Array Insertion at Beginning in DSA Java

Mental Model

To insert a new item at the start of an array, we move all existing items one step to the right to make space.

Analogy: Imagine a row of chairs where a new person wants to sit at the first chair, so everyone shifts one chair to the right to make room.

Index: 0   1   2   3   4
Array: [1]->[2]->[3]->[4]->[ ]

Dry Run Walkthrough

Input: array: [1, 2, 3, 4], insert value 0 at beginning

Goal: Add the value 0 at the start, shifting all elements right

Step 1: Shift element at index 3 (4) to index 4

[1]->[2]->[3]->[4]->[4]

Why: Make space at the beginning by moving last element right

Step 2: Shift element at index 2 (3) to index 3

[1]->[2]->[3]->[3]->[4]

Why: Continue shifting elements right to free index 0

Step 3: Shift element at index 1 (2) to index 2

[1]->[2]->[2]->[3]->[4]

Why: Keep shifting elements right

Step 4: Shift element at index 0 (1) to index 1

[1]->[1]->[2]->[3]->[4]

Why: Last shift to free index 0

Step 5: Insert new value 0 at index 0

[0]->[1]->[2]->[3]->[4]

Why: Place new value at the beginning

Result:

Final array: [0]->[1]->[2]->[3]->[4]

Annotated Code

DSA Java

public class ArrayInsertion {
    public static int[] insertAtBeginning(int[] arr, int value) {
        int n = arr.length;
        int[] newArr = new int[n + 1];
        // Shift elements right to make space at index 0
        for (int i = n - 1; i >= 0; i--) {
            newArr[i + 1] = arr[i];
        }
        // Insert new value at beginning
        newArr[0] = value;
        return newArr;
    }

    public static void main(String[] args) {
        int[] arr = {1, 2, 3, 4};
        int value = 0;
        int[] result = insertAtBeginning(arr, value);
        // Print array
        for (int i = 0; i < result.length; i++) {
            System.out.print(result[i]);
            if (i != result.length - 1) {
                System.out.print(" -> ");
            }
        }
        System.out.println();
    }
}

for (int i = n - 1; i >= 0; i--) { newArr[i + 1] = arr[i]; }

Shift all elements one position right to free index 0

newArr[0] = value;

Insert new value at the start of the array

Output

0 -> 1 -> 2 -> 3 -> 4

Complexity Analysis

Time: O(n) because we must move all n elements one step to the right

Space: O(n) because we create a new array with one extra space

vs Alternative: In-place shifting in the same array is possible but requires moving elements one by one, still O(n) time; using a linked list would allow O(1) insertion at beginning but with different memory structure

Edge Cases

empty array

Returns a new array with only the inserted value

DSA Java

int[] newArr = new int[n + 1];

array with one element

Shifts the single element to index 1 and inserts new value at index 0

DSA Java

for (int i = n - 1; i >= 0; i--) { newArr[i + 1] = arr[i]; }

When to Use This Pattern

When you need to add an element at the start of a fixed-size array, use shifting to make space because arrays have fixed positions.

Common Mistakes

Mistake: Trying to insert at index 0 without shifting elements, overwriting the first element
Fix: Always shift elements right first before inserting at index 0

Mistake: Shifting elements from left to right causing overwriting data
Fix: Shift elements from right to left to avoid overwriting

Summary

Inserts a new value at the beginning of an array by shifting existing elements right.

Use when you need to add an element at the start but arrays have fixed size and order.

The key is to shift elements from the end towards the start to avoid overwriting data.

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