Why Array Deletion at Beginning in DSA Java?

The Big Idea

What if removing the first item from a list could happen instantly without moving everything by hand?

The Scenario

Imagine you have a row of boxes lined up on a shelf, and you want to remove the very first box. You have to move every other box one step forward to fill the empty space.

The Problem

Doing this by hand is slow and tiring because you must shift all the boxes one by one. If you have many boxes, it takes a lot of time and effort, and you might accidentally drop or misplace a box.

The Solution

Using array deletion at the beginning in programming automates this shifting process. The computer quickly moves all elements one step forward, removing the first element cleanly and efficiently.

Before vs After

✗ Before

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

✓ After

int[] newArray = java.util.Arrays.copyOfRange(array, 1, array.length);

What It Enables

This operation allows programs to efficiently remove the first item from a list, making room for new data or updating the list dynamically.

Real Life Example

Think of a queue at a ticket counter where the first person leaves after buying a ticket, and everyone else moves forward to fill the gap.

Key Takeaways

Manually removing the first element requires shifting all others.

Array deletion at beginning automates this shift efficiently.

It helps manage dynamic lists where the first element changes often.

Practice

1. What happens to the elements of an array when you delete the first element (at the beginning)?

easy

A. All elements shift one position to the left, and the first element is removed.

B. All elements shift one position to the right, and the last element is removed.

C. The array size increases by one.

D. The array remains unchanged.

Solution

1. Step 1: Understand array deletion at beginning

   Deleting the first element means removing the element at index 0.

2. Step 2: Effect on remaining elements

   All other elements move one position left to fill the gap, reducing the array size by one.

3. Final Answer:

   All elements shift one position to the left, and the first element is removed. -> Option A

4. Quick Check:

   Delete first element = shift left [OK]

Hint: Deleting first element shifts all others left by one [OK]

Common Mistakes:

• Thinking elements shift right instead of left
• Assuming array size stays the same
• Believing the array remains unchanged

2. Which of the following Java code snippets correctly deletes the first element from an integer array arr and shifts the remaining elements left?

easy

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

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

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

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

Solution

1. Step 1: Identify correct loop direction and indices

   To shift elements left, start from index 0 to length-2, assigning arr[i] = arr[i+1].

2. Step 2: Check each option

   for (int i = 0; i < arr.length - 1; i++) { arr[i] = arr[i + 1]; } matches this logic; others shift right or use wrong indices causing errors.

3. Final Answer:

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

4. Quick Check:

   Shift left loop: i from 0 to length-2 [OK]

Hint: Shift left by copying arr[i+1] to arr[i] in forward loop [OK]

Common Mistakes:

• Looping backwards when shifting left
• Using wrong indices causing ArrayIndexOutOfBounds
• Assigning arr[i-1] instead of arr[i+1]

3. What is the output of the following Java code?

int[] arr = {10, 20, 30, 40, 50};
for (int i = 0; i < arr.length - 1; i++) {
    arr[i] = arr[i + 1];
}
arr[arr.length - 1] = 0;
for (int num : arr) {
    System.out.print(num + " ");
}

medium

A. 20 30 40 50 0

B. 10 20 30 40 50

C. 0 10 20 30 40

D. 50 40 30 20 10

Solution

1. Step 1: Trace the shifting loop

   Each element moves left: arr[0]=20, arr[1]=30, arr[2]=40, arr[3]=50.

2. Step 2: Set last element to 0

   arr[4] is set to 0 explicitly.

3. Final Answer:

   20 30 40 50 0 -> Option A

4. Quick Check:

   Shift left + last zero = 20 30 40 50 0 [OK]

Hint: After shift, last element set to zero to fill gap [OK]

Common Mistakes:

• Forgetting to set last element to zero
• Printing original array without shift
• Confusing shift direction

4. The following code attempts to delete the first element of an array but causes an error. What is the error?

int[] arr = {1, 2, 3, 4};
for (int i = 0; i <= arr.length - 1; i++) {
    arr[i] = arr[i + 1];
}

medium

A. Logical error: elements shift right instead of left

B. ArrayIndexOutOfBoundsException at i = 3

C. Syntax error due to missing semicolon

D. No error, code runs fine

Solution

1. Step 1: Analyze loop range

   The loop runs i from 0 to arr.length - 1 (i.e., 0 to 3).

2. Step 2: Identify out-of-bounds access

   At i=3, arr[i+1] = arr[4] is accessed, which is outside array bounds causing exception.

3. Final Answer:

   ArrayIndexOutOfBoundsException at i = 3 -> Option B

4. Quick Check:

   Accessing arr[4] in length 4 array causes exception [OK]

Hint: Loop till length-2 to avoid out-of-bounds [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Ignoring array length limits
• Assuming code runs without error

5. Given an array arr = {5, 10, 15, 20, 25}, write the output after deleting the first element twice using the standard shift-left method and setting the last element to zero each time.
What is the final state of the array?

hard

A. [5, 10, 15, 20, 25]

B. [10, 15, 20, 25, 0]

C. [0, 0, 15, 20, 25]

D. [15, 20, 25, 0, 0]

Solution

1. Step 1: First deletion

   Shift left: arr becomes [10, 15, 20, 25, 0]

2. Step 2: Second deletion

   Shift left again: arr becomes [15, 20, 25, 0, 0]

3. Final Answer:

   [15, 20, 25, 0, 0] -> Option D

4. Quick Check:

   Two shifts left + zeros at end = [15, 20, 25, 0, 0] [OK]

Hint: Each deletion shifts left and adds zero at end [OK]

Common Mistakes:

• Deleting only once instead of twice
• Not setting last element to zero after shift
• Confusing order of elements after shifts