Stack Implementation Using Array in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the array that will store stack elements.

DSA Java

int[] stack = new int[[1]];

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

A10

Bstack

Csize

Dtop

Attempts:

3 left

💡 Hint

Common Mistakes

Using variable names instead of a number for array size.

Leaving the array size blank.

The stack array must be declared with a fixed size, here 10.

2fill in blank

medium

Complete the code to initialize the top pointer for the stack.

DSA Java

int top = [1];

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

A1

B0

C-1

Dstack.length

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing top to 0 which means stack has one element.

Initializing top to stack length which is out of bounds.

The top is initialized to -1 to indicate the stack is empty.

3fill in blank

hard

Fix the error in the push method to correctly add an element to the stack.

DSA Java

public void push(int value) {
    if(top < stack.length - 1) {
        top = top [1] 1;
        stack[top] = value;
    } else {
        System.out.println("Stack Overflow");
    }
}

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

A-

B+

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using minus instead of plus causing wrong index.

Using multiplication or division which is incorrect here.

To move top up by one, use top = top + 1.

4fill in blank

hard

Fill both blanks to correctly implement the pop method that removes and returns the top element.

DSA Java

public int pop() {
    if(top >= 0) {
        int value = stack[top];
        top = top [1] 1;
        return value;
    } else {
        System.out.println("Stack Underflow");
        return [2];
    }
}

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

A-

B0

C-1

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Incrementing top instead of decrementing.

Returning 0 which could be a valid stack value.

Decrement top by 1 after popping, and return -1 if stack is empty.

5fill in blank

hard

Fill all three blanks to complete the peek method that returns the top element without removing it.

DSA Java

public int peek() {
    if(top [1] 0) {
        return stack[[2]];
    } else {
        System.out.println("Stack is empty");
        return [3];
    }
}

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

A>=

Btop

C-1

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using == instead of >= causing wrong empty check.

Returning 0 which may be a valid element.

Check if top >= 0, return stack[top], else return -1 for empty stack.

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