Min Stack Design in DSA Java - 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.

Specifically, how the time to push, pop, or get the minimum changes with more data.

Scenario Under Consideration

Analyze the time complexity of the following Min Stack implementation.

import java.util.Stack;

class MinStack {
    private Stack stack = new Stack<>();
    private Stack minStack = new Stack<>();

    public void push(int x) {
        stack.push(x);
        if (minStack.isEmpty() || x <= minStack.peek()) {
            minStack.push(x);
        }
    }

    public void pop() {
        if (stack.pop().equals(minStack.peek())) {
            minStack.pop();
        }
    }

    public int getMin() {
        return minStack.peek();
    }
}
    

This code keeps track of the minimum value in a stack efficiently using two stacks.

Identify Repeating Operations

• Primary operation: push, pop, and peek on stacks.
• How many times: Each operation runs once per call, no loops inside.

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	2-3 steps
100	2-3 steps
1000	2-3 steps

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

Final Time Complexity

Time Complexity: O(1)

This means each operation takes the same short time no matter how many items are in the stack.

Common Mistake

[X] Wrong: "Finding the minimum must check all elements, so it takes longer as the stack grows."

[OK] Correct: The min stack keeps track of minimums as we go, so getting the minimum is always quick and does not scan the whole stack.

Interview Connect

Knowing 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 used only one stack and searched for the minimum each time? How would the time complexity change?"

Practice

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

easy

A. To retrieve the minimum element in constant time along with stack operations

B. To sort elements in ascending order automatically

C. To store only the minimum elements pushed so far

D. To reverse the order of elements pushed into the stack

Solution

1. Step 1: Understand stack operations

   A stack supports push, pop, and top operations in LIFO order.

2. Step 2: Identify Min Stack feature

   Min Stack extends stack by allowing retrieval of minimum element in O(1) time.

3. Final Answer:

   To retrieve the minimum element in constant time along with stack operations -> Option A

4. Quick Check:

   Min Stack = constant time min retrieval [OK]

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

Common Mistakes:

• Thinking Min Stack sorts elements automatically
• Assuming Min Stack stores only minimum elements
• Confusing Min Stack with reversing stack order

2. Which of the following is the correct Java method signature to push an integer val onto a Min Stack?

easy

A. public int push(int val)

B. void push(int val) public

C. public void push(int val)

D. push public void(int val)

Solution

1. Step 1: Recall Java method syntax

   Java method signature starts with access modifier, return type, method name, and parameters.

2. Step 2: Check each option

   public void push(int val) follows correct syntax: public void push(int val). Others have wrong order or return type.

3. Final Answer:

   public void push(int val) -> Option C

4. Quick Check:

   Java method syntax = public void push(int val) [OK]

Hint: Java methods start with access modifier then return type [OK]

Common Mistakes:

• Placing 'public' after method name
• Using wrong return type for void methods
• Incorrect parameter placement

3. Given the following operations on a Min Stack:

MinStack minStack = new MinStack();
minStack.push(3);
minStack.push(5);
minStack.push(2);
minStack.push(1);
minStack.pop();
int min = minStack.getMin();
System.out.println(min);

What will be printed?

medium

A. 2

B. 1

C. 3

D. 5

Solution

1. Step 1: Trace push operations and min values

   Push 3 (min=3), push 5 (min=3), push 2 (min=2), push 1 (min=1).

2. Step 2: Pop removes top (1), min updates to previous min (2), then getMin returns 2.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   After pop, min = 2 [OK]

Hint: Pop updates min to previous minimum [OK]

Common Mistakes:

• Assuming min stays 1 after pop
• Confusing top element with min
• Ignoring min update on pop

4. Consider this buggy Min Stack method snippet:

public void pop() {
  if(stack.isEmpty()) return;
  int top = stack.pop();
  if(top == minStack.peek()) {
    minStack.pop();
  }
}

What is the main issue here?

medium

A. It does not update minStack when pushing

B. It pops from stack twice accidentally

C. It compares top with minStack.pop() instead of peek()

D. It does not handle empty minStack before peek

Solution

1. Step 1: Analyze pop method safety

   Before calling minStack.peek(), we must ensure minStack is not empty to avoid error.

2. Step 2: Identify missing empty check on minStack

   Code checks stack but not minStack emptiness before peek, risking exception.

3. Final Answer:

   It does not handle empty minStack before peek -> Option D

4. Quick Check:

   Check minStack empty before peek [OK]

Hint: Always check stack emptiness before peek/pop [OK]

Common Mistakes:

• Ignoring minStack empty check
• Popping stack twice
• Confusing peek and pop methods

5. You want to design a Min Stack that supports push, pop, top, and getMin all in constant time. Which approach is best?

hard

A. Use one stack and scan all elements to find min on getMin

B. Use two stacks: one for all values, one for current minimums

C. Use a queue to store elements and find min by sorting

D. Use a linked list and traverse to find min on each getMin

Solution

1. Step 1: Understand time complexity requirements

   All operations must be O(1), so scanning or traversing is not allowed.

2. Step 2: Evaluate approaches

   Two stacks approach keeps track of minimums efficiently in O(1) time.

3. Final Answer:

   Use two stacks: one for all values, one for current minimums -> Option B

4. Quick Check:

   Two stacks = O(1) min retrieval [OK]

Hint: Two stacks track values and minimums separately [OK]

Common Mistakes:

• Using scanning which is O(n)
• Using queue which is not stack
• Traversing linked list on getMin