Stack Concept and LIFO Principle in DSA Go - Time & Space Complexity

Time Complexity: Stack Concept and LIFO Principle

O(1)

Understanding Time Complexity

We want to understand how the time to add or remove items from a stack changes as the stack grows.

How does the stack handle operations when it gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

package main

import "fmt"

type Stack struct {
  items []int
}

func (s *Stack) Push(item int) {
  s.items = append(s.items, item)
}

func (s *Stack) Pop() int {
  if len(s.items) == 0 {
    return -1
  }
  last := s.items[len(s.items)-1]
  s.items = s.items[:len(s.items)-1]
  return last
}

func main() {
  s := Stack{}
  s.Push(10)
  s.Push(20)
  fmt.Println(s.Pop())
  fmt.Println(s.Pop())
}

This code shows how to add (Push) and remove (Pop) items from a stack using a slice.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Adding or removing one item at the end of the slice.
• How many times: Each Push or Pop does this once per call, no loops inside.

How Execution Grows With Input

Each Push or Pop takes about the same time no matter how many items are in the stack.

Input Size (n)	Approx. Operations
10	1 operation per Push or Pop
100	1 operation per Push or Pop
1000	1 operation per Push or Pop

Pattern observation: The time does not grow with the number of items; it stays constant.

Final Time Complexity

Time Complexity: O(1)

This means adding or removing an item from the stack takes the same short time no matter how big the stack is.

Common Mistake

[X] Wrong: "Pushing or popping takes longer as the stack grows because there are more items to check."

[OK] Correct: The stack only adds or removes the last item, so it does not look at other items. The operation time stays the same.

Interview Connect

Knowing that stack operations are quick and constant time helps you explain why stacks are useful for tasks like undo actions or expression evaluation.

Self-Check

"What if we changed the stack to remove items from the front instead of the end? How would the time complexity change?"

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