Array Insertion at Beginning in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does 'Array Insertion at Beginning' mean?

It means adding a new element at the start (index 0) of an array, shifting all other elements one position to the right.

beginner

Why do we need to shift elements when inserting at the beginning of an array?

Because arrays have fixed positions, to insert at the start, all existing elements must move one step right to make space for the new element.

intermediate

What is the time complexity of inserting an element at the beginning of an array?

The time complexity is O(n) because all n elements need to be shifted one position to the right.

beginner

Show the steps to insert 5 at the beginning of array [1, 2, 3, 4].

Step 1: Shift 4 to index 4
Step 2: Shift 3 to index 3
Step 3: Shift 2 to index 2
Step 4: Shift 1 to index 1
Step 5: Insert 5 at index 0
Result: [5, 1, 2, 3, 4]

intermediate

What happens if the array is full when inserting at the beginning?

You cannot insert directly because the array has no free space. You must create a bigger array, copy elements, then insert the new element at the start.

What is the first step when inserting an element at the beginning of an array?

A Insert the element at the last position

B Replace the first element

C Shift all elements one position to the right

D Delete the last element

To insert at the beginning, all existing elements must move right to make space.

What is the time complexity of inserting at the beginning of an array with n elements?

A O(1)

B O(n)

C O(log n)

D O(n^2)

All n elements must be shifted, so the operation takes O(n) time.

If an array is full, how can you insert a new element at the beginning?

A Overwrite the first element

B Delete the last element and insert

C Insertion is impossible

D Create a bigger array and copy elements

You must create a bigger array, copy elements, then insert the new element.

Which data structure allows faster insertion at the beginning than an array?

A Linked List

B Stack

C Queue

D Hash Table

Linked lists allow O(1) insertion at the beginning without shifting elements.

After inserting 10 at the beginning of [3, 4, 5], what is the new array?

A [10, 3, 4, 5]

B [3, 4, 5, 10]

C [3, 10, 4, 5]

D [10, 4, 5, 3]

The new element 10 is placed at index 0, shifting others right.

Explain how to insert an element at the beginning of an array and why shifting is necessary.

Think about how arrays store elements in fixed positions.

Describe what to do if the array is full and you want to insert at the beginning.

Arrays have fixed size; consider resizing.

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