Linked List vs Array When to Choose Which in DSA Java - Build Both Approaches

Linked List vs Array: When to Choose Which

📖 Scenario: Imagine you are managing a playlist of songs on your phone. Sometimes you want to add songs quickly at the end, sometimes in the middle, and sometimes you want to access a song by its position fast. You want to understand when to use an array or a linked list to store your playlist efficiently.

🎯 Goal: You will create a simple playlist using both an array and a linked list. You will add songs, access songs by position, and see how each data structure behaves. This will help you understand when to choose an array or a linked list.

📋 What You'll Learn

Create an array of songs with exact 5 song names

Create a linked list of songs with exact 5 song names

Access and print the 3rd song from both array and linked list

Add a new song at the end of the linked list

Print the final state of both data structures

💡 Why This Matters

🌍 Real World

Playlists, photo albums, and message queues often use arrays or linked lists depending on how data is accessed or changed.

💼 Career

Understanding when to use arrays or linked lists helps in writing efficient code for apps, games, and systems where data changes dynamically.

1

Create an array of songs

Create a String array called songArray with these exact songs in order: "SongA", "SongB", "SongC", "SongD", "SongE".

DSA Java

// Create the array songArray with 5 songs
// Your code here

Hint

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

2

Create a linked list of songs

Below the array, create a LinkedList<String> called songList and add these exact songs in order: "SongA", "SongB", "SongC", "SongD", "SongE". Import java.util.LinkedList at the top.

DSA Java

import java.util.LinkedList;

public class Playlist {
    public static void main(String[] args) {
        String[] songArray = {"SongA", "SongB", "SongC", "SongD", "SongE"};
        // Create the linked list songList and add 5 songs
        // Your code here
    }
}

Hint

Use songList.add("SongName") to add songs one by one.

3

Access the 3rd song from both data structures

Print the 3rd song from songArray and from songList using System.out.println. Use index 2 for the array and get(2) for the linked list.

DSA Java

import java.util.LinkedList;

public class Playlist {
    public static void main(String[] args) {
        String[] songArray = {"SongA", "SongB", "SongC", "SongD", "SongE"};
        LinkedList songList = new LinkedList();
        songList.add("SongA");
        songList.add("SongB");
        songList.add("SongC");
        songList.add("SongD");
        songList.add("SongE");
        // Print the 3rd song from songArray and songList
        // Your code here
    }
}

Hint

Remember array index starts at 0, so 3rd song is index 2.

4

Add a new song to linked list and print both

Add a new song "SongF" at the end of songList using add. Then print all songs in songArray and songList in order, each song separated by a space on its own line.

DSA Java

import java.util.LinkedList;

public class Playlist {
    public static void main(String[] args) {
        String[] songArray = {"SongA", "SongB", "SongC", "SongD", "SongE"};
        LinkedList songList = new LinkedList();
        songList.add("SongA");
        songList.add("SongB");
        songList.add("SongC");
        songList.add("SongD");
        songList.add("SongE");
        System.out.println(songArray[2]);
        System.out.println(songList.get(2));
        // Add "SongF" to songList
        // Print all songs in songArray separated by space
        // Print all songs in songList separated by space
        // Your code here
    }
}

Hint

Use a for-each loop to print all songs separated by spaces.

Practice

1. Which data structure is better when you need fast random access to elements by index?

easy

A. Linked List

B. Array

C. Stack

D. Queue

Solution

1. Step 1: Understand random access requirement

   Random access means accessing any element directly by its position quickly.

2. Step 2: Compare Array and Linked List access times

   Arrays allow direct access by index in constant time, while linked lists require traversal.

3. Final Answer:

   Array -> Option B

4. Quick Check:

   Random access = Array [OK]

Hint: Arrays provide fast index access, linked lists do not [OK]

Common Mistakes:

• Thinking linked lists allow fast random access
• Confusing stack or queue with arrays
• Ignoring access time differences

2. Which of the following is the correct way to declare an array of integers in Java?

easy

A. int[] arr = new int[5];

B. int arr = new int[5];

C. int arr[] = new int();

D. array int arr = new int[5];

Solution

1. Step 1: Recall Java array declaration syntax

   In Java, arrays are declared with type[], variable name, and size with new keyword.

2. Step 2: Check each option for syntax correctness

   int[] arr = new int[5]; matches correct syntax: int[] arr = new int[5]; others have errors.

3. Final Answer:

   int[] arr = new int[5]; -> Option A

4. Quick Check:

   Java array syntax = int[] arr = new int[5]; [OK]

Hint: Remember Java arrays use type[] name = new type[size]; [OK]

Common Mistakes:

• Missing brackets [] after type
• Using parentheses instead of brackets
• Wrong keyword 'array' instead of type

3. Consider this Java code using a linked list:

LinkedList<Integer> list = new LinkedList<>();
list.add(10);
list.add(20);
list.add(30);
System.out.println(list.get(1));

What will be the output?

medium

A. 10

B. 30

C. IndexOutOfBoundsException

D. 20

Solution

1. Step 1: Understand list contents after adds

   Elements added in order: 10 at index 0, 20 at index 1, 30 at index 2.

2. Step 2: Identify element at index 1

   list.get(1) returns the element at index 1, which is 20.

3. Final Answer:

   20 -> Option D

4. Quick Check:

   list.get(1) = 20 [OK]

Hint: Indexing starts at 0; second element is index 1 [OK]

Common Mistakes:

• Assuming index starts at 1
• Confusing add order with index
• Expecting runtime error incorrectly

4. The following Java code tries to remove an element from an array but causes an error:

int[] arr = {1, 2, 3, 4};
arr.remove(2);

What is the main issue here?

medium

A. Arrays do not have a remove method

B. Index 2 is out of bounds

C. Array elements cannot be integers

D. Syntax error in array declaration

Solution

1. Step 1: Check if arrays support remove method

   Java arrays have fixed size and do not support remove method.

2. Step 2: Identify correct way to remove elements

   To remove elements, use ArrayList or create a new array without the element.

3. Final Answer:

   Arrays do not have a remove method -> Option A

4. Quick Check:

   Arrays fixed size = no remove method [OK]

Hint: Arrays fixed size, no remove method; use ArrayList [OK]

Common Mistakes:

• Thinking arrays support remove like lists
• Assuming index 2 is invalid
• Confusing syntax errors with method errors

5. You need to frequently insert and delete elements in the middle of a large collection. Which data structure is more efficient and why?

hard

A. Array, because it uses less memory

B. Linked List, because it uses contiguous memory

C. Linked List, because it allows fast insertions and deletions without shifting elements

D. Array, because it supports random access

Solution

1. Step 1: Analyze insertion and deletion costs

   Arrays require shifting elements for insert/delete in middle, which is slow for large data.

2. Step 2: Understand linked list advantages

   Linked lists use nodes with pointers, so insert/delete in middle is fast without shifting.

3. Final Answer:

   Linked List, because it allows fast insertions and deletions without shifting elements -> Option C

4. Quick Check:

   Fast middle insert/delete = Linked List [OK]

Hint: Linked lists excel at middle insert/delete without shifting [OK]

Common Mistakes:

• Choosing array for frequent middle insert/delete
• Confusing memory layout of linked lists
• Ignoring shifting cost in arrays