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

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of inserting element at the beginning of an array

What is the output of the following Java code that inserts an element at the beginning of an array?

DSA Java

public class Main {
    public static void main(String[] args) {
        int[] arr = {2, 3, 4};
        int newElement = 1;
        int[] newArr = new int[arr.length + 1];
        newArr[0] = newElement;
        for (int i = 0; i < arr.length; i++) {
            newArr[i + 1] = arr[i];
        }
        for (int num : newArr) {
            System.out.print(num + " ");
        }
    }
}

A 1 3 4 2

B 2 3 4 1

C 1 2 3 4

D 2 1 3 4

💡 Hint

Think about how the new element is placed at index 0 and the rest are shifted.

The new element 1 is inserted at the start (index 0), and the original elements are shifted right by one position.

🧠 Conceptual

intermediate

Understanding array insertion at beginning complexity

What is the time complexity of inserting an element at the beginning of a fixed-size array in Java by creating a new array and copying elements?

A O(n) - Linear time

B O(log n) - Logarithmic time

C O(1) - Constant time

D O(n^2) - Quadratic time

💡 Hint

Consider how many elements need to be copied to the new array.

Since all elements must be copied to the new array shifted by one position, the operation takes linear time proportional to the array size.

🔧 Debug

advanced

Identify the error in array insertion at beginning code

What error will the following Java code produce when trying to insert an element at the beginning of an array?

DSA Java

int[] arr = {5, 6, 7};
int newElement = 4;
for (int i = arr.length - 1; i >= 0; i--) {
    arr[i + 1] = arr[i];
}
arr[0] = newElement;
for (int num : arr) {
    System.out.print(num + " ");
}

A NullPointerException

B No error, output: 4 5 6 7

C Compilation error due to missing array size

D ArrayIndexOutOfBoundsException

💡 Hint

Think about the size of the original array and the index accessed in the loop.

The code tries to access arr[i + 1] where i + 1 equals arr.length, which is outside the array bounds causing ArrayIndexOutOfBoundsException.

🚀 Application

advanced

Result of multiple insertions at beginning

Given an empty array and the following insertions at the beginning, what is the final array?

DSA Java

int[] arr = new int[0];
int[] temp;
int[] insertAtBeginning(int[] array, int element) {
    temp = new int[array.length + 1];
    temp[0] = element;
    for (int i = 0; i < array.length; i++) {
        temp[i + 1] = array[i];
    }
    return temp;
}
arr = insertAtBeginning(arr, 3);
arr = insertAtBeginning(arr, 2);
arr = insertAtBeginning(arr, 1);
for (int num : arr) {
    System.out.print(num + " ");
}

A 1 2 3

B 3 1 2

C 3 2 1

D 1 3 2

💡 Hint

Each insertion adds the new element at the start, pushing previous elements right.

First insert 3: [3], then insert 2 at start: [2,3], then insert 1 at start: [1,2,3].

🧠 Conceptual

expert

Why arrays are inefficient for frequent insertions at beginning

Why is inserting elements at the beginning of a fixed-size array inefficient compared to other data structures?

A Because arrays use linked nodes which slow down insertion

B Because arrays require shifting all existing elements to the right on each insertion

C Because arrays do not allow storing elements of different types

D Because arrays automatically resize and cause memory leaks

💡 Hint

Think about what happens to existing elements when you insert at the start.

Arrays have fixed size and contiguous memory, so inserting at the start requires moving all elements one position to the right, which is costly.

Practice

1. What happens when you insert an element at the beginning of a fixed-size array in Java?

easy

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

B. The array size automatically increases to fit the new element.

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

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

Solution

1. Step 1: Understand fixed-size array behavior

   Java arrays have fixed size, so to insert at the start, elements must move to make space.

2. Step 2: Describe element shifting

   All elements shift one position right to free index 0 for the new element.

3. Final Answer:

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

4. Quick Check:

   Insert at start = shift right [OK]

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

Common Mistakes:

• Thinking array size grows automatically
• Replacing first element without shifting
• Removing last element silently

2. Which of the following Java code snippets correctly inserts an element 10 at the beginning of an integer array arr of size 5?

easy

A. arr[arr.length] = 10; // Insert at end

B. for(int i = 0; i < arr.length - 1; i++) { arr[i] = arr[i+1]; } arr[arr.length-1] = 10;

C. arr[0] = 10; for(int i = 1; i < arr.length; i++) { arr[i] = arr[i-1]; }

D. for(int i = arr.length - 1; i > 0; i--) { arr[i] = arr[i-1]; } arr[0] = 10;

Solution

1. Step 1: Identify correct shifting direction

   To insert at start, shift elements right from end to start to avoid overwriting.

2. Step 2: Check code correctness

   for(int i = arr.length - 1; i > 0; i--) { arr[i] = arr[i-1]; } arr[0] = 10; shifts elements right correctly, then inserts 10 at index 0.

3. Final Answer:

   for(int i = arr.length - 1; i > 0; i--) { arr[i] = arr[i-1]; } arr[0] = 10; -> Option D

4. Quick Check:

   Shift right from end, then insert [OK]

Hint: Shift right from end to start before inserting [OK]

Common Mistakes:

• Shifting left instead of right
• Overwriting elements before shifting
• Trying to insert at arr[arr.length] (out of bounds)

3. What will be the output of the following Java code?

int[] arr = {5, 6, 7, 8, 9};
for(int i = arr.length - 1; i > 0; i--) {
  arr[i] = arr[i-1];
}
arr[0] = 1;
for(int num : arr) {
  System.out.print(num + " ");
}

medium

A. 1 6 7 8 9

B. 1 5 6 7 8

C. 5 6 7 8 9

D. 9 5 6 7 8

Solution

1. Step 1: Trace the shifting loop

   Loop shifts elements right: arr[4]=arr[3]=8, arr[3]=arr[2]=7, arr[2]=arr[1]=6, arr[1]=arr[0]=5.

2. Step 2: Insert 1 at index 0

   arr[0] = 1 replaces original 5.

3. Final Answer:

   1 5 6 7 8 -> Option B

4. Quick Check:

   Shift right + insert 1 at start = 1 5 6 7 8 [OK]

Hint: Shift right then insert new element at index 0 [OK]

Common Mistakes:

• Not shifting elements before insertion
• Overwriting elements incorrectly
• Confusing output order

4. Identify the error in the following code that tries to insert 20 at the beginning of array arr.

int[] arr = {2, 4, 6, 8, 10};
for(int i = 0; i < arr.length - 1; i++) {
  arr[i+1] = arr[i];
}
arr[0] = 20;

medium

A. The loop should start from the end to avoid overwriting.

B. The loop shifts elements left, overwriting data incorrectly.

C. The array size is too small to insert a new element.

D. No error; code works correctly.

Solution

1. Step 1: Analyze loop direction

   The loop shifts elements from start to end, causing overwriting of values before they move.

2. Step 2: Correct loop direction

   To avoid overwriting, loop must run backward from end to start.

3. Final Answer:

   The loop should start from the end to avoid overwriting. -> Option A

4. Quick Check:

   Shift right requires backward loop [OK]

Hint: Shift right means loop backward from end [OK]

Common Mistakes:

• Looping forward causes data overwrite
• Ignoring array size limits
• Assuming code works without shifting properly

5. Given a fixed-size array arr of length 6 with elements {3, 5, 7, 9, 11, 13}, you want to insert 1 at the beginning without losing any data. Which approach correctly handles this?

hard

A. Create a new array of size 7, copy all elements shifted by one, then insert 1 at index 0.

B. Shift elements right in the same array and insert 1 at index 0, overwriting the last element.

C. Replace the first element with 1 without shifting others.

D. Insert 1 at index 6 directly, ignoring array size.

Solution

1. Step 1: Understand fixed-size array limits

   Java arrays cannot grow; to keep all data, a bigger array is needed.

2. Step 2: Copy elements to new array shifted by one

   Create new array size 7, copy old elements starting at index 1, insert 1 at index 0.

3. Final Answer:

   Create a new array of size 7, copy all elements shifted by one, then insert 1 at index 0. -> Option A

4. Quick Check:

   New bigger array + shift + insert = no data loss [OK]

Hint: Use bigger array to avoid data loss when inserting [OK]

Common Mistakes:

• Overwriting last element in fixed array
• Ignoring array size limits
• Trying to insert out of bounds