Array Insertion at Beginning in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output after inserting at the beginning?

Consider an array of integers initially containing [10, 20, 30, 40]. We insert the value 5 at the beginning by shifting elements to the right. What is the final array?

DSA C

int arr[5] = {10, 20, 30, 40};
int n = 4;
int x = 5;
for (int i = n; i > 0; i--) {
    arr[i] = arr[i-1];
}
arr[0] = x;
n++;
for (int i = 0; i < n; i++) {
    printf("%d ", arr[i]);
}

A [5, 20, 30, 40, 10]

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

C [5, 10, 20, 30, 40]

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

💡 Hint

Think about shifting elements from the end towards the beginning before inserting.

To insert at the beginning, all elements must move one position right. Then the new element is placed at index 0.

🧠 Conceptual

intermediate

Why do we shift elements when inserting at the beginning?

When inserting a new element at the start of a fixed-size array, why is it necessary to shift existing elements to the right?

A To sort the array automatically after insertion.

B To make space at the beginning without losing existing elements.

C To remove duplicates from the array.

D To reverse the array order.

💡 Hint

Think about what happens if you put a new element at index 0 without moving others.

Shifting elements right creates an empty slot at index 0 so the new element can be inserted without overwriting existing data.

🔧 Debug

advanced

Find the error in this insertion code

This code attempts to insert 99 at the beginning of an array but produces wrong output. What is the error?

DSA C

int arr[5] = {1, 2, 3, 4};
int n = 4;
int x = 99;
for (int i = 0; i < n; i++) {
    arr[i+1] = arr[i];
}
arr[0] = x;
n++;
for (int i = 0; i < n; i++) {
    printf("%d ", arr[i]);
}

A The loop should run backwards from n to 0 to avoid overwriting elements.

B The array size is too small to hold the new element.

C The value x should be assigned after the loop ends.

D The loop condition should be i <= n instead of i < n.

💡 Hint

Consider what happens when you move elements forward starting from the beginning.

Shifting elements forward from the start overwrites data before it is copied. Looping backwards preserves data during shifting.

❓ Predict Output

advanced

Output after multiple insertions at beginning

Starting with an empty array of size 5, insert 3, then 7, then 1 at the beginning in that order. What is the final array?

DSA C

int arr[5];
int n = 0;
int values[] = {3, 7, 1};
for (int k = 0; k < 3; k++) {
    for (int i = n; i > 0; i--) {
        arr[i] = arr[i-1];
    }
    arr[0] = values[k];
    n++;
}
for (int i = 0; i < n; i++) {
    printf("%d ", arr[i]);
}

A [1, 7, 3]

B [3, 7, 1]

C [7, 3, 1]

D [1, 3, 7]

💡 Hint

Each new element goes to index 0, pushing others right.

Insert 3: [3] Insert 7: [7, 3] Insert 1: [1, 7, 3]

🧠 Conceptual

expert

Why is inserting at the beginning of an array costly?

Inserting an element at the beginning of a large fixed-size array requires shifting all elements. Why is this operation considered costly in terms of performance?

A Because inserting at the beginning causes the array to resize automatically.

B Because arrays cannot store elements at the beginning.

C Because the first element is always locked and cannot be changed.

D Because shifting elements requires moving many items, which takes time proportional to the array size.

💡 Hint

Think about how many elements must be moved when inserting at the start.

Each insertion at the beginning requires moving all existing elements one position right, which takes time proportional to the number of elements.