Array Insertion at Beginning in DSA Java - 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.

DSA Java

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

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

A1

B0

Carr.length

Darr.length - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting the loop from index 1 instead of 0.

Using the wrong loop condition that skips shifting the first element.

We start shifting elements from the last index down to index 1 to make space at index 0 for the new element.

2fill in blank

medium

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

DSA Java

arr[[1]] = newValue;

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

Aarr.length - 1

B1

C0

DnewValue

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning the new value to the last index instead of the first.

Using the newValue variable as an index.

The new value should be placed at index 0, which is the beginning of the array.

3fill in blank

hard

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

DSA Java

for (int i = arr.length - 1; 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 '>=' causes an index out of bounds error.

Using '<' or '<=' causes the loop to not run correctly.

We want to shift elements while i is greater than 0 to avoid accessing invalid index -1.

4fill in blank

hard

Fill both blanks to create a method that inserts a value at the beginning of an array by shifting elements.

DSA Java

public static void insertAtBeginning(int[] arr, int newValue) {
    for (int i = arr.length - 1; i [1] 0; i--) {
        arr[i] = arr[i [2] 1];
    }
    arr[0] = newValue;
}

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

A>

B-

C+

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of '-' in the index calculation.

Using '>=' in the loop condition causing errors.

The loop runs while i > 0, and elements are shifted by assigning arr[i] = arr[i - 1].

5fill in blank

hard

Fill all three blanks to complete the method that inserts a value at the beginning and prints the array.

DSA Java

public static void insertAndPrint(int[] arr, int newValue) {
    for (int i = arr.length - 1; i [1] 0; i--) {
        arr[i] = arr[i [2] 1];
    }
    arr[[3]] = newValue;
    for (int i = 0; i < arr.length; i++) {
        System.out.print(arr[i] + " ");
    }
}

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

A>

B-

C0

D+

E1

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+ 1' instead of '- 1' when shifting elements.

Assigning newValue to the wrong index.

Incorrect loop condition causing errors.

The loop condition is i > 0, elements shift by arr[i] = arr[i - 1], and newValue is assigned at index 0.

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