Why Array Deletion at Middle Index in DSA C++?

The Big Idea

What if removing one item could magically rearrange your whole list perfectly every time?

The Scenario

Imagine you have a list of your favorite songs written on paper. You want to remove the song in the middle, but to keep the list neat, you have to rewrite all the songs after it one position up.

The Problem

Doing this by hand is slow and tiring. You might skip a song or write the wrong order. It takes time to shift everything up, and mistakes can happen easily.

The Solution

Using an array deletion method in programming, the computer automatically shifts all elements after the removed one to fill the gap. This saves time and avoids errors, keeping the list organized instantly.

Before vs After

✗ Before

for (int i = index; i < size - 1; i++) {
    arr[i] = arr[i + 1];
}
size--;

✓ After

deleteAtIndex(arr, size, index); // shifts elements and updates size

What It Enables

This lets you quickly and safely remove any item from the middle of a list without messing up the order.

Real Life Example

When you delete a contact from the middle of your phone's contact list, the phone automatically moves the following contacts up to fill the empty spot.

Key Takeaways

Manual shifting is slow and error-prone.

Array deletion automates shifting elements after removal.

Keeps data organized and consistent after deletion.

Practice

1. What happens when you delete the middle element from an array of size 5?
int arr[5] = {10, 20, 30, 40, 50};
Delete the element at index 2 (middle). What is the new array?

easy

A. [20, 30, 40, 50]

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

C. [10, 20, 30, 50]

D. [10, 20, 40, 50]

Solution

1. Step 1: Identify the middle index and element

   The array size is 5, so middle index is 2. The element at index 2 is 30.

2. Step 2: Delete the middle element and shift remaining elements

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

3. Final Answer:

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

4. Quick Check:

   Deleted middle element 30, shifted left [OK]

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

Common Mistakes:

• Not shifting elements after deletion
• Deleting wrong index
• Leaving array size unchanged

2. Which of the following C++ code snippets correctly deletes the middle element from an array arr of size n?

easy

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

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

C. arr[n/2] = 0; n--;

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

Solution

1. Step 1: Understand deletion logic

   To delete middle element at index n/2, shift all elements after it one position left.

2. Step 2: Check each code snippet

   for(int i = n/2; i < n-1; i++) arr[i] = arr[i+1]; n--; correctly shifts elements from middle to end left by one and reduces size. Others either shift wrong elements or just set zero without shifting.

3. Final Answer:

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

4. Quick Check:

   Shift elements after middle left [OK]

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

Common Mistakes:

• Shifting elements before middle index
• Not reducing array size
• Assigning zero instead of shifting

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

int arr[6] = {5, 10, 15, 20, 25, 30};
int n = 6;
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++) {
  std::cout << arr[i] << " ";
}

medium

A. 5 10 15 25 30

B. 5 10 15 20 25 30

C. 5 10 15 20 30

D. 5 10 20 25 30

Solution

1. Step 1: Calculate middle index and delete element

   n=6, mid=3. Element at index 3 is 20. We delete 20 by shifting elements after index 3 left.

2. Step 2: Shift elements and print array

   Shift arr[3]=arr[4]=25, arr[4]=arr[5]=30. New array: 5,10,15,25,30. Size reduced to 5.

3. Final Answer:

   5 10 15 25 30 -> Option A

4. Quick Check:

   Deleted 20, shifted left [OK]

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

Common Mistakes:

• Not reducing array size
• Printing old array size
• Not shifting elements correctly

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

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

medium

A. Array size n should not be decremented

B. Middle index calculation is wrong

C. Loop condition should be i < n - 1, not i <= n - 1

D. Loop should start from 0, not mid

Solution

1. Step 1: Analyze loop range for shifting elements

   The loop runs from mid to n-1 inclusive, so arr[i+1] accesses arr[n], which is out of bounds.

2. Step 2: Correct loop condition

   Loop should run while i < n-1 to avoid accessing beyond array size.

3. Final Answer:

   Loop condition should be i < n - 1, not i <= n - 1 -> Option C

4. Quick Check:

   Prevent out-of-bounds by correct loop limit [OK]

Hint: Loop till one before last index to avoid overflow [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Not decrementing array size
• Wrong middle index calculation

5. Given an array arr of size n, write the logic to delete the middle element and return the new array size. What will be the size after deletion if n is even?
Choose the correct approach.

hard

A. Delete element at index (n-1)/2, shift elements left, new size = n - 1

B. Delete element at index n/2, shift elements left, new size = n - 1

C. Delete element at index n/2, no shifting needed, new size = n - 1

D. Delete element at index n/2 - 1, shift elements left, new size = n - 1

Solution

1. Step 1: Understand middle index for even-sized array

   In standard array operations, middle index is n/2 (integer division), consistent with odd sizes.

2. Step 2: Delete element at n/2 and shift elements left

   Remove element at index n/2, shift all elements after it left by one, reduce size by 1.

3. Final Answer:

   Delete element at index n/2, shift elements left, new size = n - 1 -> Option B

4. Quick Check:

   Middle index = n/2 for even n [OK]

Hint: Use n/2 as middle index regardless of parity [OK]

Common Mistakes:

• Deleting lower middle index (n/2 - 1) for even n
• Not shifting elements after deletion
• Not reducing array size