Dynamic Stack Using Resizable Array in DSA Python - Time & Space Complexity

Time Complexity: Dynamic Stack Using Resizable Array

O(1) amortized

Understanding Time Complexity

We want to understand how fast operations like push and pop run when using a stack built on a resizable array.

Specifically, how does the time to add or remove items change as the stack grows?

Scenario Under Consideration

Analyze the time complexity of the following dynamic stack implementation.

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

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

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

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

This code shows a stack that grows and shrinks its array size as needed.

Identify Repeating Operations

Look for loops and repeated steps that take time.

• Primary operation: Copying elements during resizing in _resize.
• How many times: Happens only when array is full or too empty, not every push/pop.

How Execution Grows With Input

Most pushes and pops take constant time, but resizing copies all elements.

Input Size (n)	Approx. Operations
10	About 10 simple steps, occasional 10-step copy
100	Mostly 100 simple steps, occasional 100-step copy
1000	Mostly 1000 simple steps, occasional 1000-step copy

Pattern observation: Most operations are quick, but resizing causes longer steps occasionally.

Final Time Complexity

Time Complexity: O(1) amortized

This means each push or pop usually takes a fixed small time, even though resizing sometimes takes longer.

Common Mistake

[X] Wrong: "Every push or pop takes time proportional to the stack size because of resizing."

[OK] Correct: Resizing happens rarely, so most operations are fast. Over many operations, the average time stays constant.

Interview Connect

Understanding amortized time helps you explain why dynamic arrays and stacks are efficient in real programs.

Self-Check

"What if we tripled the array size instead of doubling it? How would the time complexity change?"

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