Array Deletion at Middle Index in DSA Java - 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 we 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];
}
arr[arr.length - 1] = 0; // optional: clear last element
    

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

Identify Repeating Operations

• 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 half the array size.

How Execution Grows With Input

As the array gets bigger, the number of elements to shift grows roughly in proportion to the size of the array.

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 if the array doubles in size, the time to delete the middle element roughly doubles too.

Common Mistake

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

[OK] Correct: Because after deleting, elements must be shifted to fill the gap, which takes time proportional to how many elements are after the deleted one.

Interview Connect

Understanding how array operations scale helps you explain trade-offs in data structures and choose the right one for your needs.

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 of size 5?
int[] arr = {10, 20, 30, 40, 50};
Delete element at index 2 (middle). What is the resulting array?

easy

A. [10, 20, 40, 50]

B. [10, 20, 30, 40, 50]

C. [10, 20, 30, 50]

D. [20, 30, 40, 50]

Solution

1. Step 1: Identify the middle index and element

   The array size is 5, so middle index is 2 (0-based). Element at index 2 is 30.

2. Step 2: Remove the element and shift remaining

   Remove 30, then shift elements after index 2 left: 40 moves to index 2, 50 moves to index 3.

3. Final Answer:

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

4. Quick Check:

   Middle element 30 removed, array shrinks by one [OK]

Hint: Remove middle, shift right elements left by one [OK]

Common Mistakes:

• Not shifting elements after deletion
• Deleting wrong index (not middle)
• Leaving array size unchanged

2. Which of the following Java code snippets correctly deletes the middle element from an array arr of odd length n?

easy

A. for (int i = n - 1; i > n/2; i--) arr[i] = arr[i - 1];

B. for (int i = 0; i < n/2; i++) arr[i] = arr[i + 1];

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

D. for (int i = 0; i < n; i++) arr[i] = arr[i + 1];

Solution

1. Step 1: Understand the deletion index

   Middle index for odd length n is n/2 (integer division). We must shift elements after this index left.

2. Step 2: Check loop correctness

   Loop from n/2 to n-2 shifts elements left by copying arr[i+1] to arr[i]. This correctly removes middle element.

3. Final Answer:

   for (int i = n/2; i < n - 1; i++) arr[i] = arr[i + 1]; -> Option C

4. Quick Check:

   Shift elements after middle left [OK]

Hint: Shift elements from middle index to end left by one [OK]

Common Mistakes:

• Looping from 0 instead of middle index
• Shifting elements right instead of left
• Loop running out of array bounds

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

int[] arr = {5, 10, 15, 20, 25, 30, 35};
int n = arr.length;
int mid = n / 2;
for (int i = mid; i < n - 1; i++) {
    arr[i] = arr[i + 1];
}
n--;
for (int i = 0; i < n; i++) {
    System.out.print(arr[i] + " ");
}

medium

A. 5 10 15 20 25 35

B. 5 10 15 25 30 35

C. 5 10 15 20 25 30 35

D. 5 10 15 20 25 30

Solution

1. Step 1: Identify middle index and element

   Array length n=7, middle index mid=3, element at index 3 is 20.

2. Step 2: Shift elements left from middle

   Loop shifts elements from index 3 to 5 left by one: arr[3]=arr[4]=25, arr[4]=arr[5]=30, arr[5]=arr[6]=35.

3. Step 3: Decrease size and print

   n is decremented to 6, so print first 6 elements: 5 10 15 25 30 35.

4. Final Answer:

   5 10 15 25 30 35 -> Option B

5. Quick Check:

   Middle element 20 removed, array shifted left [OK]

Hint: Shift elements after middle left, reduce size by one [OK]

Common Mistakes:

• Printing full original array length
• Not decrementing array size after deletion
• Confusing middle index calculation

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

int mid = n / 2;
for (int i = mid; i < n; i++) {
    arr[i] = arr[i + 1];
}
n--;

medium

A. Loop should start from 0, not mid

B. Middle index calculation is wrong for odd length arrays

C. Array size n should not be decremented

D. Loop runs out of array bounds accessing arr[i+1] when i = n-1

Solution

1. Step 1: Analyze loop range and array access

   Loop runs from i = mid to i = n-1. At i = n-1, arr[i+1] = arr[n] is accessed, which is out of bounds.

2. Step 2: Correct loop boundary

   Loop should run until i < n-1 to avoid out-of-bounds access when shifting elements left.

3. Final Answer:

   Loop runs out of array bounds accessing arr[i+1] when i = n-1 -> Option D

4. Quick Check:

   Array index out of bounds error [OK]

Hint: Loop till n-1, not n, to avoid arr[i+1] error [OK]

Common Mistakes:

• Using wrong loop end condition
• Ignoring array index limits
• Incorrect middle index for even length arrays

5. Given an array arr = {2, 4, 6, 8, 10, 12} of even length 6, how do you delete the middle element(s) to keep the array balanced? Choose the correct resulting array after deleting the middle element(s).

hard

A. [2, 4, 8, 10, 12]

B. [2, 4, 6, 10, 12]

C. [2, 4, 8, 12]

D. [2, 4, 6, 8, 10]

Solution

1. Step 1: Identify middle elements for even length

   For length 6, middle indices are 2 and 3 (elements 6 and 8).

2. Step 2: Delete one or both middle elements to keep balance

   Usually, delete one middle element to keep array balanced. Deleting element at index 2 (6) shifts elements left.

3. Step 3: Resulting array after deleting element at index 2

   Array becomes [2, 4, 8, 10, 12] after shifting elements left and reducing size by one.

4. Final Answer:

   [2, 4, 8, 10, 12] -> Option A

5. Quick Check:

   Delete one middle element, shift left, size reduces [OK]

Hint: For even length, delete lower middle index element [OK]

Common Mistakes:

• Deleting both middle elements instead of one
• Deleting wrong middle index
• Not shifting elements after deletion