Why Hash Map Exists and What Problem It Solves in DSA Java - See It Work

Why Hash Map Exists and What Problem It Solves

📖 Scenario: Imagine you run a small library. You want to keep track of which book belongs to which shelf number. You need a fast way to find the shelf number when you know the book's name.

🎯 Goal: You will create a simple program that stores book names and their shelf numbers using a Java HashMap. You will then look up shelf numbers quickly by book name.

📋 What You'll Learn

Create a HashMap to store book names as keys and shelf numbers as values

Add at least 3 books with their shelf numbers to the HashMap

Use a variable to store the book name to search for

Retrieve the shelf number for the given book name from the HashMap

Print the shelf number found

💡 Why This Matters

🌍 Real World

Libraries, phone books, and databases use hash maps to find information quickly by a key like a name or ID.

💼 Career

Understanding hash maps is essential for software developers to write efficient code for data storage and retrieval.

1

Create the HashMap with books and shelf numbers

Create a HashMap called library that stores String keys and Integer values. Add these entries exactly: "Harry Potter" with 5, "Lord of the Rings" with 3, and "The Hobbit" with 7.

DSA Java

import java.util.HashMap;

// Create the HashMap called library and add the books with shelf numbers
// Your code here

Hint

Use HashMap<String, Integer> and the put method to add entries.

2

Set the book name to search for

Create a String variable called searchBook and set it to "The Hobbit".

DSA Java

import java.util.HashMap;

public class Main {
    public static void main(String[] args) {
        HashMap library = new HashMap();
        library.put("Harry Potter", 5);
        library.put("Lord of the Rings", 3);
        library.put("The Hobbit", 7);

        // Create the searchBook variable and set it to "The Hobbit"
        // Your code here
    }
}

Hint

Use String searchBook = "The Hobbit"; to store the book name.

3

Retrieve the shelf number for the book

Create an Integer variable called shelfNumber and set it to the value from library for the key searchBook using the get method.

DSA Java

import java.util.HashMap;

public class Main {
    public static void main(String[] args) {
        HashMap library = new HashMap();
        library.put("Harry Potter", 5);
        library.put("Lord of the Rings", 3);
        library.put("The Hobbit", 7);

        String searchBook = "The Hobbit";

        // Retrieve the shelf number for searchBook from library
        // Your code here
    }
}

Hint

Use library.get(searchBook) to get the shelf number.

4

Print the shelf number found

Write a System.out.println statement to print the text "Shelf number: " followed by the value of shelfNumber.

DSA Java

import java.util.HashMap;

public class Main {
    public static void main(String[] args) {
        HashMap library = new HashMap();
        library.put("Harry Potter", 5);
        library.put("Lord of the Rings", 3);
        library.put("The Hobbit", 7);

        String searchBook = "The Hobbit";

        Integer shelfNumber = library.get(searchBook);

        // Print the shelf number with the message
        // Your code here
    }
}

Hint

Use System.out.println("Shelf number: " + shelfNumber); to print the result.

Practice

1. Why do we use a HashMap in Java instead of a simple array when we want to store key-value pairs?

easy

A. Because arrays use more memory than HashMap.

B. Because HashMap allows fast lookup of values by keys.

C. Because HashMap automatically sorts the keys.

D. Because arrays can only store numbers, not keys and values.

Solution

1. Step 1: Understand the limitation of arrays for key-value storage

   Arrays store values by index, not by custom keys, so they can't quickly find a value by a key.

2. Step 2: Recognize the advantage of HashMap

   HashMap stores data as key-value pairs and uses hashing to find values quickly by keys.

3. Final Answer:

   Because HashMap allows fast lookup of values by keys. -> Option B

4. Quick Check:

   Fast key-based lookup = Because HashMap allows fast lookup of values by keys. [OK]

Hint: HashMap = fast key-value lookup, arrays = index-based [OK]

Common Mistakes:

• Thinking arrays can store key-value pairs directly
• Believing HashMap sorts keys automatically
• Confusing memory usage between arrays and HashMaps

2. Which of the following is the correct way to declare a HashMap in Java that maps String keys to Integer values?

easy

A. HashMap map = new HashMap<>();

B. HashMap map = new HashMap();

C. HashMap map = new HashMap<>();

D. HashMap map = new HashMap<String>();

Solution

1. Step 1: Identify correct generic syntax for HashMap declaration

   The correct syntax uses angle brackets with key and value types on both sides or diamond operator on the right side.

2. Step 2: Check each option for syntax correctness and type order

   HashMap map = new HashMap<>(); correctly declares a HashMap with String keys and Integer values using diamond operator. Others have wrong order or missing types.

3. Final Answer:

   HashMap map = new HashMap<>(); -> Option A

4. Quick Check:

   Correct generic syntax = HashMap map = new HashMap<>(); [OK]

Hint: Use diamond operator <> with correct key-value types [OK]

Common Mistakes:

• Swapping key and value types
• Omitting generic types on right side
• Using raw types without generics

3. What will be the output of the following Java code?

import java.util.HashMap;
public class Test {
  public static void main(String[] args) {
    HashMap map = new HashMap<>();
    map.put("apple", 10);
    map.put("banana", 20);
    map.put("apple", 30);
    System.out.println(map.get("apple"));
  }
}

medium

A. 10

B. null

C. 20

D. 30

Solution

1. Step 1: Understand how put works in HashMap with duplicate keys

   When a key already exists, put replaces the old value with the new one.

2. Step 2: Trace the code for key "apple"

   First, "apple" is set to 10, then replaced by 30. So map.get("apple") returns 30.

3. Final Answer:

   30 -> Option D

4. Quick Check:

   Last value for key "apple" = 30 [OK]

Hint: Last put value for a key overwrites previous [OK]

Common Mistakes:

• Assuming first value stays for duplicate keys
• Expecting null for existing keys
• Confusing values of different keys

4. The following code tries to check if a key exists in a HashMap and print its value. What is wrong?

HashMap map = new HashMap<>();
map.put("x", 5);
if (map.containsKey("y")) {
  System.out.println(map.get("y"));
} else {
  System.out.println("Key not found");
}

medium

A. The code throws NullPointerException when key is missing.

B. Using containsKey is wrong; should use get only.

C. No error; code works correctly.

D. The code should check for null instead of using containsKey.

Solution

1. Step 1: Understand containsKey usage

   containsKey safely checks if a key exists without throwing errors.

2. Step 2: Check code behavior for missing key "y"

   Since "y" is missing, containsKey returns false and prints "Key not found" without error.

3. Final Answer:

   No error; code works correctly. -> Option C

4. Quick Check:

   Using containsKey prevents errors = No error; code works correctly. [OK]

Hint: Use containsKey before get to avoid null issues [OK]

Common Mistakes:

• Expecting NullPointerException on missing key
• Using get without checking key existence
• Thinking containsKey is unnecessary

5. You want to count how many times each word appears in a list of words using a HashMap. Which approach correctly solves this problem?

hard

A. Use a HashMap<String, Integer>, for each word check if it exists; if yes, increment count; else, add with count 1.

B. Use a HashMap<Integer, String> and store counts as keys and words as values.

C. Use an array to store counts and loop through words without a map.

D. Use a HashMap<String, String> and store words as both keys and values.

Solution

1. Step 1: Identify the correct key-value types for counting words

   Keys should be words (String), values should be counts (Integer).

2. Step 2: Understand the counting logic using HashMap

   For each word, check if it exists in map; if yes, increment count; else add with count 1.

3. Final Answer:

   Use a HashMap, increment counts per word. -> Option A

4. Quick Check:

   Word count needs String keys and Integer values [OK]

Hint: Map words to counts; increment existing or add new [OK]

Common Mistakes:

• Swapping key and value types
• Using arrays without keys for counting
• Storing words as values instead of counts