Array Deletion at Middle Index in DSA Rust - Time & Space Complexity

Time Complexity: Array Deletion at Middle Index

O(n)

Understanding Time Complexity

We want to understand how long it takes to remove an element from the middle of an array.

Specifically, how the work grows as the array gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn delete_middle(arr: &mut Vec<i32>) {
    let mid = arr.len() / 2;
    for i in mid..arr.len() - 1 {
        arr[i] = arr[i + 1];
    }
    arr.pop();
}
    

This code removes the middle element by shifting all elements after it one step left, then removes the last duplicate.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop shifting elements left after the middle index.
• How many times: It runs roughly n/2 times, where n is the array size.

How Execution Grows With Input

As the array size grows, the number of shifts grows roughly half as fast.

Input Size (n)	Approx. Operations
10	About 5 shifts
100	About 50 shifts
1000	About 500 shifts

Pattern observation: The work grows linearly with the size of the array.

Final Time Complexity

Time Complexity: O(n)

This means the time to delete grows directly with the number of elements in the array.

Common Mistake

[X] Wrong: "Deleting an element from the middle of an array is always fast and constant time."

[OK] Correct: Because after removing the element, all later elements must move left, which takes time proportional to the number of elements after the deleted one.

Interview Connect

Understanding this helps you explain why arrays are not always the best choice for frequent middle deletions, showing your grasp of data structure trade-offs.

Self-Check

"What if we changed the array to a linked list? How would the time complexity of deleting the middle element change?"

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