Array Insertion at Beginning in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to insert an element at the beginning of the array by shifting elements right.

DSA C++

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

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

A1

B0

Cn

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting the loop at i > 1 instead of i > 0 causes the first element not to shift.

Using i >= 0 causes out-of-bound errors.

We start shifting elements from the end (index n) down to index 1, moving each element one position right. The loop stops when i > 0, so the first element at index 0 is free for insertion.

2fill in blank

medium

Complete the code to assign the new element at the beginning of the array after shifting.

DSA C++

arr[[1]] = new_element;

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

A0

Bn-1

C1

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning the new element at index 1 or n causes incorrect placement.

Using n or n-1 places the element beyond the intended start.

After shifting elements right, the first position at index 0 is free to hold the new element.

3fill in blank

hard

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

DSA C++

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

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

A>=

B<

C<=

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' or '<=' causes the loop to run incorrectly or infinitely.

Using '>=' causes out-of-bound access at i=0.

The loop must run while i is greater than 0 to avoid accessing arr[-1]. Using '>' ensures safe shifting.

4fill in blank

hard

Fill both blanks to complete the function that inserts an element at the beginning of the array.

DSA C++

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

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

A0

B1

Cn

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using different indices for shifting and assignment causes data overwrite.

Incorrect loop condition leads to out-of-bound errors.

The loop shifts elements while i > 0, freeing index 0 for the new element. Then the new element is assigned at index 0.

5fill in blank

hard

Fill all three blanks to create a function that inserts an element at the beginning and updates the size.

DSA C++

void insertAtBeginning(int arr[], int& [1], int [2]) {
    for (int i = [3]; i > 0; i--) {
        arr[i] = arr[i - 1];
    }
    arr[0] = [2];
    [1]++;
}

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

An

Bnew_element

Dsize

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable names causes compilation errors.

Not passing size by reference prevents updating the array size.

The function takes the array size 'n' by reference to update it, the new element to insert, and shifts elements from index n down to 1. Then it inserts the new element at index 0 and increments n.

Practice

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

easy

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

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

C. The array size decreases by one.

D. The array elements are sorted automatically.

Solution

1. Step 1: Understand array structure

   An array stores elements in continuous memory locations.

2. Step 2: Effect of insertion at beginning

   To insert at the start, all elements must move right to free the first position.

3. Final Answer:

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

4. Quick Check:

   Insertion at start shifts elements right = B [OK]

Hint: Insertion at start shifts elements right to free first slot [OK]

Common Mistakes:

• Thinking new element replaces first without shifting
• Assuming array size decreases
• Believing array sorts automatically

2. Which of the following is the correct syntax to insert an element `x` at the beginning of an array `arr` of size `n` in C++ (assuming enough space)?

easy

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

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

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

D. arr[n] = x; n++;

Solution

1. Step 1: Shift elements right from end

   Loop from last index `n` down to 1, moving each element to the right by one.

2. Step 2: Insert new element at index 0 and increase size

   Place `x` at `arr[0]` and increment `n`.

3. Final Answer:

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

4. Quick Check:

   Shift right from end, insert at 0 = C [OK]

Hint: Shift from end to start, then insert at zero index [OK]

Common Mistakes:

• Shifting from start to end overwrites data
• Not incrementing array size
• Inserting without shifting elements

3. What is the output of the following C++ code?

int arr[5] = {2, 4, 6, 8, 10};
int n = 5;
int x = 1;
for(int i = n; i > 0; i--) {
  arr[i] = arr[i-1];
}
arr[0] = x;
n++;
for(int i = 0; i < n; i++) {
  std::cout << arr[i] << " ";
}

medium

A. 2 4 6 8 10 1

B. 1 2 4 6 8 10

C. 1 2 4 6 8

D. 2 4 6 8 10

Solution

1. Step 1: Shift elements right to make space

   Elements move 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.

2. Step 2: Insert 1 at arr[0] and print

   arr[0]=1, array becomes [1,2,4,6,8,10], printed with space.

3. Final Answer:

   1 2 4 6 8 10 -> Option B

4. Quick Check:

   Shift right + insert 1 at start = A [OK]

Hint: Shift right, insert new at zero, print all elements [OK]

Common Mistakes:

• Not shifting elements correctly
• Printing old array size
• Inserting at wrong index

4. Identify the error in this code snippet that inserts `x` at the beginning of array `arr` of size `n`:

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

medium

A. Loop shifts elements left instead of right, overwriting data.

B. Array size `n` is not incremented after insertion.

C. New element is inserted at wrong index.

D. Loop should start from `n` and go down to 1 to avoid overwrite.

Solution

1. Step 1: Analyze loop direction

   Loop moves from 0 to n-1, shifting elements forward, overwriting original data.

2. Step 2: Correct loop direction

   Loop must run backward from n to 1 to shift elements safely without overwrite.

3. Final Answer:

   Loop should start from `n` and go down to 1 to avoid overwrite. -> Option D

4. Quick Check:

   Backward loop prevents overwrite = D [OK]

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

Common Mistakes:

• Shifting elements forward overwrites data
• Forgetting to increment array size
• Inserting element at wrong position

5. Given an array `arr` of size `n` with elements sorted in ascending order, how can you insert a new element `x` at the beginning and keep the array sorted?

hard

A. If `x` is smaller than `arr[0]`, shift elements right and insert `x` at start; else find correct position.

B. Insert `x` at the beginning and then sort the entire array again.

C. Always insert `x` at the beginning without shifting elements.

D. Replace the last element with `x` if `x` is smaller than last element.

Solution

1. Step 1: Check if `x` is smaller than first element

   If yes, shift all elements right and insert `x` at index 0 to keep order.

2. Step 2: If `x` is not smallest, find correct position

   Find proper index to insert `x` to maintain sorted order, not necessarily at start.

3. Final Answer:

   If `x` is smaller than `arr[0]`, shift elements right and insert `x` at start; else find correct position. -> Option A

4. Quick Check:

   Insert at start only if smaller, else find position = A [OK]

Hint: Insert at start only if smaller; else find correct sorted position [OK]

Common Mistakes:

• Always inserting at start breaks sorting
• Sorting entire array after insertion wastes time
• Replacing last element loses data