Array Deletion at Middle Index in DSA Rust - Build from Scratch

Array Deletion at Middle Index

📖 Scenario: You are managing a list of daily tasks. Sometimes, you need to remove the task that is exactly in the middle of your list to keep your schedule balanced.

🎯 Goal: Build a Rust program that creates a list of tasks, finds the middle task, removes it, and then shows the updated list.

📋 What You'll Learn

Create a vector called tasks with these exact strings: "Email", "Meeting", "Coding", "Review", "Lunch"

Create a variable called middle_index that holds the middle index of the tasks vector

Remove the task at middle_index from the tasks vector

Print the updated tasks vector

💡 Why This Matters

🌍 Real World

Managing lists of tasks, events, or items often requires removing elements from the middle to keep balance or update priorities.

💼 Career

Understanding how to manipulate arrays or lists by removing elements at specific positions is a common task in software development, especially in user interface programming and data processing.

1

Create the initial list of tasks

Create a vector called tasks with these exact strings: "Email", "Meeting", "Coding", "Review", "Lunch"

DSA Rust

# Create a vector called tasks with the given strings
# Your code here

Hint

Use let mut tasks = vec![...] to create a mutable vector with the exact strings.

2

Find the middle index

Create a variable called middle_index that holds the middle index of the tasks vector using integer division

DSA Rust

fn main() {
    let mut tasks = vec!["Email", "Meeting", "Coding", "Review", "Lunch"];
    
    // Calculate the middle index of tasks
    // Your code here
}

Hint

Use tasks.len() to get the length and divide by 2 to find the middle index.

3

Remove the middle task

Remove the task at middle_index from the tasks vector using the remove method

DSA Rust

fn main() {
    let mut tasks = vec!["Email", "Meeting", "Coding", "Review", "Lunch"];
    let middle_index = tasks.len() / 2;
    
    // Remove the middle task
    // Your code here
}

Hint

Use tasks.remove(middle_index) to delete the middle element.

4

Print the updated list

Print the tasks vector using println! to show the updated list after removal

DSA Rust

fn main() {
    let mut tasks = vec!["Email", "Meeting", "Coding", "Review", "Lunch"];
    let middle_index = tasks.len() / 2;
    tasks.remove(middle_index);
    
    // Print the updated tasks vector
    // Your code here
}

Hint

Use println!("{:?}", tasks); to print the vector in debug format.

Practice

1. What happens when you delete the middle element from an array with an odd number of elements?

easy

A. The array size remains the same but the middle element is set to zero.

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

C. The last element is removed instead of the middle element.

D. The array is reversed after deletion.

Solution

1. Step 1: Identify the middle index in an odd-length array

   The middle index is calculated as length divided by 2 (integer division).

2. Step 2: Remove the element at the middle index

   Deleting this element reduces the array size by one and shifts elements after it left.

3. Final Answer:

   The middle element is removed and the array size decreases by one. -> Option B

4. Quick Check:

   Deleting middle element shrinks array [OK]

Hint: Middle index = length / 2 for odd arrays [OK]

Common Mistakes:

• Thinking array size stays the same after deletion
• Removing last element instead of middle
• Not shifting elements after deletion

2. Which Rust code snippet correctly deletes the middle element from a vector v with odd length?

easy

A. v.pop(v.len() / 2);

B. v.delete(v.len() / 2);

C. v.remove(v.len() / 2);

D. v.erase(v.len() / 2);

Solution

1. Step 1: Identify Rust method to remove element by index

   Rust's Vec type uses remove(index) to delete element at given index.

2. Step 2: Calculate middle index and call remove

   Middle index is v.len() / 2. Calling v.remove(v.len() / 2) deletes middle element.

3. Final Answer:

   v.remove(v.len() / 2); -> Option C

4. Quick Check:

   Use remove(index) to delete element [OK]

Hint: Use Vec::remove(index) to delete element [OK]

Common Mistakes:

• Using non-existent methods like delete or erase
• Using pop which removes last element only
• Passing index to pop which takes no arguments

3. What is the output of this Rust code?

let mut v = vec![10, 20, 30, 40, 50];
v.remove(v.len() / 2);
println!("{:?}", v);

medium

A. [10, 30, 40, 50]

B. [10, 20, 30, 40]

C. [20, 30, 40, 50]

D. [10, 20, 40, 50]

Solution

1. Step 1: Calculate middle index of vector

   Vector length is 5, so middle index = 5 / 2 = 2.

2. Step 2: Remove element at index 2 and print vector

   Element at index 2 is 30, removed. Remaining vector is [10, 20, 40, 50].

3. Final Answer:

   [10, 20, 40, 50] -> Option D

4. Quick Check:

   Removed middle element 30, vector shrinks [OK]

Hint: Remove element at len()/2 index for middle deletion [OK]

Common Mistakes:

• Removing wrong index element
• Confusing index with value
• Expecting vector size unchanged

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

let mut v = vec![1, 2];
v.remove(v.len() / 2 + 1);
println!("{:?}", v);

medium

A. Index used for removal is off by one, causing panic.

B. Vector cannot be mutable for removal.

C. remove() method does not exist for vectors.

D. The println! macro is used incorrectly.

Solution

1. Step 1: Calculate middle index for even-length vector

   Vector length is 2, middle index should be 2 / 2 = 1.

2. Step 2: Check removal index used in code

   Code uses index 1 + 1 = 2, which is out of bounds.

3. Step 3: Understand consequences

   Index 2 >= length 2 causes runtime panic (out of bounds).

4. Final Answer:

   Index used for removal is off by one, causing panic. -> Option A

5. Quick Check:

   Use correct middle index = len()/2 [OK]

Hint: Middle index is len()/2, no +1 for removal [OK]

Common Mistakes:

• Adding +1 to middle index incorrectly
• Trying to remove from immutable vector
• Misusing remove method name

5. Given a vector v of even length, how can you delete the lower middle element (the one at index len()/2 - 1) and then print the updated vector in Rust?

hard

A. v.remove(v.len() / 2 - 1); println!("{:?}", v);

B. v.remove(v.len() / 2); println!("{:?}", v);

C. v.pop(); println!("{:?}", v);

D. v.remove(v.len() / 2 + 1); println!("{:?}", v);

Solution

1. Step 1: Identify lower middle index for even-length vector

   For even length, lower middle index is len()/2 - 1.

2. Step 2: Remove element at lower middle index and print vector

   Use v.remove(v.len() / 2 - 1) to delete lower middle element, then print with println!("{:?}", v);.

3. Final Answer:

   v.remove(v.len() / 2 - 1); println!("{:?}", v); -> Option A

4. Quick Check:

   Remove lower middle with len()/2 - 1 [OK]

Hint: For even length, lower middle = len()/2 - 1 [OK]

Common Mistakes:

• Removing upper middle instead of lower middle
• Using pop which removes last element
• Adding +1 to index incorrectly