: Push to minStack only if new value ≤ current min [OK]

Common Mistakes:

• Pushing all elements to minStack
• Not pushing duplicates of min
• Confusing minStack with main stack

4. The following code snippet is intended to implement the pop() operation of a Min Stack. Identify the error:

void pop() {
  if (s.top() == minStack.top()) {
    minStack.pop();
  }
  s.pop();
}

medium

A. It does not check if stacks are empty before popping

B. It pops from minStack before popping from s, causing mismatch

C. It compares values incorrectly using == instead of <=

D. It should pop from s twice when values match

Solution

1. Step 1: Analyze pop operation safety

   The code pops from stacks without checking if they are empty, risking runtime errors.

2. Step 2: Identify missing safety checks

   Before calling top() or pop(), stacks should be checked for emptiness.

3. Final Answer:

   It does not check if stacks are empty before popping -> Option A

4. Quick Check:

   Pop safety requires empty checks [OK]

Hint: Always check stack empty before pop or top [OK]

Common Mistakes:

• Ignoring empty stack checks
• Misusing comparison operators
• Popping stacks in wrong order

5. You want to design a Min Stack that supports push, pop, top, and getMin all in O(1) time. Which approach below correctly achieves this with minimal extra space?

hard

A. Use one stack to store values and a second stack to store minimums only when a new minimum is pushed

B. Use one stack and sort it after each push to find minimum

C. Use two stacks: one for values and one for all values pushed, then find min by scanning

D. Use a queue to store values and a stack to store minimums

Solution

1. Step 1: Understand O(1) min retrieval requirement

   To get minimum in constant time, we must track minimums during push/pop without scanning.

2. Step 2: Evaluate approaches

   Use one stack to store values and a second stack to store minimums only when a new minimum is pushed uses a second stack to store minimums only when a new minimum appears, saving space and ensuring O(1) operations.

3. Final Answer:

   Use one stack to store values and a second stack to store minimums only when a new minimum is pushed -> Option A

4. Quick Check:

   Min Stack with minimal space = two stacks, min pushed selectively [OK]

Hint: Push min only when new min found to save space [OK]

Common Mistakes:

• Sorting stack after each push (slow)
• Scanning stack to find min (not O(1))
• Using queue which breaks stack order