Stack Concept and LIFO Principle in DSA Javascript - Build from Scratch

Stack Concept and LIFO Principle

📖 Scenario: Imagine you have a stack of books on your desk. You can only add or remove the top book. This is how a stack data structure works using the Last In, First Out (LIFO) principle.

🎯 Goal: You will create a stack using a JavaScript array, add books to it, remove the top book, and then display the current stack.

📋 What You'll Learn

Create an empty stack using a JavaScript array called stack.

Add three books to the stack in this order: 'Book1', 'Book2', 'Book3'.

Create a variable called removedBook to store the book removed from the top of the stack.

Remove the top book from the stack using the correct stack operation.

Print the removedBook and the current state of the stack.

💡 Why This Matters

🌍 Real World

Stacks are used in real life for things like undo buttons in apps, browser history, and managing tasks where the last added task is done first.

💼 Career

Understanding stacks helps in software development, especially in algorithms, parsing expressions, and managing function calls.

1

Create an empty stack

Create an empty array called stack to represent the stack.

DSA Javascript

// Your code here

Hint

Use [] to create an empty array and assign it to stack.

2

Add books to the stack

Add 'Book1', 'Book2', and 'Book3' to the stack using the push method in this exact order.

DSA Javascript

const stack = [];
// Your code here

Hint

Use stack.push('Book1') to add 'Book1' to the stack.

3

Remove the top book

Create a variable called removedBook and assign it the value returned by removing the top book from the stack using the pop method.

DSA Javascript

const stack = [];
stack.push('Book1');
stack.push('Book2');
stack.push('Book3');
// Your code here

Hint

Use stack.pop() to remove the top book and assign it to removedBook.

4

Print the removed book and current stack

Print the value of removedBook and then print the current state of the stack using console.log.

DSA Javascript

const stack = [];
stack.push('Book1');
stack.push('Book2');
stack.push('Book3');
const removedBook = stack.pop();
// Your code here

Hint

Use console.log(removedBook) and console.log(stack) to print the results.

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