Array Deletion at Middle Index in DSA C++ - Time & Space Complexity

Time Complexity: Array Deletion at Middle Index

O(n)

Understanding Time Complexity

We want to understand how long it takes to remove an element from the middle of an array.

How does the time needed change as the array gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code removes 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 shifting elements left after the middle index.
• How many times: Approximately n/2 times, where 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 operations grow linearly with input size, about half the size.

Final Time Complexity

Time Complexity: O(n)

This means the time to delete the middle element grows in direct proportion to the array size.

Common Mistake

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

[OK] Correct: Because elements after the deleted one must be shifted left, which takes time proportional to how many elements follow.

Interview Connect

Knowing this helps you explain why arrays are not always the best choice for frequent middle deletions, showing your understanding of data structure trade-offs.

Self-Check

"What if we used a linked list instead of an array? How would the time complexity of deleting the middle element change?"

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