Array Deletion at End in DSA Rust - Execution Trace

Concept Flow - Array Deletion at End

Start with array

↓

Check if array is empty?

Yes→Stop: nothing to delete

No↓

Remove last element

↓

Update array size

↓

Done

This flow shows checking if the array is empty, then removing the last element and updating the size.

Execution Sample

DSA Rust

let mut arr = vec![10, 20, 30, 40];
arr.pop();
println!("{:?}", arr);

This code removes the last element from the array and prints the updated array.

Execution Table

Step	Operation	Array Before	Pointer Changes	Array After	Visual State
1	Start with array	[10, 20, 30, 40]	head -> index 0, tail -> index 3	[10, 20, 30, 40]	10 -> 20 -> 30 -> 40 -> null
2	Check if array is empty	[10, 20, 30, 40]	tail index 3 valid	[10, 20, 30, 40]	10 -> 20 -> 30 -> 40 -> null
3	Remove last element (pop)	[10, 20, 30, 40]	tail moves from index 3 to 2	[10, 20, 30]	10 -> 20 -> 30 -> null
4	Print updated array	[10, 20, 30]	tail at index 2	[10, 20, 30]	10 -> 20 -> 30 -> null
5	End	[10, 20, 30]	tail at index 2	[10, 20, 30]	10 -> 20 -> 30 -> null

💡 Array size reduced by one after pop; tail pointer updated; no elements left to remove at index 3.

Variable Tracker

Variable	Start	After Step 3	Final
arr	[10, 20, 30, 40]	[10, 20, 30]	[10, 20, 30]
tail_index	3	2	2
size	4	3	3

Key Moments - 3 Insights

Why do we check if the array is empty before deleting?

Because deleting from an empty array would cause an error. Step 2 in the execution_table shows the check to avoid this.

What happens to the last element after pop?

It is removed and no longer accessible. Step 3 shows the tail pointer moving back, effectively removing the last element.

Does the array memory shrink after pop?

In Rust, pop removes the element and updates size, but underlying capacity may remain. The visual state shows the logical array after deletion.

Visual Quiz - 3 Questions

Look at the execution_table, what is the array state after step 3?

A [10, 20, 30, 40]

B [20, 30, 40]

C [10, 20, 30]

D []

💡 Hint

Check the 'Array After' column in row with Step 3.

At which step does the tail pointer move?

A Step 1

B Step 3

C Step 2

D Step 4

💡 Hint

Look at the 'Pointer Changes' column to find when tail index changes.

If the array was empty at start, what would happen at step 2?

A Check fails and stops deletion

B Pop removes last element

C Tail pointer moves

D Array prints normally

💡 Hint

Step 2 checks if array is empty before deletion.

Concept Snapshot

Array Deletion at End (pop):
- Check if array is empty
- Remove last element if not empty
- Update size and tail pointer
- Underlying capacity may stay same
- Safe operation to avoid errors

Full Transcript

This visual trace shows how deleting the last element from an array works. We start with an array of four elements. First, we check if the array is empty to avoid errors. Then we remove the last element by moving the tail pointer back and reducing the size. The array now has three elements. The visual state shows the array elements linked with arrows ending in null. This process safely updates the array without removing other elements or causing errors.

Practice

1. What happens when you remove the last element from a Rust Vec using pop()?

easy

A. The last element is removed and returned as Some(value).

B. The first element is removed and returned.

C. The entire vector is cleared.

D. The vector becomes immutable.

Solution

1. Step 1: Understand pop() behavior

   The pop() method removes the last element from the vector and returns it wrapped in Some(value) if the vector is not empty.

2. Step 2: Confirm it does not affect other elements

   Only the last element is removed; the rest of the vector remains unchanged.

3. Final Answer:

   The last element is removed and returned as Some(value). -> Option A

4. Quick Check:

   pop() removes last element [OK]

Hint: pop() removes and returns last element wrapped in Some [OK]

Common Mistakes:

• Thinking pop() removes the first element
• Assuming pop() clears the entire vector
• Believing pop() makes vector immutable

2. Which of the following is the correct syntax to remove the last element from a mutable vector v in Rust?

easy

A. v.erase_last();

B. v.remove_last();

C. v.delete_end();

D. v.pop();

Solution

1. Step 1: Recall Rust vector methods

   The standard method to remove the last element from a vector is pop().

2. Step 2: Check method names

   Methods like remove_last(), delete_end(), and erase_last() do not exist in Rust's Vec API.

3. Final Answer:

   v.pop(); -> Option D

4. Quick Check:

   Correct method to remove last element is pop() [OK]

Hint: Use pop() to remove last element from vector [OK]

Common Mistakes:

• Using non-existent methods like remove_last()
• Confusing pop() with remove() which removes by index
• Trying to use erase_last() or delete_end() which are invalid

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

let mut v = vec![10, 20, 30];
let x = v.pop();
println!("{:?}", v);
println!("{:?}", x);

medium

A. [10, 20]
None

B. [10, 20, 30]
None

C. [10, 20]
Some(30)

D. [20, 30]
Some(10)

Solution

1. Step 1: Understand pop() effect on vector

   The pop() removes the last element, which is 30, and returns Some(30).

2. Step 2: Check vector after pop

   After removal, the vector contains [10, 20].

3. Final Answer:

   [10, 20] Some(30) -> Option C

4. Quick Check:

   pop() removes last element 30, vector left with [10, 20] [OK]

Hint: pop() returns last element and shortens vector [OK]

Common Mistakes:

• Expecting pop() to remove first element
• Assuming pop() returns None when vector is not empty
• Confusing vector contents after pop

4. Identify the error in the following Rust code that tries to remove the last element from a vector:

let v = vec![1, 2, 3];
v.pop();

medium

A. pop() method does not exist for vectors.

B. Cannot call pop() on an immutable vector.

C. pop() removes the first element, not the last.

D. The vector must be empty before calling pop().

Solution

1. Step 1: Check vector mutability

   The vector v is declared immutable with let. To modify it, it must be mutable.

2. Step 2: Understand pop() requirements

   The pop() method requires a mutable vector to remove the last element.

3. Final Answer:

   Cannot call pop() on an immutable vector. -> Option B

4. Quick Check:

   pop() needs mutable vector [OK]

Hint: Make vector mutable with 'mut' before pop() [OK]

Common Mistakes:

• Trying to pop from immutable vector
• Thinking pop() removes first element
• Assuming pop() works on empty vector without error

5. Given a vector v with elements [5, 10, 15, 20], write Rust code to remove the last two elements and print the resulting vector. What will be printed?

hard

A. [5, 10]

B. [5, 10, 15]

C. [15, 20]

D. [5, 10, 15, 20]

Solution

1. Step 1: Remove last element twice using pop()

   Calling pop() twice removes 20 then 15, leaving [5, 10].

2. Step 2: Print the vector after removals

   Printing v now shows [5, 10].

3. Final Answer:

   [5, 10] -> Option A

4. Quick Check:

   Two pops remove last two elements [OK]

Hint: Call pop() twice to remove last two elements [OK]

Common Mistakes:

• Removing only one element instead of two
• Confusing pop() with remove(index)
• Expecting vector to be empty after two pops