Why Array Deletion at Middle Index in DSA Java?

The Big Idea

What if removing one item from a list could be done perfectly every time without any mistakes or extra work?

The Scenario

Imagine you have a long list of your favorite songs written on paper. You want to remove one song from the middle. You have to erase it and then shift all the songs after it one step up to fill the empty space.

The Problem

Doing this by hand is slow and tiring. You might accidentally skip a song or write the wrong order. If the list is very long, it takes a lot of time and effort to fix the order after removing one song.

The Solution

Using array deletion at a middle index in programming automatically shifts all elements after the removed one to the left. This keeps the list neat and saves you from doing the hard work manually.

Before vs After

✗ Before

int[] songs = {1, 2, 3, 4, 5};
// Remove song at index 2 manually
// Shift elements left by one
songs[2] = songs[3];
songs[3] = songs[4];
songs[4] = 0; // empty

✓ After

int[] songs = {1, 2, 3, 4, 5};
int deleteIndex = 2;
for (int i = deleteIndex; i < songs.length - 1; i++) {
    songs[i] = songs[i + 1];
}
songs[songs.length - 1] = 0;

What It Enables

This lets programs quickly and correctly remove items from anywhere in a list, keeping data organized and ready for use.

Real Life Example

When you delete a contact from the middle of your phone's contact list, the phone automatically moves the contacts below up to fill the gap, so your list stays tidy.

Key Takeaways

Manual removal and shifting is slow and error-prone.

Array deletion automates shifting elements left after removal.

Keeps data organized and easy to manage.

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