Stack Implementation Using Array in DSA Java - Time & Space Complexity

Time Complexity: Stack Implementation Using Array

O(1)

Understanding Time Complexity

We want to understand how fast stack operations run when using an array.

How does the time to push or pop change as the stack grows?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public class Stack {
    private int[] arr;
    private int top;

    public Stack(int size) {
        arr = new int[size];
        top = -1;
    }

    public void push(int x) {
        arr[++top] = x;
    }

    public int pop() {
        return arr[top--];
    }
}

This code shows how to add (push) and remove (pop) items from a stack using an array.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Each push or pop accesses one array element.
• 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, no matter how big the stack is.

Input Size (n)	Approx. Operations
10	1 operation per push/pop
100	1 operation per push/pop
1000	1 operation per push/pop

Pattern observation: The time stays the same regardless of stack size.

Final Time Complexity

Time Complexity: O(1)

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

Common Mistake

[X] Wrong: "Push or pop takes longer as the stack grows because the array is bigger."

[OK] Correct: The array size does not affect push or pop time because we only access one position directly.

Interview Connect

Knowing that stack operations run in constant time helps you explain efficient data handling in many problems.

Self-Check

"What if the array needs to grow dynamically when full? How would the time complexity of push change then?"

Practice

1. What is the main purpose of using an array in stack implementation?

easy

A. To allow random access to elements anywhere in the stack

B. To store elements in a fixed-size linear structure

C. To automatically resize the stack when full

D. To store elements in a linked manner with pointers

Solution

1. Step 1: Understand stack storage using array

   An array provides a fixed-size linear block of memory to store stack elements sequentially.

2. Step 2: Compare with other storage methods

   Unlike linked lists, arrays do not use pointers and have fixed size, so they store elements linearly.

3. Final Answer:

   To store elements in a fixed-size linear structure -> Option B

4. Quick Check:

   Array = fixed-size linear storage [OK]

Hint: Stack array stores elements linearly with fixed size [OK]

Common Mistakes:

• Thinking array allows automatic resizing
• Confusing array with linked list storage
• Assuming random access is main stack feature

2. Which of the following is the correct way to declare a stack array of size 10 in Java?

easy

A. int[] stack = new int[10];

B. int stack = new int[10];

C. int stack[] = int[10];

D. int stack = int[10];

Solution

1. Step 1: Recall Java array declaration syntax

   In Java, arrays are declared with type[], variable = new type[size];

2. Step 2: Match correct syntax

   int[] stack = new int[10]; matches this syntax: int[] stack = new int[10]; others have syntax errors.

3. Final Answer:

   int[] stack = new int[10]; -> Option A

4. Quick Check:

   Java array declaration = int[] name = new int[size]; [OK]

Hint: Use 'type[] name = new type[size];' for arrays in Java [OK]

Common Mistakes:

• Missing 'new' keyword
• Wrong placement of brackets
• Assigning array without 'new'

3. What will be the output after the following operations on a stack of size 3?

Stack s = new Stack(3);
s.push(10);
s.push(20);
s.pop();
s.push(30);
s.push(40);

Assuming push prints "Pushed: value" and pop prints "Popped: value".

medium

A. "Pushed: 10\nPushed: 20\nPopped: 20\nStack Overflow\nPushed: 40"

B. "Pushed: 10\nPushed: 20\nPopped: 20\nPushed: 30"

C. "Pushed: 10\nPushed: 20\nPopped: 20\nPushed: 30\nStack Overflow"

D. "Pushed: 10\nPushed: 20\nPopped: 20\nPushed: 30\nPushed: 40"

Solution

1. Step 1: Trace push and pop operations on stack size 3

   push(10) prints "Pushed: 10" stack:[10]; push(20) "Pushed: 20" [10,20]; pop() "Popped: 20" [10]; push(30) "Pushed: 30" [10,30]; push(40) "Pushed: 40" [10,30,40] (capacity 3, net 3 elements).

2. Step 2: Identify outputs printed

   All pushes and pop print their messages; no overflow since 3 <= capacity 3.

3. Final Answer:

   "Pushed: 10\nPushed: 20\nPopped: 20\nPushed: 30\nPushed: 40" -> Option D

4. Quick Check:

   Net 3 elements <= size 3 = no overflow [OK]

Hint: Net pushes (4 push - 1 pop = 3) <= size 3: no overflow [OK]

Common Mistakes:

• Thinking stack doesn't reuse space after pop
• Miscalculating net elements (3 <= 3)
• Expecting overflow on 3rd net push

4. Identify the error in the following stack push method and select the correct fix:

public void push(int x) {
  if (top == size) {
    System.out.println("Stack Overflow");
  } else {
    stack[top] = x;
    top++;
  }
}

medium

A. Change condition to if (top >= size) to correctly detect overflow

B. Change condition to if (top == size - 1) to avoid overflow

C. Initialize top to 1 instead of 0 to fix indexing

D. Replace top++ with ++top to fix increment order

Solution

1. Step 1: Analyze overflow condition

   Since array indices go from 0 to size-1, top == size means stack is full; condition should catch top >= size to prevent overflow.

2. Step 2: Check other options

   Changing to size - 1 is incorrect because top points to next free slot; initializing top to 1 or changing increment order is unrelated.

3. Final Answer:

   Change condition to if (top >= size) to correctly detect overflow -> Option A

4. Quick Check:

   Overflow check = top >= size [OK]

Hint: Overflow when top index reaches or exceeds size [OK]

Common Mistakes:

• Using top == size - 1 instead of top >= size
• Incorrect top initialization
• Misunderstanding increment order

5. Given a stack implemented using an array with size 5, which sequence of operations will result in the stack containing elements [5, 10, 15] from bottom to top?

Operations:
1) push(5)
2) push(10)
3) pop()
4) push(15)
5) push(10)

hard

A. Perform operations in order: 1, 2, 3, 4, 5

B. Perform operations in order: 1, 3, 2, 4, 5

C. Perform operations in order: 1, 2, 5, 3, 4

D. Perform operations in order: 1, 2, 4, 3, 5

Solution

1. Step 1: Trace operations in order 1,2,3,4,5

   push(5) -> [5]; push(10) -> [5,10]; pop() removes 10 -> [5]; push(15) -> [5,15]; push(10) -> [5,15,10]

2. Step 2: Check final stack order bottom to top

   Stack is [5,15,10], which is not [5,10,15]. So Perform operations in order: 1, 2, 3, 4, 5 is incorrect. Check other options similarly.

3. Step 3: Trace Perform operations in order: 1, 3, 2, 4, 5 (1,3,2,4,5)

   push(5) -> [5]; pop() removes 5 -> []; push(10) -> [10]; push(15) -> [10,15]; push(10) -> [10,15,10] not correct.

4. Step 4: Trace Perform operations in order: 1, 2, 4, 3, 5 (1,2,4,3,5)

   push(5) -> [5]; push(10) -> [5,10]; push(15) -> [5,10,15]; pop() removes 15 -> [5,10]; push(10) -> [5,10,10] not correct.

5. Step 5: Trace Perform operations in order: 1, 2, 5, 3, 4 (1,2,5,3,4)

   push(5) -> [5]; push(10) -> [5,10]; push(10) -> [5,10,10]; pop() removes 10 -> [5,10]; push(15) -> [5,10,15] correct final stack.

6. Final Answer:

   Perform operations in order: 1, 2, 5, 3, 4 -> Option C

7. Quick Check:

   Final stack = [5,10,15] after Perform operations in order: 1, 2, 5, 3, 4 [OK]

Hint: Simulate stack step-by-step to verify final order [OK]

Common Mistakes:

• Assuming order of operations does not affect stack
• Not simulating pop effects correctly
• Confusing push order with final stack order