Array Deletion at End in DSA Go - Build from Scratch

Array Deletion at End

📖 Scenario: You are managing a list of tasks to do today. Sometimes, you finish the last task and want to remove it from your list.

🎯 Goal: Build a simple Go program that creates a list of tasks, sets up a count of tasks, removes the last task from the list, and then prints the updated list.

📋 What You'll Learn

Create a slice of strings called tasks with exactly these values: "Laundry", "Dishes", "Homework", "Shopping"

Create an integer variable called count that stores the number of tasks in the tasks slice

Remove the last task from the tasks slice by slicing it

Print the updated tasks slice

💡 Why This Matters

🌍 Real World

Managing lists of tasks, items, or data where you often add or remove the last element, like a to-do list or recent files list.

💼 Career

Understanding how to manipulate slices is essential for Go developers working on applications that handle dynamic lists, such as task managers, inventory systems, or user activity logs.

1

Create the tasks list

Create a slice of strings called tasks with these exact values: "Laundry", "Dishes", "Homework", "Shopping"

DSA Go

# Your code here

Hint

Use tasks := []string{} to create the list with the exact tasks inside.

2

Count the number of tasks

Create an integer variable called count that stores the number of tasks in the tasks slice using the len function

DSA Go

package main

import "fmt"

func main() {
    tasks := []string{"Laundry", "Dishes", "Homework", "Shopping"}
    
    // Your code here
}

Hint

Use count := len(tasks) to get the number of tasks.

3

Remove the last task

Remove the last task from the tasks slice by slicing it to keep only the first count - 1 elements

DSA Go

package main

import "fmt"

func main() {
    tasks := []string{"Laundry", "Dishes", "Homework", "Shopping"}
    count := len(tasks)
    
    // Your code here
}

Hint

Use slicing like tasks = tasks[:count-1] to remove the last task.

4

Print the updated tasks list

Print the updated tasks slice using fmt.Println(tasks)

DSA Go

package main

import "fmt"

func main() {
    tasks := []string{"Laundry", "Dishes", "Homework", "Shopping"}
    count := len(tasks)
    tasks = tasks[:count-1]
    
    // Your code here
}

Hint

Use fmt.Println(tasks) to show the updated list.

Practice

1. What happens when you delete an element from the end of an array in Go?

easy

A. The last element is removed and the array size decreases by one.

B. The first element is removed and the array size decreases by one.

C. The array size stays the same but the last element becomes zero.

D. The array is completely cleared and becomes empty.

Solution

1. Step 1: Understand array deletion at end

   Deleting at the end means removing the last element only, not the first or all.

2. Step 2: Effect on array size

   After removal, the array size decreases by one because one element is gone.

3. Final Answer:

   The last element is removed and the array size decreases by one. -> Option A

4. Quick Check:

   Deleting end element reduces size by one [OK]

Hint: Deleting at end removes last element and shrinks array [OK]

Common Mistakes:

• Thinking first element is removed
• Assuming array size stays same
• Believing array clears completely

2. Which of the following Go code snippets correctly deletes the last element from a slice named arr?

easy

A. arr = arr[len(arr):]

B. arr = arr[:len(arr)-1]

C. arr = arr[1:]

D. arr = arr[:len(arr)+1]

Solution

1. Step 1: Understand slice slicing syntax

   To remove last element, slice up to length minus one: arr[:len(arr)-1].

2. Step 2: Check other options

   arr[len(arr):] results in empty slice, arr[1:] removes first element, arr[:len(arr)+1] causes runtime error.

3. Final Answer:

   arr = arr[:len(arr)-1] -> Option B

4. Quick Check:

   Slice up to len-1 removes last element [OK]

Hint: Use arr[:len(arr)-1] to drop last element [OK]

Common Mistakes:

• Using arr[1:] removes first element
• Using arr[len(arr):] gives empty slice
• Using arr[:len(arr)+1] causes error

3. What is the output of this Go code?

arr := []int{10, 20, 30, 40}
arr = arr[:len(arr)-1]
fmt.Println(arr)

medium

A. [10 20 30 40]

B. []

C. [10 20 30]

D. [20 30 40]

Solution

1. Step 1: Initial array and slicing

   arr starts as [10 20 30 40]. Slicing arr[:len(arr)-1] means arr[:3].

2. Step 2: Result after slicing

   arr[:3] gives first three elements: [10 20 30].

3. Final Answer:

   [10 20 30] -> Option C

4. Quick Check:

   Slice up to len-1 excludes last element [OK]

Hint: Slicing arr[:len(arr)-1] drops last element [OK]

Common Mistakes:

• Expecting last element to remain
• Confusing arr[1:] with arr[:len(arr)-1]
• Thinking output is empty slice

4. Identify the error in this Go code that tries to delete the last element from arr:

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

medium

A. It deletes all elements making the array empty.

B. It causes a runtime panic due to invalid slice bounds.

C. It deletes the first element instead of the last.

D. It does not delete any element; array remains unchanged.

Solution

1. Step 1: Analyze slicing bounds

   arr[:len(arr)] means slice from start to full length, so no element is removed.

2. Step 2: Effect on array

   Since slice covers entire array, arr remains [1 2 3] unchanged.

3. Final Answer:

   It does not delete any element; array remains unchanged. -> Option D

4. Quick Check:

   Slicing arr[:len(arr)] keeps full array [OK]

Hint: arr[:len(arr)] keeps full array, no deletion [OK]

Common Mistakes:

• Assuming last element is deleted
• Expecting runtime error
• Thinking first element is removed

5. Given a slice arr := []int{5, 10, 15, 20, 25}, write the Go code to delete the last two elements and print the resulting slice.

hard

A. arr = arr[:len(arr)-2]
fmt.Println(arr)

B. arr = arr[2:]
fmt.Println(arr)

C. arr = arr[:len(arr)-1]
fmt.Println(arr)

D. arr = arr[1:len(arr)-1]
fmt.Println(arr)

Solution

1. Step 1: Understand deletion of last two elements

   To remove last two elements, slice up to length minus two: arr[:len(arr)-2].

2. Step 2: Apply slicing and print

   arr[:3] gives [5 10 15]. Printing arr shows these three elements.

3. Final Answer:

   arr = arr[:len(arr)-2] fmt.Println(arr) -> Option A

4. Quick Check:

   Slice up to len-2 removes last two elements [OK]

Hint: Use arr[:len(arr)-2] to remove last two elements [OK]

Common Mistakes:

• Using arr[2:] removes first two elements
• Removing only last element instead of two
• Incorrect slice bounds causing errors