Min Stack Design in DSA C++ - 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 we add more elements.

Specifically, how does the time to push, pop, or get the minimum change with more data?

Scenario Under Consideration

Analyze the time complexity of the following Min Stack implementation.

class MinStack {
  std::stack<int> mainStack, minStack;
public:
  void push(int x) {
    mainStack.push(x);
    if(minStack.empty() || x <= minStack.top())
      minStack.push(x);
  }
  void pop() {
    if(mainStack.top() == minStack.top())
      minStack.pop();
    mainStack.pop();
  }
  int getMin() {
    return minStack.top();
  }
};
    

This code keeps track of the minimum element using an extra stack alongside the main stack.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Each push, pop, and getMin call does a fixed number of stack operations.
• How many times: Each operation runs once per call, no loops inside these methods.

How Execution Grows With Input

Each operation does a constant amount of work regardless of how many elements are in the stack.

Input Size (n)	Approx. Operations per call
10	2-3 stack operations
100	2-3 stack operations
1000	2-3 stack operations

Pattern observation: The time per operation stays the same no matter how big the stack grows.

Final Time Complexity

Time Complexity: O(1)

This means each push, pop, or getMin operation takes the same short time no matter how many elements are stored.

Common Mistake

[X] Wrong: "Finding the minimum must scan the whole stack, so it takes longer as the stack grows."

[OK] Correct: The extra minStack keeps track of minimums so getMin just looks at the top, no scanning needed.

Interview Connect

Understanding how to keep track of minimums efficiently shows you can design smart data structures that save time.

This skill is useful for many problems where quick access to special values is needed.

Self-Check

"What if we stored minimums only when they change, instead of every push? How would that affect time complexity?"

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