Array Deletion at Beginning in DSA Rust - Execution Trace

Concept Flow - Array Deletion at Beginning

Start with array

↓

Remove first element

↓

Shift all elements left by 1

↓

Reduce array size by 1

↓

Done

This flow shows removing the first element from an array by shifting all other elements left and reducing the size.

Execution Sample

DSA Rust

let mut arr = vec![10, 20, 30, 40];
arr.remove(0);
println!("{:?}", arr);

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

Execution Table

Step	Operation	Nodes in Array	Pointer Changes	Visual State
1	Start with array	[10, 20, 30, 40]	head -> index 0	10 -> 20 -> 30 -> 40 -> null
2	Remove element at index 0	[10, 20, 30, 40]	head -> index 0 (to be removed)	10 -> 20 -> 30 -> 40 -> null
3	Shift elements left	[20, 30, 40, 40]	head -> index 0 now 20	20 -> 30 -> 40 -> 40 -> null
4	Reduce array size by 1	[20, 30, 40]	head -> index 0	20 -> 30 -> 40 -> null
5	Print final array	[20, 30, 40]	head -> index 0	20 -> 30 -> 40 -> null

💡 Array size reduced by 1, first element removed, operation complete.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	Final
arr	[10, 20, 30, 40]	[10, 20, 30, 40]	[20, 30, 40, 40]	[20, 30, 40]	[20, 30, 40]
head_index	0	0 (to remove)	0 (now 20)	0 (now 20)	0 (now 20)
size	4	4	4	3	3

Key Moments - 3 Insights

Why do we shift elements left after removing the first element?

Because arrays have fixed positions, removing the first element leaves a gap at index 0. Shifting left fills this gap so the array remains continuous, as shown in step 3 of the execution_table.

Why does the array size reduce only after shifting elements?

Shifting moves elements left but does not change the array length. The size reduces after shifting to reflect the removal, as seen in step 4 where the last duplicate element is discarded.

What happens to the last element after shifting left?

It becomes duplicated temporarily during shifting (step 3), but is removed when the array size reduces (step 4), ensuring no leftover duplicates.

Visual Quiz - 3 Questions

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

A [10, 20, 30, 40]

B [20, 30, 40, 40]

C [20, 30, 40]

D [30, 40, 40]

💡 Hint

Check the 'Nodes in Array' column at step 3 in execution_table.

At which step does the array size reduce?

A Step 2

B Step 3

C Step 4

D Step 5

💡 Hint

Look at the 'Operation' and 'Nodes in Array' columns to see when size changes.

If we remove the element at index 1 instead of 0, how would the visual state at step 3 change?

A Elements from index 2 shift left, first element stays same

B All elements shift left including first element

C No shifting happens

D Array size reduces before shifting

💡 Hint

Consider which elements move when removing index 1, referencing the concept_flow.

Concept Snapshot

Array Deletion at Beginning:
- Remove element at index 0
- Shift all elements left by one
- Reduce array size by one
- Result: array without first element
- Maintains continuous memory layout

Full Transcript

This visual trace shows how to delete the first element from an array. We start with the full array. Then we remove the element at index zero. Next, all elements shift left to fill the gap. After shifting, the array size reduces by one to remove the duplicate last element. Finally, the updated array is printed. This process keeps the array continuous and correctly sized.

Practice

1. What happens when you delete the first element from an array in Rust using remove(0)?

easy

A. The array elements remain unchanged.

B. The last element is removed instead of the first.

C. The array becomes empty regardless of its original size.

D. The first element is removed and all other elements shift left by one position.

Solution

1. Step 1: Understand remove(0) method

   The remove(0) method removes the element at index 0, which is the first element in the array.

2. Step 2: Effect on array elements

   After removal, all elements to the right shift left by one to fill the gap, reducing the array length by one.

3. Final Answer:

   The first element is removed and all other elements shift left by one position. -> Option D

4. Quick Check:

   Deleting at start shifts elements left [OK]

Hint: Deleting at index 0 shifts all elements left by one [OK]

Common Mistakes:

• Thinking last element is removed instead of first
• Assuming array size stays the same
• Believing elements do not shift after deletion

2. Which of the following Rust code snippets correctly deletes the first element from a vector v?

easy

A. v.remove(0);

B. v.delete(0);

C. v.pop_front();

D. v.shift();

Solution

1. Step 1: Identify correct Rust method for deletion at index

   Rust vectors use remove(index) to delete element at a specific position.

2. Step 2: Check other options for validity

   delete, pop_front, and shift are not valid Rust vector methods.

3. Final Answer:

   v.remove(0); -> Option A

4. Quick Check:

   Use remove(0) to delete first element [OK]

Hint: Use remove(0) to delete first element in Rust vector [OK]

Common Mistakes:

• Using non-existent methods like pop_front or shift
• Confusing Rust with other languages' syntax
• Trying to use delete which is invalid in Rust

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

let mut v = vec![10, 20, 30, 40];
v.remove(0);
println!("{:?}", v);

medium

A. [20, 30, 40]

B. [10, 20, 30, 40]

C. [30, 40]

D. [40]

Solution

1. Step 1: Initial vector elements

   The vector v starts as [10, 20, 30, 40].

2. Step 2: Remove first element

   v.remove(0) removes 10, shifting remaining elements left.

3. Step 3: Resulting vector

   After removal, v becomes [20, 30, 40].

4. Final Answer:

   [20, 30, 40] -> Option A

5. Quick Check:

   Remove first element shifts left [OK]

Hint: Removing index 0 drops first element, shifts others left [OK]

Common Mistakes:

• Expecting original vector unchanged
• Thinking multiple elements removed
• Confusing output format

4. Identify the error in this Rust code that tries to delete the first element of a vector:

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

medium

A. Cannot remove elements from a vector.

B. Missing index argument in remove method.

C. remove method does not exist for vectors.

D. Vector cannot be mutable.

Solution

1. Step 1: Check method signature

   The remove method requires an index argument specifying which element to remove.

2. Step 2: Identify missing argument

   The code calls v.remove() without any index, causing a compile error.

3. Final Answer:

   Missing index argument in remove method. -> Option B

4. Quick Check:

   remove needs index argument [OK]

Hint: Always provide index to remove method [OK]

Common Mistakes:

• Calling remove without index
• Thinking remove works like pop() without arguments
• Assuming vector immutability causes error

5. Given a vector v = vec![5, 10, 15, 20, 25], write Rust code to delete the first two elements and print the resulting vector. What will be the output?

hard

A. [5, 10, 15]

B. [10, 15, 20, 25]

C. [15, 20, 25]

D. [20, 25]

Solution

1. Step 1: Remove first element twice

   Calling v.remove(0) twice removes elements at index 0 each time, first 5 then 10.

2. Step 2: Resulting vector after removals

   After removals, vector becomes [15, 20, 25].

3. Final Answer:

   [15, 20, 25] -> Option C

4. Quick Check:

   Two removals at index 0 drop first two elements [OK]

Hint: Remove index 0 repeatedly to delete from start [OK]

Common Mistakes:

• Removing only once instead of twice
• Removing wrong indices after first removal
• Expecting original vector unchanged