Array Insertion at Beginning in DSA Java - Build from Scratch

Array Insertion at Beginning

📖 Scenario: You are managing a list of daily tasks in a simple to-do app. You want to add a new urgent task at the beginning of your task list.

🎯 Goal: Build a Java program that inserts a new task at the beginning of an existing array of tasks.

📋 What You'll Learn

Create an array of tasks with exact values

Create a new task string to insert

Write code to insert the new task at the beginning of the array

Print the updated array showing the new task at the start

💡 Why This Matters

🌍 Real World

Managing task lists or ordered data where new important items need to be added at the start.

💼 Career

Understanding array manipulation is fundamental for programming jobs, especially when working with fixed-size data structures.

1

Create the initial array of tasks

Create a String array called tasks with these exact values: "Email client", "Write report", "Team meeting"

DSA Java

# Create the tasks array with the given tasks
// Your code here

Hint

Use curly braces {} to list the tasks inside the array.

2

Create the new task to insert

Create a String variable called newTask and set it to "Urgent: Fix bug"

DSA Java

String[] tasks = {"Email client", "Write report", "Team meeting"};
// Create the newTask string
// Your code here

Hint

Use double quotes "" to set the string value.

3

Insert the new task at the beginning

Create a new String array called updatedTasks with length one more than tasks. Set the first element of updatedTasks to newTask. Then copy all elements from tasks into updatedTasks starting at index 1 using a for loop with variable i.

DSA Java

String[] tasks = {"Email client", "Write report", "Team meeting"};
String newTask = "Urgent: Fix bug";
// Insert newTask at the beginning of updatedTasks
// Your code here

Hint

Remember arrays have fixed size, so create a new array bigger by one.

Use a for loop to copy old tasks starting at index 1.

4

Print the updated array

Use a for loop with variable j to print each element of updatedTasks on its own line.

DSA Java

String[] tasks = {"Email client", "Write report", "Team meeting"};
String newTask = "Urgent: Fix bug";
String[] updatedTasks = new String[tasks.length + 1];
updatedTasks[0] = newTask;
for (int i = 0; i < tasks.length; i++) {
    updatedTasks[i + 1] = tasks[i];
}
// Print each task in updatedTasks
// Your code here

Hint

Use System.out.println() inside the loop to print each task.

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