Array Deletion at Middle Index in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What happens to the elements in an array when you delete an element at the middle index?

All elements after the deleted element shift one position to the left to fill the gap, reducing the array's effective size by one.

beginner

Why can't we physically reduce the size of a fixed-size array in Java after deletion?

Because arrays in Java have a fixed size once created; we can only logically reduce the size by ignoring the last element or creating a new smaller array.

intermediate

In Java, what is the time complexity of deleting an element at the middle index of an array?

O(n) because all elements after the deleted index must be shifted one position to the left.

beginner

Show the array state after deleting the element at index 2 from [10, 20, 30, 40, 50].

After deletion, the array elements shift left: [10, 20, 40, 50, _] (where _ is ignored or overwritten). The logical size is now 4.

intermediate

What is the main difference between deleting an element in an array vs. a linked list?

In an array, elements after the deleted one must be shifted, causing O(n) time. In a linked list, deletion is O(1) if you have the node reference, as you just change pointers.

What is the first step when deleting an element at the middle index in an array?

A Shift all elements after the index one position to the left

B Resize the array physically

C Replace the element with zero

D Swap the element with the last element

To delete an element, we shift all elements after it one position left to fill the gap.

What is the time complexity of deleting an element at the middle index in a fixed-size array?

A O(n)

B O(1)

C O(log n)

D O(n^2)

Because shifting elements requires moving up to n-1 elements, the time complexity is O(n).

After deleting an element at index 3 in an array of size 6, what is the new logical size?

A 6

B 3

C 4

D 5

Deleting one element reduces the logical size by one, so from 6 to 5.

In Java, can you reduce the physical size of an array after deletion?

A Yes, arrays resize automatically

B No, arrays have fixed size

C Yes, by calling array.resize()

D Only if the array is empty

Java arrays have fixed size; you cannot resize them after creation.

Which data structure allows faster deletion at the middle without shifting elements?

A Array

B Stack

C Linked List

D Queue

Linked lists allow O(1) deletion if you have the node reference, without shifting elements.

Explain step-by-step how to delete an element at the middle index of an array and what happens to the other elements.

Think about how to fill the gap left by the deleted element.

Compare deleting an element at the middle index in an array versus a linked list in terms of process and efficiency.

Focus on how elements are stored and accessed.

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