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

Time Complexity: Array Deletion at Middle Index

O(n)

Understanding Time Complexity

When we delete an element from the middle of an array, the time it takes depends on how many elements need to move.

We want to know how the work grows as the array gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

def delete_middle(arr):
    mid = len(arr) // 2
    for i in range(mid, len(arr) - 1):
        arr[i] = arr[i + 1]
    arr.pop()

# Deletes the middle element by shifting elements left

This code removes the middle element by shifting all elements after it one step to the left.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

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

How Execution Grows With Input

As the array grows, the number of shifts grows roughly in half.

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 in a straight line as the array gets bigger.

Common Mistake

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

[OK] Correct: Because after deletion, elements must move to fill the gap, which takes time proportional to how many elements are shifted.

Interview Connect

Understanding how array operations scale helps you explain trade-offs in data structures clearly and confidently.

Self-Check

"What if we deleted the last element instead of the middle? How would the time complexity change?"

Practice

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

easy

A. The element at the exact middle index is removed, and the array size decreases by one.

B. The first element is removed instead of the middle one.

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

D. The last element is removed instead of the middle one.

Solution

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

   For an array with odd length n, the middle index is n//2 (integer division).

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

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

3. Final Answer:

   The element at the exact middle index is removed, and the array size decreases by one. -> Option A

4. Quick Check:

   Middle element removed = The element at the exact middle index is removed, and the array size decreases by one. [OK]

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

Common Mistakes:

• Removing first or last element instead of middle
• Not reducing array size after deletion
• Confusing middle index calculation

2. Which of the following Python statements correctly deletes the middle element from a list named arr with odd length?

easy

A. arr.pop(len(arr))

B. del arr[len(arr)//2]

C. arr.remove(len(arr)//2)

D. arr.delete(len(arr)//2)

Solution

1. Step 1: Understand Python deletion syntax

   The del statement removes an element at a specific index from a list.

2. Step 2: Calculate middle index and delete

   Using len(arr)//2 gives the middle index for odd length. So del arr[len(arr)//2] deletes the middle element correctly.

3. Final Answer:

   del arr[len(arr)//2] -> Option B

4. Quick Check:

   Correct syntax for deletion = del arr[len(arr)//2] [OK]

Hint: Use del with index: del arr[len(arr)//2] [OK]

Common Mistakes:

• Using remove() which deletes by value, not index
• Using pop() with wrong index
• Using non-existent delete() method

3. What is the output of the following code?

arr = [10, 20, 30, 40, 50]
del arr[len(arr)//2]
print(arr)

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

   Length of arr is 5, so middle index = 5 // 2 = 2.

2. Step 2: Delete element at index 2 and print

   Element at index 2 is 30. After deletion, arr becomes [10, 20, 40, 50].

3. Final Answer:

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

4. Quick Check:

   Deleted arr[2] = 30, remaining = [10, 20, 40, 50] [OK]

Hint: Middle index 2 in 5 elements; delete arr[2] [OK]

Common Mistakes:

• Deleting wrong index (like 3 or 1)
• Confusing element value with index
• Printing before deletion

4. The following code is intended to delete the middle element from an array arr. What is the error?

arr = [1, 2, 3, 4, 5]
arr.remove(len(arr)//2)
print(arr)

medium

A. It deletes the last element instead of the middle one.

B. It causes an IndexError because remove expects an index.

C. It deletes the element with value 2 instead of the middle index element.

D. There is no error; it deletes the middle element correctly.

Solution

1. Step 1: Understand remove() method behavior

   remove(x) deletes the first element with value x, not at index x.

2. Step 2: Analyze the argument passed to remove()

   len(arr)//2 = 5//2 = 2, so arr.remove(2) deletes element with value 2, not element at index 2.

3. Final Answer:

   It deletes the element with value 2 instead of the middle index element. -> Option C

4. Quick Check:

   remove deletes by value, not index = It deletes the element with value 2 instead of the middle index element. [OK]

Hint: remove() deletes by value, del deletes by index [OK]

Common Mistakes:

• Confusing remove() with del or pop()
• Expecting remove() to delete by index
• Not calculating middle index correctly

5. Given an array arr = [5, 10, 15, 20, 25, 30] with even length, which code correctly deletes the middle element(s) so that after deletion the array is [5, 10, 25, 30]?

hard

A. del arr[len(arr)//2 - 1]; del arr[len(arr)//2]

B. del arr[len(arr)//2]; del arr[len(arr)//2 - 1]

C. arr.pop(len(arr)//2); arr.pop(len(arr)//2 - 1)

D. arr.remove(len(arr)//2); arr.remove(len(arr)//2 - 1)

Solution

1. Step 1: Identify middle indices for even length array

   Length is 6, middle indices are 2 and 3 (0-based), values 15 and 20.

2. Step 2: Delete elements at indices 2 and 3 accounting for length change

   del arr[len(arr)//2 - 1] deletes index 2 (15), array becomes [5, 10, 20, 25, 30]. Then del arr[len(arr)//2] deletes new index 2 (20), resulting in [5, 10, 25, 30].

3. Final Answer:

   del arr[len(arr)//2 - 1]; del arr[len(arr)//2] -> Option A

4. Quick Check:

   del arr[2] then del arr[2] (after shift) = [5, 10, 25, 30] [OK]

Hint: For even length, delete lower middle index first then the new middle index [OK]

Common Mistakes:

• Deleting higher index first without adjusting for length change
• Using pop() in wrong order causing incorrect shifts
• Using remove() which deletes by value