Stack Concept and LIFO Principle in DSA Javascript - Execution Trace

Concept Flow - Stack Concept and LIFO Principle

Start with empty stack

↓

Push element on top

→Top points to new element

↓

Push more elements?

Yes→Repeat push

No↓

Pop element from top

→Remove top element

↓

Stack empty?

No→Repeat pop

Yes↓

End

Stack works by adding (push) and removing (pop) elements only at the top, following Last In First Out order.

Execution Sample

DSA Javascript

let stack = [];
stack.push(1);
stack.push(2);
stack.pop();
stack.push(3);

This code adds 1 and 2 to the stack, removes the top (2), then adds 3.

Execution Table

Step	Operation	Nodes in Stack	Pointer Changes	Visual State
1	Initialize empty stack	[]	top = null	null
2	Push 1	[1]	top -> 1	1 -> null
3	Push 2	[1, 2]	top -> 2	2 -> 1 -> null
4	Pop	[1]	top -> 1	1 -> null
5	Push 3	[1, 3]	top -> 3	3 -> 1 -> null

💡 No more operations; stack ends with elements 3 on top of 1.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5
stack	[]	[1]	[1, 2]	[1]	[1, 3]
top	null	1	2	1	3

Key Moments - 3 Insights

Why does pop remove the last pushed element, not the first?

Because the stack follows LIFO: last in is first out. See execution_table step 4 where pop removes 2, the last pushed.

What happens to the pointer 'top' after a pop?

The 'top' pointer moves to the next element below. In step 4, top moves from 2 to 1 after pop.

Why can't we remove elements from the bottom of the stack?

Stack only allows push/pop at the top to keep order simple and predictable, as shown in concept_flow.

Visual Quiz - 3 Questions

Look at the execution_table, what is the stack state after step 3?

A [1, 2]

B [2, 1]

C [1]

D [2]

💡 Hint

Check the 'Nodes in Stack' column at step 3 in execution_table.

At which step does the 'top' pointer change from 2 to 1?

A Step 2

B Step 4

C Step 3

D Step 5

💡 Hint

Look at 'Pointer Changes' column in execution_table for when top moves from 2 to 1.

If we push 4 after step 5, what will be the new visual state?

A 1 -> 3 -> 4 -> null

B 3 -> 1 -> 4 -> null

C 4 -> 3 -> 1 -> null

D 4 -> 1 -> 3 -> null

💡 Hint

New pushes add on top, so 4 will be on top of 3 and 1, as per concept_flow.

Concept Snapshot

Stack stores elements in order.
Push adds to top.
Pop removes from top.
Follows Last In First Out (LIFO).
Only top element accessible.
Used for undo, backtracking, call stacks.

Full Transcript

A stack is a data structure where elements are added and removed only from the top. This follows the Last In First Out principle, meaning the last element added is the first to be removed. We start with an empty stack. When we push 1, it becomes the top. Pushing 2 puts it on top of 1. Popping removes the top element, which is 2, leaving 1 on top. Pushing 3 adds it on top of 1. The pointer 'top' always points to the current top element. This behavior is shown step-by-step in the execution table and variable tracker. Understanding that pop removes the last pushed element and that only the top can be accessed is key to using stacks correctly.

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