Array Deletion at Middle Index in DSA C# - 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.

int[] arr = {1, 2, 3, 4, 5};
int deleteIndex = arr.Length / 2; // middle index
for (int i = deleteIndex; i < arr.Length - 1; i++)
{
    arr[i] = arr[i + 1];
}
// last element is now duplicate or ignored

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop that shifts elements left after the deleted index.
• How many times: It runs from the middle index to the end, about n/2 times if 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	5 shifts
100	50 shifts
1000	500 shifts

Pattern observation: The work grows linearly with input size, about half the size.

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 of an array is always fast and constant time."

[OK] Correct: Because after deletion, elements must move to fill the gap, which takes time proportional to how many elements are 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 deleted the first element instead of the middle? How would the time complexity change?"

Practice

1. What happens when you delete the middle element from an array in C#?

easy

A. The element is removed and all elements after it shift left by one position.

B. The array size automatically decreases by one without shifting elements.

C. The middle element is replaced by zero but array size stays the same.

D. The array throws an error because arrays have fixed size.

Solution

1. Step 1: Understand array behavior on deletion

   In C#, arrays have fixed size, so to delete an element, we shift elements after it left to overwrite it.

2. Step 2: Effect on array elements

   After shifting, the last element is duplicated or ignored logically, but the middle element is effectively removed from the sequence.

3. Final Answer:

   The element is removed and all elements after it shift left by one position. -> Option A

4. Quick Check:

   Deletion shifts elements left [OK]

Hint: Deleting means shifting elements left to fill the gap [OK]

Common Mistakes:

• Thinking array size changes automatically
• Assuming deletion replaces element with zero
• Believing arrays throw error on deletion

2. Which of the following C# code snippets correctly deletes the middle element from an integer array arr of odd length?

easy

A. int mid = arr.Length / 2; arr[mid] = 0;

B. int mid = arr.Length / 2; for(int i = mid; i < arr.Length - 1; i++) arr[i] = arr[i + 1];

C. int mid = arr.Length / 2; arr.RemoveAt(mid);

D. int mid = arr.Length / 2; Array.Delete(arr, mid);

Solution

1. Step 1: Identify valid deletion method in arrays

   Arrays in C# do not have RemoveAt or Delete methods; we must shift elements manually.

2. Step 2: Check code correctness

   int mid = arr.Length / 2; for(int i = mid; i < arr.Length - 1; i++) arr[i] = arr[i + 1]; calculates middle index and shifts elements left from middle to end-1, correctly deleting the middle element.

3. Final Answer:

   int mid = arr.Length / 2; for(int i = mid; i < arr.Length - 1; i++) arr[i] = arr[i + 1]; -> Option B

4. Quick Check:

   Manual shift after mid index [OK]

Hint: Use loop to shift elements left from middle index [OK]

Common Mistakes:

• Using non-existent array methods like RemoveAt
• Just setting middle element to zero without shifting
• Trying to delete without shifting elements

3. What is the output of the following C# code?

int[] arr = {10, 20, 30, 40, 50};
int mid = arr.Length / 2;
for(int i = mid; i < arr.Length - 1; i++)
    arr[i] = arr[i + 1];
for(int i = 0; i < arr.Length - 1; i++)
    Console.Write(arr[i] + " ");

medium

A. 10 20 30 50

B. 10 20 30 40 50

C. 10 20 40 50

D. 10 20 40 50 50

Solution

1. Step 1: Calculate middle index and shift elements

   Array length is 5, mid = 2. Loop shifts arr[2] = arr[3] (40), arr[3] = arr[4] (50).

2. Step 2: Print elements except last

   Loop prints arr[0]=10, arr[1]=20, arr[2]=40, arr[3]=50. Last element ignored.

3. Final Answer:

   10 20 40 50 -> Option C

4. Quick Check:

   Shifted middle element out = 10 20 40 50 [OK]

Hint: Shift elements left, print all but last [OK]

Common Mistakes:

• Printing full array including last duplicate
• Not shifting elements correctly
• Confusing middle index calculation

4. Find the error in this code that tries to delete the middle element from an array arr of length 6:

int mid = arr.Length / 2;
for(int i = mid; i <= arr.Length - 1; i++)
    arr[i] = arr[i + 1];

medium

A. Array elements are not shifted correctly.

B. Middle index calculation is wrong for even length arrays.

C. No error; code works fine.

D. Loop condition causes index out of range error.

Solution

1. Step 1: Analyze loop range and array indexing

   Loop runs from mid to arr.Length - 1 inclusive, accessing arr[i + 1]. For i = arr.Length - 1, arr[i + 1] is out of bounds.

2. Step 2: Identify cause of error

   Accessing arr[arr.Length] causes IndexOutOfRangeException.

3. Final Answer:

   Loop condition causes index out of range error. -> Option D

4. Quick Check:

   Loop index out of bounds [OK]

Hint: Loop must stop before last index to avoid out of range [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Ignoring array bounds when accessing i+1
• Misunderstanding middle index for even length

5. You have an array arr = {5, 10, 15, 20, 25, 30, 35}. Write the steps to delete the middle element and print the resulting array with correct size handling.

hard

A. Create a list from arr; remove element at mid; convert back to array; print.

B. Calculate mid = arr.Length / 2; shift elements left from mid; print first arr.Length - 1 elements.

C. Use Array.RemoveAt(mid); print arr.

D. Set arr[mid] = 0; print arr as is.

Solution

1. Step 1: Convert array to list for flexible size

   Arrays have fixed size, so convert to List<int> to remove element easily.

2. Step 2: Remove middle element and convert back

   Remove element at mid index, then convert list back to array for printing.

3. Step 3: Print resulting array

   Print elements of new array which has one less element.

4. Final Answer:

   Create a list from arr; remove element at mid; convert back to array; print. -> Option A

5. Quick Check:

   Use List for deletion, then convert back [OK]

Hint: Use List to delete middle element easily, then convert back [OK]

Common Mistakes:

• Trying to resize arrays directly
• Only shifting elements without resizing
• Using non-existent Array.RemoveAt method