Pop Operation on Stack in DSA Go - Build from Scratch

Pop Operation on Stack

📖 Scenario: You are managing a stack of books. You want to remove the top book from the stack to read it.

🎯 Goal: Build a program that creates a stack of books, sets up a variable to hold the top book, removes the top book using the pop operation, and then prints the stack after removal.

📋 What You'll Learn

Create a stack as a slice of strings with exact books: "Book1", "Book2", "Book3", "Book4"

Create a variable called topBook to hold the book popped from the stack

Use the pop operation to remove the last book from the stack and assign it to topBook

Print the stack after the pop operation

💡 Why This Matters

🌍 Real World

Stacks are used in many real-world applications like undo features in text editors, browser history, and managing function calls.

💼 Career

Understanding stack operations is essential for software developers, especially when working with algorithms, parsing, and memory management.

1

Create the stack of books

Create a slice of strings called stack with these exact books in order: "Book1", "Book2", "Book3", "Book4"

DSA Go

# Your code here

Hint

Use a slice literal to create stack with the given books.

2

Create a variable to hold the popped book

Create a variable called topBook of type string to hold the book popped from the stack

DSA Go

package main

func main() {
    stack := []string{"Book1", "Book2", "Book3", "Book4"}
    // Create variable topBook here
}

Hint

Declare topBook as a string variable without assigning a value yet.

3

Pop the top book from the stack

Use the pop operation to remove the last book from stack and assign it to topBook. Update stack to remove the popped book.

DSA Go

package main

func main() {
    stack := []string{"Book1", "Book2", "Book3", "Book4"}
    var topBook string
    // Pop the top book from stack and assign to topBook
}

Hint

Use len(stack)-1 to get the last element and slice to remove it.

4

Print the stack after pop

Print the stack after the pop operation using fmt.Println(stack). Import fmt package at the top.

DSA Go

package main

func main() {
    stack := []string{"Book1", "Book2", "Book3", "Book4"}
    var topBook string
    topBook = stack[len(stack)-1]
    stack = stack[:len(stack)-1]
    // Print the stack here
}

Hint

Use fmt.Println(stack) to print the current stack.

Practice

1. What does the pop operation do in a stack?

easy

A. Clears all elements from the stack

B. Adds an element to the bottom of the stack

C. Returns the bottom element without removing it

D. Removes and returns the top element of the stack

Solution

1. Step 1: Understand stack behavior

   A stack follows Last In First Out (LIFO) principle, so the last added element is removed first.

2. Step 2: Define pop operation

   The pop operation removes and returns the top element of the stack, which is the last inserted item.

3. Final Answer:

   Removes and returns the top element of the stack -> Option D

4. Quick Check:

   Pop = remove top element [OK]

Hint: Pop always removes the last added item from the stack [OK]

Common Mistakes:

• Thinking pop removes from bottom
• Confusing pop with peek
• Assuming pop clears the stack

2. Which of the following is the correct Go syntax to pop an element from a stack represented as a slice named stack?

easy

A. element := stack[0]; stack = stack[1:]

B. element := stack[len(stack)-1]; stack = stack[:len(stack)-1]

C. element := stack[len(stack)]; stack = stack[:len(stack)-1]

D. element := stack[len(stack)-1]; stack = stack[1:]

Solution

1. Step 1: Identify top element index in slice

   In Go, the top of the stack is the last element: index len(stack)-1.

2. Step 2: Correctly slice to remove top element

   To pop, get element at len(stack)-1, then slice stack up to len(stack)-1 to remove it.

3. Final Answer:

   element := stack[len(stack)-1]; stack = stack[:len(stack)-1] -> Option B

4. Quick Check:

   Pop syntax = element last, slice off last [OK]

Hint: Pop last element with slice[:len(slice)-1] [OK]

Common Mistakes:

• Using index len(stack) which is out of range
• Removing from front instead of end
• Not updating the stack slice after pop

3. What is the output of the following Go code?

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

medium

A. 30
[10 20]

B. 20
[10 20 30]

C. 30
[10 20 30]

D. 10
[20 30]

Solution

1. Step 1: Identify popped element

   element = stack[len(stack)-1] = stack[2] = 30

2. Step 2: Update stack after pop

   stack = stack[:2] = [10, 20]

3. Final Answer:

   30 [10 20] -> Option A

4. Quick Check:

   Pop returns 30, stack left [10 20] [OK]

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

Common Mistakes:

• Printing old stack without slicing
• Confusing index for popped element
• Not updating stack slice after pop

4. Consider this Go code snippet for popping from a stack:

stack := []int{5, 15, 25}
element := stack[len(stack)]
stack = stack[:len(stack)-1]
fmt.Println(element)

What is the error in this code?

medium

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

B. Stack slice is not updated correctly after pop

C. Variable element is not declared

D. No error, code runs fine

Solution

1. Step 1: Check index used for popping

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

2. Step 2: Identify runtime error

   This causes an index out of range runtime error when trying to access stack[len(stack)].

3. Final Answer:

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

4. Quick Check:

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

Hint: Slice indices max at len(slice)-1, not len(slice) [OK]

Common Mistakes:

• Using len(stack) as index
• Ignoring runtime errors
• Assuming slice indexing is inclusive of len

5. You have a stack implemented as a slice in Go: stack := []int{1, 2, 3, 4, 5}. You want to pop two elements and then push 10 and 20. What will be the final stack?

hard

A. [10 20]

B. [1 2 3 4 5 10 20]

C. [1 2 3 10 20]

D. [1 2 10 20]

Solution

1. Step 1: Pop two elements from stack

   Pop removes last two elements: 5 and 4. Stack becomes [1, 2, 3].

2. Step 2: Push 10 and 20 onto stack

   Appending 10 and 20 results in [1, 2, 3, 10, 20].

3. Final Answer:

   [1 2 3 10 20] -> Option C

4. Quick Check:

   Pop twice, then push twice = [1 2 3 10 20] [OK]

Hint: Pop removes last; push adds at end [OK]

Common Mistakes:

• Not popping twice before pushing
• Appending instead of replacing popped elements
• Confusing order of push and pop