Min Stack Design in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to push an element onto the stack.

DSA Python

def push(self, val):
    self.stack.append([1])

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

Aval

Bself.min_stack[-1]

Cself.stack[-1]

Dmin(val)

Attempts:

3 left

💡 Hint

Common Mistakes

Using self.min_stack[-1] instead of val.

Trying to call min(val) which is incorrect.

Using self.stack[-1] which is the last element, not the new one.

We push the given value 'val' onto the main stack.

2fill in blank

medium

Complete the code to update the min_stack when pushing a new value.

DSA Python

def push(self, val):
    self.stack.append(val)
    if not self.min_stack or val [1] self.min_stack[-1]:
        self.min_stack.append(val)
    else:
        self.min_stack.append(self.min_stack[-1])

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

A>

B>=

C==

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' instead of '<=' which breaks min tracking.

Using '==' which misses smaller values.

Using '>=' which is incorrect for minimum.

We add val to min_stack if it is smaller or equal to the current minimum.

3fill in blank

hard

Fix the error in the pop method to correctly remove elements from both stacks.

DSA Python

def pop(self):
    if self.stack:
        self.stack.pop()
        [1]

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

Aself.min_stack.push()

Bself.min_stack.pop()

Cself.min_stack.append()

Dself.min_stack.remove()

Attempts:

3 left

💡 Hint

Common Mistakes

Using append() instead of pop().

Using push() which is not a list method.

Using remove() which removes by value, not last element.

We must pop from min_stack to keep stacks in sync.

4fill in blank

hard

Fill both blanks to return the current minimum element from the stack.

DSA Python

def getMin(self):
    if self.min_stack:
        return self.min_stack[1][2]

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

A[-1]

B[0]

C.pop()

D.append()

Attempts:

3 left

💡 Hint

Common Mistakes

Using [0] which returns the first element, not the minimum.

Using .pop() which removes the element.

Using .append() which adds an element.

We return the last element of min_stack which is the current minimum.

5fill in blank

hard

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

DSA Python

class MinStack:
    def __init__(self):
        self.stack = []
        self.min_stack = []

    def push(self, val):
        self.stack.append(val)
        if not self.min_stack or val [1] self.min_stack[-1]:
            self.min_stack.append(val)
        else:
            self.min_stack.append(self.min_stack[-1])

    def pop(self):
        if self.stack:
            self.stack.pop()
            self.min_stack.[2]()

    def getMin(self):
        if self.min_stack:
            return self.min_stack[3]

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

A<=

Bpop

C[-1]

Dappend

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' instead of '<=' in push.

Using append() instead of pop() in pop method.

Using [0] or pop() in getMin which removes element.

The push uses '<=' to update min_stack, pop calls pop(), and getMin returns last element.

Practice

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

easy

A. To store only the minimum elements pushed so far

B. To sort elements in ascending order automatically

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

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

Solution

1. Step 1: Understand the Min Stack concept

   A Min Stack is designed to support normal stack operations and also return the minimum element quickly.

2. Step 2: Identify the key feature

   The key feature is retrieving the minimum element in constant time, not sorting or reversing.

3. Final Answer:

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

4. Quick Check:

   Min Stack = fast minimum retrieval [OK]

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

Common Mistakes:

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

2. Which of the following is the correct way to initialize a Min Stack class in Python?

easy

A. class MinStack:
def __init__(self):
stack = []
min_stack = []

B. class MinStack:
def __init__(self):
self = []

C. class MinStack:
def __init__(self):
self.stack = None
self.min_stack = None

D. class MinStack:
def __init__(self):
self.stack = []
self.min_stack = []

Solution

1. Step 1: Check correct attribute initialization

   Attributes must be initialized with self to be instance variables.

2. Step 2: Verify list initialization

   Both stack and min_stack should be empty lists initialized with self.

3. Final Answer:

   class MinStack:\n def __init__(self):\n self.stack = []\n self.min_stack = [] -> Option D

4. Quick Check:

   Use self for instance variables [OK]

Hint: Always use self to define instance variables in __init__ [OK]

Common Mistakes:

• Not using self for attributes
• Initializing attributes as None instead of lists
• Assigning self to a list directly

3. What will be the output of the following code snippet?

min_stack = MinStack()
min_stack.push(3)
min_stack.push(5)
min_stack.push(2)
min_stack.push(1)
min_stack.pop()
print(min_stack.get_min())

medium

A. 2

B. 1

C. 3

D. 5

Solution

1. Step 1: Trace push operations and min tracking

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

2. Step 2: Pop removes 1, min updates to 2

   After popping 1, the minimum is the next smallest, which is 2.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Pop removes min 1, next min = 2 [OK]

Hint: Pop min, next min is previous min in min stack [OK]

Common Mistakes:

• Assuming min stays 1 after pop
• Confusing push order with min updates
• Ignoring min stack updates on pop

4. Identify the bug in this Min Stack pop method and choose the correct fix:

def pop(self):
    if self.stack:
        self.stack.pop()
        self.min_stack.pop()

medium

A. No bug; code is correct

B. Pop from min_stack only if popped element equals current min

C. Pop from min_stack before popping from stack

D. Check if min_stack is empty before popping it

Solution

1. Step 1: Understand pop logic in Min Stack

   Pop from min_stack only if the popped element is the current minimum to keep min_stack accurate.

2. Step 2: Identify bug and fix

   Current code always pops min_stack, which can cause errors if min_stack is out of sync.

3. Final Answer:

   Pop from min_stack only if popped element equals current min -> Option B

4. Quick Check:

   Pop min_stack conditionally to keep min correct [OK]

Hint: Pop min_stack only if popped value equals current min [OK]

Common Mistakes:

• Always popping min_stack without condition
• Not checking min_stack emptiness
• Changing pop order incorrectly

5. You want to design a Min Stack that supports push, pop, top, and get_min all in O(1) time. Which approach is best?

hard

A. Use two stacks: one for all elements, one for current minimums

B. Use a queue to store elements and a variable for min

C. Use one stack and scan all elements to find min on get_min

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

Solution

1. Step 1: Analyze time complexity requirements

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

2. Step 2: Choose data structure approach

   Using two stacks, one for elements and one for minimums, allows constant time min retrieval.

3. Final Answer:

   Use two stacks: one for all elements, one for current minimums -> Option A

4. Quick Check:

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

Hint: Two stacks track elements and minimums for O(1) min [OK]

Common Mistakes:

• Using scanning which is O(n)
• Using queue which doesn't support stack operations
• Using linked list traversal for min