Stack Concept and LIFO Principle in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to add an element to the top of the stack.

DSA Javascript

stack.[1](5);

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

Apop

Bpush

Cshift

Dunshift

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of push removes an element instead of adding.

Using shift or unshift changes the start of the array, not the top of the stack.

The push method adds an element to the top of the stack, following the LIFO principle.

2fill in blank

medium

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

DSA Javascript

const top = stack.[1]();

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

Aunshift

Bpush

Cshift

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using push instead of pop adds an element instead of removing.

Using shift or unshift removes or adds elements at the start, not the top.

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

3fill in blank

hard

Fix the error in the code to correctly check if the stack is empty.

DSA Javascript

if (stack.[1] === 0) {
  console.log('Stack is empty');
}

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

Alength

Bsize

Clength()

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

Using length() calls a function that does not exist, causing an error.

Using size or count are not valid properties for arrays in JavaScript.

The length property (without parentheses) gives the number of elements in the stack array.

4fill in blank

hard

Fill both blanks to peek at the top element without removing it.

DSA Javascript

const top = stack[stack.[1] [2] 1];

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

Alength

B-

C+

Dsize

Attempts:

3 left

💡 Hint

Common Mistakes

Using size instead of length causes undefined errors.

Adding 1 to length goes out of array bounds.

Accessing stack[stack.length - 1] gets the top element without removing it.

5fill in blank

hard

Fill all three blanks to create a function that pushes a value and returns the new top.

DSA Javascript

function pushAndPeek(stack, value) {
  stack.[1](value);
  return stack[stack.[2] [3] 1];
}

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

Apush

Blength

C-

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using plus instead of minus causes out-of-bounds errors.

Using pop removes the element instead of peeking.

The function uses push to add the value, then returns the last element using stack.length - 1.

Practice

1. What does the LIFO principle mean in the context of a stack?

easy

A. Elements are removed based on their value

B. First In, First Out - the first element added is the first to be removed

C. Elements are removed randomly

D. Last In, First Out - the last element added is the first to be removed

Solution

1. Step 1: Understand the LIFO principle

   LIFO means the last element you put in is the first one you take out.

2. Step 2: Apply to stack behavior

   Stacks follow LIFO, so the newest item is removed first.

3. Final Answer:

   Last In, First Out - the last element added is the first to be removed -> Option D

4. Quick Check:

   LIFO = Last In, First Out [OK]

Hint: Remember stack like a stack of plates, last added removed first [OK]

Common Mistakes:

• Confusing LIFO with FIFO
• Thinking elements are removed randomly
• Assuming removal depends on element value

2. Which of the following is the correct way to push an element '5' onto a stack named stack in JavaScript?

easy

A. stack.insert(5);

B. stack.add(5);

C. stack.push(5);

D. stack.enqueue(5);

Solution

1. Step 1: Recall JavaScript stack operations

   JavaScript arrays use push() to add elements at the end (top of stack).

2. Step 2: Match method to stack push

   push() adds element to stack top, so stack.push(5); is correct.

3. Final Answer:

   stack.push(5); -> Option C

4. Quick Check:

   Push method adds element to stack [OK]

Hint: Use push() to add elements to stack in JavaScript [OK]

Common Mistakes:

• Using add() or insert() which are not array methods
• Confusing enqueue() with queue operation
• Trying to use non-existent methods

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

const stack = [];
stack.push(10);
stack.push(20);
const popped = stack.pop();
console.log(popped);
console.log(stack);

medium

A. 20
[10]

B. 10
[20]

C. 10
[10, 20]

D. 20
[20]

Solution

1. Step 1: Trace push operations

   stack.push(10) adds 10, stack.push(20) adds 20. Stack is now [10, 20].

2. Step 2: Trace pop operation

   stack.pop() removes last element 20 and returns it. Stack becomes [10].

3. Final Answer:

   20 [10] -> Option A

4. Quick Check:

   Pop returns last pushed element 20 [OK]

Hint: Pop returns last pushed element, stack shrinks by one [OK]

Common Mistakes:

• Thinking pop returns first element
• Confusing push and pop order
• Expecting stack unchanged after pop

4. Identify the error in the following code snippet that tries to implement a stack using an array:

const stack = [];
stack.push(1);
stack.push(2);
const top = stack.pop;
console.log(top);

medium

A. pop is used without parentheses, so it returns the function, not the popped value

B. push method is incorrect, should be pushElement

C. stack should be initialized as an object, not an array

D. console.log cannot print stack elements

Solution

1. Step 1: Check pop usage

   stack.pop is a function reference; missing () means no call, so top is function, not value.

2. Step 2: Understand effect on output

   console.log(top) prints function code, not popped element. Correct usage is stack.pop().

3. Final Answer:

   pop is used without parentheses, so it returns the function, not the popped value -> Option A

4. Quick Check:

   Call pop() with () to get value [OK]

Hint: Always use parentheses to call pop() method [OK]

Common Mistakes:

• Forgetting parentheses on pop()
• Thinking push method is wrong
• Confusing array with object for stack

5. You have a stack initially empty. You perform these operations in order:

push(1), push(2), pop(), push(3), pop(), pop()

What is the final output of each pop operation in sequence?

hard

A. [2, 3, 3]

B. [2, 3, 1]

C. [2, 3, undefined]

D. [1, 3, 2]

Solution

1. Step 1: Perform push(1) and push(2)

   Stack becomes [1, 2].

2. Step 2: pop() removes last pushed element

   pop() returns 2, stack is now [1].

3. Step 3: push(3)

   Stack becomes [1, 3].

4. Step 4: pop() returns 3, stack is [1]

5. Step 5: pop() returns 1, stack is empty

6. Final Answer:

   [2, 3, 1] -> Option B

7. Quick Check:

   Pop returns last pushed elements in reverse order [OK]

Hint: Track stack after each operation to find pop outputs [OK]

Common Mistakes:

• Mixing order of popped elements
• Assuming pop returns undefined when stack not empty
• Confusing push and pop effects