Why Stack Exists and What Problems It Solves in DSA Java - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to push an element onto the stack.

DSA Java

stack.[1](value);

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

Apush

Bremove

Cpeek

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of push will remove an element instead of adding.

peek only looks at the top element without removing or adding.

The push method adds an element to the top of the stack.

2fill in blank

medium

Complete the code to remove the top element from the stack.

DSA Java

stack.[1]();

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

Aadd

Bpush

Cpeek

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using push will add an element instead of removing.

peek only looks at the top element without removing it.

The pop method removes and returns the top element from the stack.

3fill in blank

hard

Fix the error in the code to check the top element without removing it.

DSA Java

Element top = stack.[1]();

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

Apush

Bpop

Cpeek

Dremove

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop removes the element, which is not desired here.

push adds an element, not what we want.

The peek method returns the top element without removing it.

4fill in blank

hard

Fill both blanks to create a stack and add an element to it.

DSA Java

Stack<String> stack = new [1]<>();
stack.[2]("apple");

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

AStack

Bpush

Cpop

DQueue

Attempts:

3 left

💡 Hint

Common Mistakes

Using Queue instead of Stack for the data structure.

Using pop instead of push to add an element.

We create a new Stack object and use push to add an element.

5fill in blank

hard

Fill all three blanks to check if the stack is empty and print a message.

DSA Java

if (stack.[1]()) {
    System.out.println("Stack is [2].");
} else {
    System.out.println("Stack is [3].");
}

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

AisEmpty

Bempty

Cnot empty

Dfull

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'empty()' as a method which does not exist.

Confusing 'full' with 'empty' in the messages.

The isEmpty() method checks if the stack has no elements. We print 'empty' if true, otherwise 'not empty'.

Practice

1. Why do we use a stack data structure in programming?

easy

A. To store large amounts of data permanently

B. To store data in random order

C. To sort numbers automatically

D. To keep track of function calls and return addresses

Solution

1. Step 1: Understand the role of stack in function calls

   Stacks help keep track of where each function should return after finishing execution, storing return addresses in order.

2. Step 2: Compare other options

   Stacks do not sort data automatically, nor store data randomly or permanently; their main use is managing function calls.

3. Final Answer:

   To keep track of function calls and return addresses -> Option D

4. Quick Check:

   Stack manages function calls = A [OK]

Hint: Stack remembers last task first, like a pile of plates [OK]

Common Mistakes:

• Thinking stack sorts data automatically
• Confusing stack with permanent storage
• Assuming stack stores data randomly

2. Which of the following is the correct way to push an element onto a stack in Java?

easy

A. stack.push(element);

B. stack.put(element);

C. stack.insert(element);

D. stack.add(element);

Solution

1. Step 1: Recall Java Stack method names

   The Java Stack class uses the method push() to add elements on top of the stack.

2. Step 2: Check other method names

   Methods like add(), insert(), or put() are not standard for Stack push operation in Java.

3. Final Answer:

   stack.push(element); -> Option A

4. Quick Check:

   Push method adds element = D [OK]

Hint: Use push() to add on top of stack in Java [OK]

Common Mistakes:

• Using add() instead of push()
• Confusing insert() with push()
• Using put() which is not a Stack method

3. What will be the output of the following Java code?

Stack<Integer> stack = new Stack<>();
stack.push(10);
stack.push(20);
stack.push(30);
System.out.println(stack.pop());
System.out.println(stack.peek());

medium

A. 20
30

B. 10
20

C. 30
20

D. 30
10

Solution

1. Step 1: Trace push operations

   Elements pushed in order: 10 (bottom), 20, 30 (top).

2. Step 2: Trace pop and peek

   pop() removes and returns top element 30; peek() returns new top element 20 without removing it.

3. Final Answer:

   30 20 -> Option C

4. Quick Check:

   pop()=30, peek()=20 [OK]

Hint: pop() removes top, peek() shows top without removing [OK]

Common Mistakes:

• Confusing pop() and peek() results
• Assuming pop() returns bottom element
• Mixing order of pushed elements

4. Identify the error in the following Java code using Stack:

Stack<String> stack = new Stack<>();
stack.push("A");
stack.push("B");
System.out.println(stack.pop());
System.out.println(stack.pop());
System.out.println(stack.pop());

medium

A. Stack underflow error at third pop()

B. No error, code runs fine

C. Syntax error in stack declaration

D. push() method is incorrect

Solution

1. Step 1: Count push and pop operations

   Two elements pushed: "A", "B"; three pop() calls made.

2. Step 2: Analyze pop() calls

   First two pop() calls remove "B" and "A"; third pop() tries to remove from empty stack causing underflow error.

3. Final Answer:

   Stack underflow error at third pop() -> Option A

4. Quick Check:

   More pops than pushes = underflow error [OK]

Hint: Pop more times than push causes error [OK]

Common Mistakes:

• Ignoring stack underflow on extra pop()
• Thinking push() syntax is wrong
• Assuming no error occurs

5. You want to reverse a string using a stack in Java. Which sequence of operations correctly achieves this?

hard

A. Push all characters, then push again to reverse

B. Push each character, then pop all to build reversed string

C. Pop each character, then push all to build reversed string

D. Peek each character, then pop all to build reversed string

Solution

1. Step 1: Understand stack reversal logic

   Pushing characters stores them in order; popping them reverses order because stack is last-in-first-out.

2. Step 2: Evaluate options

   Only pushing then popping all characters correctly reverses the string; popping before pushing or peeking does not reverse.

3. Final Answer:

   Push each character, then pop all to build reversed string -> Option B

4. Quick Check:

   Push then pop reverses string = C [OK]

Hint: Push all, then pop all to reverse order [OK]

Common Mistakes:

• Trying to pop before pushing
• Using peek instead of pop to build reversed string
• Pushing twice without popping