Stack Concept and LIFO Principle in DSA Go

Mental Model

A stack is a collection where the last item added is the first one removed. This is called Last In, First Out (LIFO).

Analogy: Think of a stack of plates: you put a new plate on top and take plates from the top only.

Top -> [  ] -> [  ] -> [  ] -> Bottom
          ↑

Dry Run Walkthrough

Input: Start with empty stack, push 1, push 2, push 3, then pop once

Goal: Show how items are added and removed following LIFO

Step 1: Push 1 onto stack

Top -> [1] -> Bottom
          ↑

Why: First item goes on top, stack was empty

Step 2: Push 2 onto stack

Top -> [2] -> [1] -> Bottom
          ↑

Why: New item goes on top, above previous items

Step 3: Push 3 onto stack

Top -> [3] -> [2] -> [1] -> Bottom
          ↑

Why: Keep adding on top, last pushed is on top

Step 4: Pop top item (3) from stack

Top -> [2] -> [1] -> Bottom
          ↑

Why: Remove the last added item first, following LIFO

Result:

Top -> [2] -> [1] -> Bottom
          ↑
Popped value: 3

Annotated Code

DSA Go

package main

import "fmt"

// Stack holds integers with LIFO behavior
type Stack struct {
	items []int
}

// Push adds an item on top of the stack
func (s *Stack) Push(val int) {
	s.items = append(s.items, val) // add to end (top)
}

// Pop removes and returns the top item
func (s *Stack) Pop() (int, bool) {
	if len(s.items) == 0 {
		return 0, false // empty stack
	}
	val := s.items[len(s.items)-1] // get last item
	s.items = s.items[:len(s.items)-1] // remove last item
	return val, true
}

// String returns stack as string from top to bottom
func (s *Stack) String() string {
	result := "Top -> "
	for i := len(s.items) - 1; i >= 0; i-- {
		result += fmt.Sprintf("[%d] -> ", s.items[i])
	}
	result += "Bottom"
	return result
}

func main() {
	stack := Stack{}
	stack.Push(1) // step 1
	stack.Push(2) // step 2
	stack.Push(3) // step 3
	fmt.Println(stack.String())
	val, ok := stack.Pop() // step 4
	if ok {
		fmt.Println("Popped value:", val)
	}
	fmt.Println(stack.String())
}

s.items = append(s.items, val) // add to end (top)

add new item to top of stack

if len(s.items) == 0 { return 0, false // empty stack }

check if stack is empty before popping

val := s.items[len(s.items)-1] // get last item

get top item to pop

s.items = s.items[:len(s.items)-1] // remove last item

remove top item from stack

Output

Top -> [3] -> [2] -> [1] -> Bottom

Popped value: 3

Top -> [2] -> [1] -> Bottom

Complexity Analysis

Time: O(1) for both push and pop because we add or remove only the last element

Space: O(n) because stack stores all pushed elements in a slice

vs Alternative: Compared to a queue which removes from front (O(n) if array), stack operations are faster and simpler with O(1) time

Edge Cases

Pop from empty stack

Returns false and zero value without error

DSA Go

if len(s.items) == 0 {
	return 0, false // empty stack
}

When to Use This Pattern

When you see a problem requiring last added item to be accessed or removed first, use stack and LIFO principle for efficient solution.

Common Mistakes

Mistake: Trying to remove from the front of the slice instead of the end
Fix: Always remove from the end of the slice to maintain O(1) pop operation

Summary

Stack stores items so the last added is the first removed (LIFO).

Use stack when you need to reverse order or track recent items first.

Remember: push adds on top, pop removes from top, like a plate stack.

Practice

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

easy

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

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

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

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

Solution

1. Step 1: Understand LIFO meaning

   LIFO stands for Last In, First Out, meaning the last item added to the stack is the first one removed.

2. Step 2: Match with stack behavior

   Stacks follow LIFO, so the last pushed element is popped first.

3. Final Answer:

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

4. Quick Check:

   LIFO = Last In, First Out [OK]

Hint: Remember stack removes newest item first [OK]

Common Mistakes:

• Confusing LIFO with FIFO
• Thinking first added is removed first
• Mixing up stack with queue behavior

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

easy

A. stack.push(5)

B. stack = append(stack, 5)

C. stack.add(5)

D. stack.insert(5)

Solution

1. Step 1: Recall Go slice push syntax

   In Go, stacks are often slices; to push, use append(slice, value).

2. Step 2: Check options for correct syntax

   Only stack = append(stack, 5) correctly adds 5 to the stack slice.

3. Final Answer:

   stack = append(stack, 5) -> Option B

4. Quick Check:

   Go stack push uses append [OK]

Hint: Use append to add elements to Go slices [OK]

Common Mistakes:

• Using non-existent push method on slice
• Using add or insert which are invalid for slices
• Forgetting to assign append result back

3. What will be the output of the following Go code snippet?

stack := []int{}
stack = append(stack, 10)
stack = append(stack, 20)
stack = append(stack, 30)
last := stack[len(stack)-1]
stack = stack[:len(stack)-1]
fmt.Println(last)

medium

A. 30

B. 20

C. 10

D. Error

Solution

1. Step 1: Trace stack push operations

   Elements 10, 20, 30 are pushed in order; stack is [10, 20, 30].

2. Step 2: Pop last element

   last = stack[len(stack)-1] gets 30; then stack is sliced to remove last element.

3. Final Answer:

   30 -> Option A

4. Quick Check:

   Last pushed element popped = 30 [OK]

Hint: Last appended element is popped first [OK]

Common Mistakes:

• Printing first element instead of last
• Forgetting to slice stack after pop
• Confusing index for pop operation

4. Identify the error in this Go code that tries to pop an element from a stack:

stack := []int{1, 2, 3}
last := stack[len(stack)]
stack = stack[:len(stack)-1]
fmt.Println(last)

medium

A. Cannot print variable last

B. Syntax error in slice operation

C. Index out of range error due to accessing stack[len(stack)]

D. No error, code runs correctly

Solution

1. Step 1: Check index used for popping

   Indexing with stack[len(stack)] is invalid because slice indices go from 0 to len(stack)-1.

2. Step 2: Identify error type

   This causes an index out of range runtime error.

3. Final Answer:

   Index out of range error due to accessing stack[len(stack)] -> Option C

4. Quick Check:

   Index must be len(stack)-1, not len(stack) [OK]

Hint: Use len(stack)-1 for last index [OK]

Common Mistakes:

• Using len(stack) as index
• Ignoring zero-based indexing
• Assuming code runs without error

5. Given a stack implemented as a slice in Go, how would you reverse the order of elements using only stack operations (push and pop)?

hard

A. Pop elements and discard them to reverse the stack

B. Push elements onto the same stack again without popping

C. Use a queue instead of a stack to reverse elements

D. Pop all elements into a temporary slice, then push them back in order

Solution

1. Step 1: Pop all elements from original stack

   Pop elements one by one and store them in a temporary slice to hold reversed order.

2. Step 2: Push elements back to original stack

   Push elements from temporary slice back to stack; this reverses the order due to LIFO.

3. Final Answer:

   Pop all elements into a temporary slice, then push them back in order -> Option D

4. Quick Check:

   Reversing stack needs pop then push back [OK]

Hint: Pop all, then push back reverses stack [OK]

Common Mistakes:

• Pushing without popping does not reverse
• Discarding elements loses data
• Using queue is not stack operation