Dynamic Stack Using Resizable Array in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Push and Pop Operations on Dynamic Stack

What is the printed state of the stack after the following operations?

DSA Python

class DynamicStack:
    def __init__(self):
        self.array = [None] * 2
        self.size = 0

    def push(self, val):
        if self.size == len(self.array):
            self.resize(2 * len(self.array))
        self.array[self.size] = val
        self.size += 1

    def pop(self):
        if self.size == 0:
            return None
        self.size -= 1
        val = self.array[self.size]
        self.array[self.size] = None
        if self.size > 0 and self.size == len(self.array) // 4:
            self.resize(len(self.array) // 2)
        return val

    def resize(self, capacity):
        new_array = [None] * capacity
        for i in range(self.size):
            new_array[i] = self.array[i]
        self.array = new_array

    def __str__(self):
        return ' -> '.join(str(self.array[i]) for i in range(self.size)) + ' -> null'

stack = DynamicStack()
stack.push(10)
stack.push(20)
stack.push(30)
stack.pop()
stack.push(40)
print(stack)

A 10 -> 20 -> null

B 10 -> 20 -> 30 -> 40 -> null

C 10 -> 20 -> 40 -> null

D 20 -> 30 -> 40 -> null

💡 Hint

Remember that pop removes the last pushed element and resizing happens when size is one-fourth of capacity.

After pushing 10, 20, 30, the stack is 10 -> 20 -> 30 -> null. Popping removes 30. Then pushing 40 adds it on top. So stack is 10 -> 20 -> 40 -> null.

❓ Predict Output

intermediate

Capacity of Array After Pushes and Pops

What is the capacity of the internal array after these operations?

DSA Python

stack = DynamicStack()
stack.push(1)
stack.push(2)
stack.push(3)
stack.push(4)
stack.pop()
stack.pop()
stack.pop()
print(len(stack.array))

A 2

B 4

C 1

D 8

💡 Hint

Check when the array resizes down after popping elements.

Initially capacity is 2. After pushing 4 elements, capacity grows to 4. After popping 3 elements, size is 1 which triggers resize down to capacity 2.

🔧 Debug

advanced

Identify the Error in Resize Method

What error will occur when running this resize method in the DynamicStack class?

DSA Python

def resize(self, capacity):
    new_array = [None] * capacity
    for i in range(self.size + 1):
        new_array[i] = self.array[i]
    self.array = new_array

A IndexError because loop goes out of bounds

B TypeError because new_array is not a list

C No error, works correctly

D ValueError because capacity is negative

💡 Hint

Check the loop range compared to size of array.

Loop runs from 0 to size inclusive, but valid indices are 0 to size-1. This causes IndexError.

🧠 Conceptual

advanced

Why Does Dynamic Stack Resize Down?

Why does the dynamic stack reduce its array size when the number of elements is one-fourth of the capacity?

A To prevent stack overflow errors

B To increase speed of push operations

C To keep the array size always even

D To save memory and avoid wasting space when stack shrinks

💡 Hint

Think about memory usage when many elements are removed.

Reducing array size when stack shrinks saves memory by not holding unused space.

❓ Predict Output

expert

Final Stack State After Complex Operations

What is the printed stack after these operations?

DSA Python

stack = DynamicStack()
for i in range(1, 9):
    stack.push(i)
for _ in range(5):
    stack.pop()
stack.push(100)
stack.push(200)
print(stack)

A 100 -> 200 -> null

B 1 -> 2 -> 3 -> 100 -> 200 -> null

C 1 -> 2 -> 3 -> 4 -> 5 -> 100 -> 200 -> null

D 4 -> 5 -> 6 -> 7 -> 8 -> 100 -> 200 -> null

💡 Hint

Track pushes and pops carefully and remember pop removes last pushed element.

Pushed 1 to 8, then popped 5 times removing 8,7,6,5,4. Left with 1,2,3. Then pushed 100 and 200 on top.

Practice

1. What is the main advantage of using a dynamic stack implemented with a resizable array over a fixed-size array stack?

easy

A. It always has a fixed capacity.

B. It uses less memory than a linked list stack.

C. It does not require resizing when full.

D. It can grow or shrink automatically as elements are pushed or popped.

Solution

1. Step 1: Understand fixed-size array stack limitation

   A fixed-size array stack has a set capacity and cannot grow beyond it, causing overflow if exceeded.

2. Step 2: Analyze dynamic stack behavior

   A dynamic stack using a resizable array automatically increases or decreases its size when needed, avoiding overflow and wasted space.

3. Final Answer:

   It can grow or shrink automatically as elements are pushed or popped. -> Option D

4. Quick Check:

   Dynamic resizing = automatic growth/shrink [OK]

Hint: Dynamic stacks resize arrays automatically when full or empty [OK]

Common Mistakes:

• Thinking dynamic stack never resizes
• Confusing dynamic stack with linked list stack
• Assuming fixed capacity for dynamic stack

2. Which of the following is the correct way to double the size of the internal array in a dynamic stack implemented in Python?

easy

A. self.array.extend([None] * len(self.array))

B. self.array = self.array * 2

C. self.array = [None] * (len(self.array) * 2)

D. self.array.append(self.array)

Solution

1. Step 1: Understand array resizing in Python list

   To double the size, we add extra None elements equal to current length to the list.

2. Step 2: Check each option

   self.array.extend([None] * len(self.array)) correctly extends the list by adding None elements equal to current length, doubling size.

3. Final Answer:

   self.array.extend([None] * len(self.array)) -> Option A

4. Quick Check:

   Extend with None * current length = double size [OK]

Hint: Use extend with None times current length to double array [OK]

Common Mistakes:

• Using multiplication on list copies elements, not size
• Assigning new list loses old elements
• Appending array inside itself causes nested list

3. Consider this Python code for a dynamic stack push operation:

def push(self, item):
    if self.top == len(self.array):
        self.array.extend([None] * len(self.array))
    self.array[self.top] = item
    self.top += 1

stack = DynamicStack()
stack.push(10)
stack.push(20)
stack.push(30)
print(stack.array[:stack.top])

What will be the printed output?

medium

A. [10, 20, 30]

B. [10, 20, 30, None, None, None]

C. [None, None, None]

D. IndexError

Solution

1. Step 1: Trace push calls and array state

   Initially array is empty or small. On first push, item 10 is placed at index 0, top becomes 1. Similarly for 20 and 30 at indices 1 and 2.

2. Step 2: Print array slice up to top

   Printing array[:top] shows only elements at indices 0,1,2 which are [10, 20, 30].

3. Final Answer:

   [10, 20, 30] -> Option A

4. Quick Check:

   Print array slice up to top = pushed elements [OK]

Hint: Print only up to top index to see current stack elements [OK]

Common Mistakes:

• Printing full array including None values
• Confusing top index with length
• Expecting error due to resizing

4. Given this buggy pop method for a dynamic stack:

def pop(self):
    if self.top == 0:
        return None
    self.top -= 1
    item = self.array[self.top]
    self.array[self.top] = None
    if self.top < len(self.array) // 4:
        self.array = self.array[:len(self.array) // 2]
    return item

What is the main issue with this code?

medium

A. It does not set popped element to None.

B. It does not check for empty stack before popping.

C. It shrinks the array too early, causing frequent resizing.

D. It doubles the array size instead of shrinking.

Solution

1. Step 1: Understand shrinking condition

   The array shrinks when top is less than one-fourth of array length, which can cause frequent resizing if elements fluctuate near this threshold.

2. Step 2: Identify problem with frequent resizing

   Frequent shrinking and expanding wastes time and resources, so better to shrink less aggressively.

3. Final Answer:

   It shrinks the array too early, causing frequent resizing. -> Option C

4. Quick Check:

   Shrink threshold too high = frequent resize [OK]

Hint: Shrink array only when usage is very low to avoid thrashing [OK]

Common Mistakes:

• Ignoring empty stack check (it is present)
• Confusing shrinking with doubling
• Not clearing popped element

5. You want to implement a dynamic stack that minimizes memory usage by shrinking the internal array only when the number of elements is less than 1/8 of the array size. Which change should you make to the pop method's shrinking condition?

def pop(self):
    if self.top == 0:
        return None
    self.top -= 1
    item = self.array[self.top]
    self.array[self.top] = None
    if ???:
        self.array = self.array[:len(self.array) // 2]
    return item

hard

A. if self.top == 0

B. if self.top < len(self.array) // 8

C. if self.top <= len(self.array) // 2

D. if self.top < len(self.array) // 4

Solution

1. Step 1: Understand shrinking threshold

   To shrink only when usage is less than 1/8, the condition must check if top is less than array length divided by 8.

2. Step 2: Match condition to code

   if self.top < len(self.array) // 8 correctly uses 'if self.top < len(self.array) // 8' to trigger shrinking at 1/8 usage.

3. Final Answer:

   if self.top < len(self.array) // 8 -> Option B

4. Quick Check:

   Shrink when usage < 1/8 = condition in if self.top < len(self.array) // 8 [OK]

Hint: Use 1/8 threshold to shrink less often and save memory [OK]

Common Mistakes:

• Using 1/4 threshold shrinks too early
• Using equality instead of less than
• Shrinking when stack is empty only