What if you could remove the last item from a list instantly without any hassle or mistakes?
Why Array Deletion at End in DSA Go?
Imagine you have a list of your favorite songs written on paper. When you want to remove the last song you added, you have to carefully erase it without messing up the rest. Doing this by hand every time can be slow and tiring.
Manually removing the last item means you must find it first, then erase it carefully. If the list is long, this takes time and you might accidentally erase the wrong song or leave a messy gap. It's easy to make mistakes and waste effort.
Using array deletion at the end lets you quickly remove the last item without touching the rest. It's like having a magic eraser that only removes the last song instantly and cleanly, saving time and avoiding errors.
songs := []string{"Song1", "Song2", "Song3"}
// Manually find last index and remove
lastIndex := len(songs) - 1
songs = songs[:lastIndex]songs := []string{"Song1", "Song2", "Song3"}
// Delete last element by slicing
songs = songs[:len(songs)-1]This lets you manage lists efficiently, quickly removing the last item whenever needed without errors or delays.
When you use a music app and remove the last song from your playlist, the app instantly updates the list without slowing down or messing up your other songs.
Manual removal of last item is slow and error-prone.
Array deletion at end is fast and clean using slicing.
It helps manage lists efficiently in many real-world apps.
Practice
Solution
Step 1: Understand array deletion at end
Deleting at the end means removing the last element only, not the first or all.Step 2: Effect on array size
After removal, the array size decreases by one because one element is gone.Final Answer:
The last element is removed and the array size decreases by one. -> Option AQuick Check:
Deleting end element reduces size by one [OK]
- Thinking first element is removed
- Assuming array size stays same
- Believing array clears completely
arr?Solution
Step 1: Understand slice slicing syntax
To remove last element, slice up to length minus one: arr[:len(arr)-1].Step 2: Check other options
arr[len(arr):] results in empty slice, arr[1:] removes first element, arr[:len(arr)+1] causes runtime error.Final Answer:
arr = arr[:len(arr)-1] -> Option BQuick Check:
Slice up to len-1 removes last element [OK]
- Using arr[1:] removes first element
- Using arr[len(arr):] gives empty slice
- Using arr[:len(arr)+1] causes error
arr := []int{10, 20, 30, 40}
arr = arr[:len(arr)-1]
fmt.Println(arr)Solution
Step 1: Initial array and slicing
arr starts as [10 20 30 40]. Slicing arr[:len(arr)-1] means arr[:3].Step 2: Result after slicing
arr[:3] gives first three elements: [10 20 30].Final Answer:
[10 20 30] -> Option CQuick Check:
Slice up to len-1 excludes last element [OK]
- Expecting last element to remain
- Confusing arr[1:] with arr[:len(arr)-1]
- Thinking output is empty slice
arr:arr := []int{1, 2, 3}
arr = arr[:len(arr)]
fmt.Println(arr)Solution
Step 1: Analyze slicing bounds
arr[:len(arr)] means slice from start to full length, so no element is removed.Step 2: Effect on array
Since slice covers entire array, arr remains [1 2 3] unchanged.Final Answer:
It does not delete any element; array remains unchanged. -> Option DQuick Check:
Slicing arr[:len(arr)] keeps full array [OK]
- Assuming last element is deleted
- Expecting runtime error
- Thinking first element is removed
arr := []int{5, 10, 15, 20, 25}, write the Go code to delete the last two elements and print the resulting slice.Solution
Step 1: Understand deletion of last two elements
To remove last two elements, slice up to length minus two: arr[:len(arr)-2].Step 2: Apply slicing and print
arr[:3] gives [5 10 15]. Printing arr shows these three elements.Final Answer:
arr = arr[:len(arr)-2] fmt.Println(arr) -> Option AQuick Check:
Slice up to len-2 removes last two elements [OK]
- Using arr[2:] removes first two elements
- Removing only last element instead of two
- Incorrect slice bounds causing errors
Ready to prove it?
Pass the test and this pattern moves unknown
