Array Deletion at Middle Index in DSA C++ - Build from Scratch

Array Deletion at Middle Index

📖 Scenario: You have a list of daily temperatures recorded for a week. Sometimes, a wrong temperature is recorded in the middle of the list. You want to remove that wrong temperature from the list.

🎯 Goal: Build a program that deletes the middle element from an array of temperatures and shows the updated list.

📋 What You'll Learn

Create an array called temperatures with exactly 7 integer values: 23, 25, 22, 30, 28, 26, 24

Create an integer variable called middleIndex that stores the middle index of the array

Remove the element at middleIndex from the temperatures array by shifting elements

Print the updated temperatures array elements separated by spaces

💡 Why This Matters

🌍 Real World

Removing incorrect or unwanted data from a list is common in data cleaning and processing tasks.

💼 Career

Understanding array manipulation is fundamental for software developers working with low-level data structures and performance-critical applications.

1

Create the temperatures array

Create an integer array called temperatures with these exact values: 23, 25, 22, 30, 28, 26, 24.

DSA C++

// Create the temperatures array with 7 values
// Your code here

Hint

Use int temperatures[7] = {23, 25, 22, 30, 28, 26, 24}; to create the array.

2

Calculate the middle index

Create an integer variable called middleIndex and set it to the middle index of the temperatures array (which has 7 elements).

DSA C++

int temperatures[7] = {23, 25, 22, 30, 28, 26, 24};
// Calculate the middle index of the array
// Your code here

Hint

The middle index of an array with 7 elements is 7 / 2.

3

Delete the middle element by shifting

Use a for loop with variable i starting from middleIndex to 6 (last index) to shift elements left in the temperatures array, overwriting the middle element.

DSA C++

int temperatures[7] = {23, 25, 22, 30, 28, 26, 24};
int middleIndex = 7 / 2;
// Shift elements left to delete middle element
// Your code here

Hint

Start the loop at middleIndex and go up to index 6 (last index). Move each element one step left.

4

Print the updated array

Use a for loop with variable i from 0 to 5 to print the updated temperatures array elements separated by spaces.

DSA C++

int temperatures[7] = {23, 25, 22, 30, 28, 26, 24};
int middleIndex = 7 / 2;
for (int i = middleIndex; i < 6; i++) {
    temperatures[i] = temperatures[i + 1];
}
// Print updated array elements
// Your code here

Hint

Print elements from index 0 to 5 separated by spaces. Use std::cout.

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