Dynamic Stack Using Resizable Array in DSA Typescript - 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 work in a dynamic stack that grows as needed.

How does the time to add or remove items change as the stack gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class DynamicStack {
  private arr: number[] = [];
  private capacity: number = 2;
  private top: number = -1;

  push(value: number): void {
    if (this.top + 1 === this.capacity) {
      this.capacity *= 2;
      const newArr = new Array(this.capacity);
      for (let i = 0; i <= this.top; i++) {
        newArr[i] = this.arr[i];
      }
      this.arr = newArr;
    }
    this.arr[++this.top] = value;
  }

  pop(): number | undefined {
    if (this.top === -1) return undefined;
    return this.arr[this.top--];
  }
}
    

This code shows a stack that doubles its array size when full, copying old elements to a bigger array before adding new ones.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Copying all elements to a new larger array when resizing.
• How many times: This copying happens only when the array is full, which occurs after doubling the capacity each time.
• Other operations: Push and pop themselves are simple and happen once per call.

How Execution Grows With Input

Most push operations just add an item quickly, but occasionally the stack must copy all items to a bigger array.

Input Size (n)	Approx. Operations
10	About 10 simple pushes + 8 copies during resizes
100	About 100 simple pushes + 196 copies during resizes
1000	About 1000 simple pushes + 2044 copies during resizes

Pattern observation: Most pushes are fast, but copying happens less often as the stack grows, spreading the cost over many pushes.

Final Time Complexity

Time Complexity: O(1) amortized

This means each push usually takes constant time, but sometimes takes longer when resizing, yet averaged out it stays fast.

Common Mistake

[X] Wrong: "Every push takes a long time because of copying."

[OK] Correct: Copying happens only occasionally when the array is full, so most pushes are quick and the average time stays low.

Interview Connect

Understanding amortized time helps you explain why dynamic arrays and stacks are efficient in real programs, showing you know how data structures handle growth smoothly.

Self-Check

What if we increased the array size by only 1 each time instead of doubling? How would the time complexity change?

Practice

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

easy

A. It can grow or shrink in size as needed during runtime.

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

C. It always has a fixed maximum size.

D. It does not require resizing or copying elements.

Solution

1. Step 1: Understand fixed vs dynamic size

   A fixed-size array stack has a set size and cannot grow beyond it, limiting capacity.

2. Step 2: Understand dynamic resizing

   A dynamic stack with resizable array can increase or decrease its size during runtime to fit more or fewer elements.

3. Final Answer:

   It can grow or shrink in size as needed during runtime. -> Option A

4. Quick Check:

   Dynamic resizing = flexible size [OK]

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

Common Mistakes:

• Thinking dynamic stack has fixed max size
• Confusing memory use with linked list
• Assuming no resizing needed

2. Which of the following TypeScript code snippets correctly doubles the size of the internal array when the stack is full?

easy

A. this.array = [...this.array, ...new Array(this.array.length)];

B. this.array.push(...new Array(this.array.length));

C. this.array.length = this.array.length * 2;

D. const newArray = new Array(this.array.length * 2); for(let i=0; i

Solution

1. Step 1: Understand resizing by copying

   To resize, create a new larger array and copy existing elements into it.

2. Step 2: Check each option

   A uses spread but adds empty elements, not doubling size properly; B creates new array, copies elements, then replaces original array; C only changes length property but does not copy elements; D pushes empty elements, not resizing properly.

3. Final Answer:

   const newArray = new Array(this.array.length * 2); for(let i=0; i -> Option D

4. Quick Check:

   Copy elements to new larger array = B [OK]

Hint: Resize by copying elements to new bigger array [OK]

Common Mistakes:

• Only changing length without copying
• Using push to resize array
• Adding empty elements without copying

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

push(value: number) {
  if (this.size === this.array.length) {
    const newArray = new Array(this.array.length * 2);
    for (let i = 0; i < this.array.length; i++) {
      newArray[i] = this.array[i];
    }
    this.array = newArray;
  }
  this.array[this.size] = value;
  this.size++;
}

If the initial array length is 2 and we push 3 elements: 10, 20, 30, what is the internal array state after these pushes?

medium

A. [10, 20, 30, undefined]

B. [10, 20, undefined, undefined]

C. [10, 20, 30, 0]

D. [10, 20, 30]

Solution

1. Step 1: Trace initial array and pushes

   Initial array length is 2, so after pushing 10 and 20, array is full.

2. Step 2: On pushing 30, array resizes to length 4 and copies elements 10, 20.

   Then 30 is added at index 2, size becomes 3.

3. Final Answer:

   [10, 20, 30, undefined] -> Option A

4. Quick Check:

   Resize doubles length, elements copied, new slot added [OK]

Hint: Resize doubles array, new elements added at next index [OK]

Common Mistakes:

• Forgetting to resize before adding third element
• Assuming array length stays 2
• Confusing undefined with 0

4. Given this pop method for a dynamic stack:

pop(): number | undefined {
  if (this.size === 0) return undefined;
  this.size--;
  const value = this.array[this.size];
  this.array[this.size] = undefined;
  if (this.size > 0 && this.size === Math.floor(this.array.length / 4)) {
    const newArray = new Array(Math.floor(this.array.length / 2));
    for (let i = 0; i < this.size; i++) {
      newArray[i] = this.array[i];
    }
    this.array = newArray;
  }
  return value;
}

What is the bug in this code?

medium

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

B. It sets popped element to undefined before copying, losing data.

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

D. It does not update the size after popping.

Solution

1. Step 1: Analyze order of operations

   The code sets this.array[this.size] = undefined before copying elements to newArray.

2. Step 2: Understand effect on copying

   When copying, the popped element is already undefined, so data is lost if resizing happens now.

3. Final Answer:

   It sets popped element to undefined before copying, losing data. -> Option B

4. Quick Check:

   Clear after copying to avoid data loss [OK]

Hint: Clear popped slot after resizing to keep data safe [OK]

Common Mistakes:

• Ignoring order of clearing and copying
• Shrinking array too often is not the main bug here
• Missing empty stack check is false here

5. You want to implement a dynamic stack using a resizable array that shrinks its size when the number of elements falls below one-quarter of the array length. Which approach below correctly avoids frequent resizing when pushing and popping many elements?

hard

A. Resize array to double size on push when full, and halve size on pop when size is exactly one-quarter of array length.

B. Resize array to double size on push when full, and halve size on pop when size is less than or equal to one-quarter of array length.

C. Resize array to double size on push when full, and halve size on pop only when size is less than one-quarter of array length and array length is greater than initial capacity.

D. Resize array to double size on push when full, and halve size on pop whenever size decreases.

Solution

1. Step 1: Understand resizing conditions

   Doubling on push when full is standard. Halving on pop should avoid frequent resizing.

2. Step 2: Avoid frequent resizing by checking size and minimum capacity

   Only halve when size is less than one-quarter and array length is above initial capacity to prevent shrinking too small.

3. Final Answer:

   Resize array to double size on push when full, and halve size on pop only when size is less than one-quarter of array length and array length is greater than initial capacity. -> Option C

4. Quick Check:

   Resize carefully to avoid thrashing [OK]

Hint: Shrink only if size < 1/4 and array bigger than initial [OK]

Common Mistakes:

• Shrinking too early causing many resizes
• Shrinking when size equals one-quarter causing thrashing
• Shrinking below initial capacity