Array Insertion at Middle Index in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output after inserting element in the middle of an array

What is the output of the following C code after inserting the value 99 at the middle index of the array?

DSA C

int main() {
    int arr[6] = {10, 20, 30, 40, 50};
    int n = 5;
    int val = 99;
    int mid = n / 2;
    for (int i = n; i > mid; i--) {
        arr[i] = arr[i - 1];
    }
    arr[mid] = val;
    n++;
    for (int i = 0; i < n; i++) {
        printf("%d ", arr[i]);
    }
    return 0;
}

A 10 20 30 40 99 50

B 10 20 30 99 40 50

C 10 20 99 30 40 50

D 10 20 30 40 50 99

💡 Hint

Remember that integer division truncates, so middle index is n/2 = 2 for n=5.

The middle index for an array of size 5 is 2 (0-based). The code shifts elements from the end to the middle index to the right by one position, then inserts 99 at index 2. The resulting array is: 10 20 99 30 40 50.

🧠 Conceptual

intermediate

Understanding middle index calculation for array insertion

Given an array of length n, which formula correctly calculates the middle index to insert a new element so that the array remains balanced?

A middle = (n + 1) / 2

B middle = (n - 1) / 2

C middle = n - 1

D middle = n / 2

💡 Hint

Consider zero-based indexing and integer division truncation.

For zero-based arrays, the middle index is calculated as n / 2 using integer division. This places the new element roughly in the center, balancing elements on both sides.

🔧 Debug

advanced

Identify the error in array insertion code

What error will occur when running this code snippet that inserts an element at the middle of an array of size 5? int arr[5] = {1, 2, 3, 4, 5}; int n = 5; int val = 10; int mid = n / 2; for (int i = n; i > mid; i--) { arr[i] = arr[i - 1]; } arr[mid] = val; n++;

A Buffer overflow (writing outside array bounds)

B No error, runs correctly

C Segmentation fault due to null pointer

D Compilation error due to syntax

💡 Hint

Check the array size and the index used in the loop.

The array size is 5, but the loop writes to arr[5], which is outside the allocated memory, causing buffer overflow.

❓ Predict Output

advanced

Output after multiple insertions at middle index

What is the output after inserting values 99 and then 77 at the middle index of the array?

DSA C

int main() {
    int arr[8] = {1, 2, 3, 4, 5, 6};
    int n = 6;
    int val1 = 99;
    int val2 = 77;
    int mid = n / 2;
    // Insert 99
    for (int i = n; i > mid; i--) {
        arr[i] = arr[i - 1];
    }
    arr[mid] = val1;
    n++;
    // Insert 77
    mid = n / 2;
    for (int i = n; i > mid; i--) {
        arr[i] = arr[i - 1];
    }
    arr[mid] = val2;
    n++;
    for (int i = 0; i < n; i++) {
        printf("%d ", arr[i]);
    }
    return 0;
}

A 1 2 3 77 99 4 5 6

B 1 2 3 99 77 4 5 6

C 1 2 3 99 4 77 5 6

D 1 2 3 4 77 99 5 6

💡 Hint

Remember to recalculate middle index after first insertion.

After first insertion, array is: 1 2 3 99 4 5 6 New size n=7, middle index = 7/2=3 Second insertion shifts elements from index 3, inserts 77 at index 3. Final array: 1 2 3 77 99 4 5 6

🧠 Conceptual

expert

Time complexity of inserting an element at the middle of an array

What is the time complexity of inserting an element at the middle index of an array of size n, assuming shifting elements is required?

A O(1)

B O(n)

C O(log n)

D O(n^2)

💡 Hint

Consider how many elements need to be moved to make space.

Inserting at the middle requires shifting about half the elements, which is proportional to n, so time complexity is O(n).

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