Array Insertion at Middle Index in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare an integer array of size 5.

DSA C

int arr[[1]];

Drag options to blanks, or click blank then click option'

A7

B10

C3

D5

Attempts:

3 left

💡 Hint

Common Mistakes

Using a size different than 5.

Forgetting to specify the size.

The array size should be 5 to hold 5 elements.

2fill in blank

medium

Complete the code to find the middle index of the array of size 5.

DSA C

int mid = [1] / 2;

Drag options to blanks, or click blank then click option'

A6

B5

C4

D10

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong size number.

Using floating point division instead of integer.

The middle index is size divided by 2, so 5 / 2 = 2 (integer division).

3fill in blank

hard

Fix the error in the loop that shifts elements to the right before insertion.

DSA C

for (int i = n - 1; i >= [1]; i--) {
    arr[i + 1] = arr[i];
}

Drag options to blanks, or click blank then click option'

Amid

B0

Cmid + 1

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Starting loop at mid + 1 causes skipping the middle element.

Starting loop at 0 shifts all elements unnecessarily.

We start shifting from the middle index to make space for the new element.

4fill in blank

hard

Fill both blanks to insert a new value at the middle index and update the size.

DSA C

arr[[1]] = new_value;
n = [2];

Drag options to blanks, or click blank then click option'

Amid

Bn + 1

Cn - 1

Dmid + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Inserting at wrong index.

Not updating the size after insertion.

Insert at middle index and increase size by 1 after insertion.

5fill in blank

hard

Fill all three blanks to print the array elements after insertion.

DSA C

for (int i = [1]; i < [2]; i[3]) {
    printf("%d ", arr[i]);
}

Drag options to blanks, or click blank then click option'

A0

Bn

C++i

D--i

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong start or end index.

Using decrement instead of increment in loop.

Start from 0, go to n (size), and increment i to print all elements.

Practice

1. What is the correct middle index to insert an element in an array of size 6?

easy

A. 5

B. 2

C. 4

D. 3

Solution

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

   For an array of size 6, middle index is size/2 = 6/2 = 3.

2. Step 2: Confirm insertion position

   Inserting at index 3 places the element exactly in the middle.

3. Final Answer:

   3 -> Option D

4. Quick Check:

   Middle index = size/2 = 3 [OK]

Hint: Middle index = array size divided by 2 [OK]

Common Mistakes:

• Using size-1 as middle index
• Using size/2 - 1 for even size
• Confusing zero-based indexing

2. Which of the following C code snippets correctly inserts an element at the middle index of an array arr of size n?

easy

A. int mid = n / 2; for(int i = 0; i < mid; i++) arr[i] = arr[i+1]; arr[mid] = x;

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

C. int mid = n / 2; arr[mid] = x;

D. int mid = n / 2; for(int i = mid; i < n; i++) arr[i] = arr[i+1]; arr[mid] = x;

Solution

1. Step 1: Identify correct shifting direction

   To insert at middle, elements from middle to end must shift right to make space.

2. Step 2: Check code correctness

   int mid = n / 2; for(int i = n; i > mid; i--) arr[i] = arr[i-1]; arr[mid] = x; shifts elements right from end to middle, then inserts x at mid. Others shift incorrectly or overwrite.

3. Final Answer:

   int mid = n / 2; for(int i = n; i > mid; i--) arr[i] = arr[i-1]; arr[mid] = x; -> Option B

4. Quick Check:

   Shift right from end to mid, then insert [OK]

Hint: Shift elements right from end to middle before insertion [OK]

Common Mistakes:

• Shifting elements left instead of right
• Not shifting elements before insertion
• Overwriting elements without shifting

3. What will be the content of the array after inserting 99 at the middle index of arr = {10, 20, 30, 40, 50} using the correct insertion method?

medium

A. [10, 20, 30, 40, 99, 50]

B. [10, 20, 99, 30, 40, 50]

C. [10, 20, 30, 99, 40, 50]

D. [10, 20, 30, 40, 50, 99]

Solution

1. Step 1: Calculate middle index

   Array size is 5, middle index = 5/2 = 2 (integer division).

2. Step 2: Shift elements right and insert 99

   Shift elements from index 4 to 2 right by one position. Insert 99 at index 2.

3. Final Answer:

   [10, 20, 99, 30, 40, 50] -> Option C

4. Quick Check:

   Insert at index 2 shifts elements right [OK]

Hint: Middle index = size/2; shift right, then insert [OK]

Common Mistakes:

• Inserting at wrong index
• Not shifting elements before insertion
• Overwriting elements

4. Identify the error in the following code snippet that inserts an element x at the middle of array arr of size n:

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

medium

A. Loop should run backwards from n to mid

B. Index mid is incorrect for insertion

C. No need to shift elements before insertion

D. Array size n should be incremented before insertion

Solution

1. Step 1: Analyze loop direction

   The loop shifts elements from mid to n forward, but does so from low to high index, causing overwrite.

2. Step 2: Correct loop direction

   Loop must run backwards from n to mid to shift elements right safely without overwriting.

3. Final Answer:

   Loop should run backwards from n to mid -> Option A

4. Quick Check:

   Shift right must be done backwards [OK]

Hint: Shift elements right by looping backwards from end [OK]

Common Mistakes:

• Looping forward causing data overwrite
• Not updating array size after insertion
• Incorrect middle index calculation

5. Given an array arr = {5, 10, 15, 20, 25, 30} of size 6, write the final array after inserting 99 at the middle index and then inserting 77 at the new middle index of the updated array.

hard

A. [5, 10, 15, 77, 99, 20, 25, 30]

B. [5, 10, 77, 99, 15, 20, 25, 30]

C. [5, 10, 99, 15, 77, 20, 25, 30]

D. [5, 10, 15, 99, 77, 20, 25, 30]

Solution

1. Step 1: Insert 99 at middle of original array

   Original size = 6, middle index = 3. Insert 99 at index 3 shifting elements right: [5, 10, 15, 99, 20, 25, 30]

2. Step 2: Insert 77 at middle of updated array

   New size = 7, middle index = 7/2 = 3. Insert 77 at index 3 shifting elements right: [5, 10, 15, 77, 99, 20, 25, 30]

3. Final Answer:

   [5, 10, 15, 77, 99, 20, 25, 30] -> Option A

4. Quick Check:

   Insert twice at middle recalculating index each time [OK]

Hint: Recalculate middle after each insertion before next insert [OK]

Common Mistakes:

• Not recalculating middle index after first insertion
• Overwriting elements without shifting
• Inserting at wrong index second time