Array Insertion at Beginning in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to shift elements right before inserting at the beginning.

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'

A1

B0

Cn

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting the loop from 1 instead of 0 causes missing the first element.

Using n instead of 0 causes out-of-bound errors.

We shift elements starting from the last index down to 0 to make space at the beginning.

2fill in blank

medium

Complete the code to insert the new element at the beginning of the array.

DSA C

arr[[1]] = new_element;

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

An

B1

C0

Dn-1

Attempts:

3 left

💡 Hint

Common Mistakes

Inserting at index n or n-1 overwrites the last element.

Inserting at index 1 misses the true beginning.

The new element is inserted at index 0, the beginning of the array.

3fill in blank

hard

Fix the error in the loop condition to correctly shift elements for insertion.

DSA C

for(int i = n - 1; i [1] 0; i--) {
    arr[i + 1] = arr[i];
}

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

A>

B<

C<=

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' causes the loop to never run.

Using '<=' causes incorrect shifting direction.

The loop must run while i is greater than or equal to 0 to shift all elements.

4fill in blank

hard

Fill both blanks to correctly shift elements and insert at the beginning.

DSA C

for(int i = n - 1; i [1] [2]; i--) {
    arr[i + 1] = arr[i];
}

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

A>=

B>

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' and 1 misses shifting the first element.

Using '<' or '<=' causes wrong loop direction.

The loop runs while i is greater than or equal to 0 to shift all elements starting from the last index.

5fill in blank

hard

Fill all three blanks to complete the insertion function at the beginning of the array.

DSA C

void insertAtBeginning(int arr[], int n, int new_element) {
    for(int i = n - 1; i [1] [2]; i--) {
        arr[i + 1] = arr[i];
    }
    arr[[3]] = new_element;
}

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

A>=

B0

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' without '=' misses shifting the first element.

Inserting at index other than 0 overwrites wrong element.

The loop shifts elements while i >= 0, and the new element is inserted at index 0.

Practice

1. What happens when you insert an element at the beginning of an array in C?

easy

A. The new element replaces the first element without shifting others.

B. All existing elements shift one position to the right to make space.

C. The array size automatically increases without shifting elements.

D. The last element is removed to insert the new element at the start.

Solution

1. Step 1: Understand array memory layout

   Arrays store elements in continuous memory. To insert at the start, existing elements must move to free space.

2. Step 2: Shift elements to the right

   Each element moves one position right to open index 0 for the new element.

3. Final Answer:

   All existing elements shift one position to the right to make space. -> Option B

4. Quick Check:

   Insert at start = shift right [OK]

Hint: Insert at start means shift all elements right [OK]

Common Mistakes:

• Assuming new element overwrites first element
• Thinking array size grows automatically
• Removing last element without reason

2. Which of the following is the correct syntax to insert an element at the beginning of an array in C (assuming space is available)?

easy

A. arr[0] = new_element; for(int i = 1; i <= n; i++) arr[i] = arr[i-1];

B. for(int i = 0; i < n; i++) arr[i+1] = arr[i]; arr[0] = new_element;

C. for(int i = n; i > 0; i--) arr[i] = arr[i-1]; arr[0] = new_element;

D. arr[n] = new_element; for(int i = n-1; i >= 0; i--) arr[i] = arr[i+1];

Solution

1. Step 1: Identify correct loop direction for shifting

   To avoid overwriting, shift elements from end to start (i = n down to 1).

2. Step 2: Assign new element at index 0

   After shifting, place new_element at arr[0].

3. Final Answer:

   for(int i = n; i > 0; i--) arr[i] = arr[i-1]; arr[0] = new_element; -> Option C

4. Quick Check:

   Shift right from end, then insert [OK]

Hint: Shift from end to start to avoid overwrite [OK]

Common Mistakes:

• Shifting from start to end causing overwrite
• Assigning new element before shifting
• Incorrect loop boundaries

3. What will be the content of the array after executing this code?

int arr[5] = {10, 20, 30, 40, 0};
int n = 4;
int new_element = 5;
for(int i = n; i > 0; i--) {
  arr[i] = arr[i-1];
}
arr[0] = new_element;

medium

A. [0, 10, 20, 30, 40]

B. [10, 5, 20, 30, 40]

C. [5, 10, 20, 30, 0]

D. [5, 10, 20, 30, 40]

Solution

1. Step 1: Shift elements right from index 4 to 1

   arr[4]=arr[3]=40, arr[3]=arr[2]=30, arr[2]=arr[1]=20, arr[1]=arr[0]=10

2. Step 2: Insert new_element at arr[0]

   arr[0] = 5

3. Final Answer:

   [5, 10, 20, 30, 40] -> Option D

4. Quick Check:

   Shift right + insert 5 at start = [5,10,20,30,40] [OK]

Hint: Trace shifts from end, then insert new element at start [OK]

Common Mistakes:

• Not shifting last element correctly
• Overwriting elements during shift
• Forgetting to insert new element at index 0

4. Identify the error in this code snippet that tries to insert an element at the beginning of an array:

int arr[5] = {1, 2, 3, 4, 0};
int n = 4;
int new_element = 0;
for(int i = 0; i < n; i++) {
  arr[i+1] = arr[i];
}
arr[0] = new_element;

medium

A. Loop direction causes overwriting of elements during shift.

B. Array size is too small to insert new element.

C. New element is assigned after shifting elements.

D. Loop condition should be i <= n instead of i < n.

Solution

1. Step 1: Analyze loop direction and assignments

   Loop shifts elements from start to end, causing arr[i+1] to overwrite arr[i+1] before it is copied.

2. Step 2: Correct approach is to shift from end to start

   Shifting from end to start prevents overwriting original values during shift.

3. Final Answer:

   Loop direction causes overwriting of elements during shift. -> Option A

4. Quick Check:

   Shift must go backward to avoid overwrite [OK]

Hint: Shift elements backward to avoid overwriting [OK]

Common Mistakes:

• Shifting forward overwrites data
• Ignoring array size limits
• Incorrect loop boundaries

5. Given an array int arr[7] = {2, 4, 6, 8, 10, 0, 0}; with n = 5 elements, write the steps to insert 1 at the beginning and then 0 at the beginning again. What will be the final array content?

hard

A. [0, 1, 2, 4, 6, 8, 10]

B. [0, 1, 2, 4, 6, 8]

C. [0, 1, 2, 4, 6, 8, 10, 0]

D. [1, 0, 2, 4, 6, 8]

Solution

1. Step 1: Insert 1 at beginning

   Shift elements right: arr[5]=arr[4]=10, arr[4]=arr[3]=8, arr[3]=arr[2]=6, arr[2]=arr[1]=4, arr[1]=arr[0]=2; then arr[0]=1. Array: [1, 2, 4, 6, 8, 10]

2. Step 2: Insert 0 at beginning

   Shift elements right again: arr[6]=arr[5]=10, arr[5]=arr[4]=8, arr[4]=arr[3]=6, arr[3]=arr[2]=4, arr[2]=arr[1]=2, arr[1]=arr[0]=1; then arr[0]=0. Array: [0, 1, 2, 4, 6, 8, 10]

3. Final Answer:

   [0, 1, 2, 4, 6, 8, 10] -> Option A

4. Quick Check:

   Two shifts + inserts = [0,1,2,4,6,8,10] [OK]

Hint: Shift right twice, insert each new element at index 0 [OK]

Common Mistakes:

• Not updating array size after insertions
• Overwriting elements during second insertion
• Assuming array can hold more than 6 elements