Frequency Counter Pattern Using Hash Map in DSA Java - Build from Scratch

Frequency Counter Pattern Using Hash Map

📖 Scenario: You are working in a small bookstore. You want to count how many copies of each book title are currently in stock. This will help you know which books are popular and need restocking.

🎯 Goal: Build a program that counts the number of copies for each book title using a hash map (dictionary). You will create the data, set up a counter map, count the frequencies, and then print the results.

📋 What You'll Learn

Create a list of book titles (each entry represents one copy in stock)

Create an empty hash map to store frequency counts

Use a for loop to count frequencies of each book title

Print the frequency map showing book titles and their counts

💡 Why This Matters

🌍 Real World

Counting frequencies is useful in inventory management, word counting, and data analysis.

💼 Career

Many programming jobs require counting and grouping data efficiently using hash maps or dictionaries.

1

Create the initial book stock data

Create a List<String> called books with these exact titles in order: "The Hobbit", "1984", "Dune", "The Hobbit", "Dune". Each entry represents one copy in stock.

DSA Java

// Create a List called books with the exact titles
// Your code here

Hint

Use Arrays.asList to create the list with the exact titles in order.

2

Create an empty frequency map

Create an empty HashMap<String, Integer> called frequencyMap to store the count of each book title.

DSA Java

import java.util.*;

public class FrequencyCounter {
    public static void main(String[] args) {
        List books = Arrays.asList("The Hobbit", "1984", "Dune", "The Hobbit", "Dune");
        // Create an empty HashMap called frequencyMap
        // Your code here
    }
}

Hint

Use new HashMap<>() to create an empty map.

3

Count the frequency of each book title

Use a for loop with variable book to iterate over books. Inside the loop, update frequencyMap to count how many times each book appears. If the book is not in the map, add it with count 1. If it is already there, increase its count by 1.

DSA Java

import java.util.*;

public class FrequencyCounter {
    public static void main(String[] args) {
        List books = Arrays.asList("The Hobbit", "1984", "Dune", "The Hobbit", "Dune");
        HashMap frequencyMap = new HashMap();
        // Use a for loop to count frequencies of each book title
        // Your code here
    }
}

Hint

Use containsKey to check if the book is already counted.

4

Print the frequency map

Print the frequencyMap to show the count of each book title.

DSA Java

import java.util.*;

public class FrequencyCounter {
    public static void main(String[] args) {
        List books = Arrays.asList("The Hobbit", "1984", "Dune", "The Hobbit", "Dune");
        HashMap frequencyMap = new HashMap();
        for (String book : books) {
            if (frequencyMap.containsKey(book)) {
                frequencyMap.put(book, frequencyMap.get(book) + 1);
            } else {
                frequencyMap.put(book, 1);
            }
        }
        // Print the frequencyMap
        // Your code here
    }
}

Hint

Use System.out.println(frequencyMap); to print the map.

Practice

1. What is the main purpose of using a hash map in the Frequency Counter Pattern?

easy

A. To count how many times each element appears in a collection

B. To sort elements in ascending order

C. To store elements without duplicates

D. To reverse the order of elements

Solution

1. Step 1: Understand the role of a hash map

   A hash map stores key-value pairs, allowing quick lookup and update of counts.

2. Step 2: Apply to frequency counting

   By using elements as keys and counts as values, it tracks how many times each element appears.

3. Final Answer:

   To count how many times each element appears in a collection -> Option A

4. Quick Check:

   Frequency counting = count elements [OK]

Hint: Hash map keys track elements, values track counts [OK]

Common Mistakes:

• Thinking hash map sorts elements
• Confusing frequency counting with removing duplicates
• Using hash map to reverse order

2. Which of the following is the correct way to declare a HashMap to count frequencies of characters in Java?

easy

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

B. HashMap freq = new HashMap<>();

C. HashMap<String> freq = new HashMap<>();

D. HashMap<Character> freq = new HashMap<>();

Solution

1. Step 1: Identify correct generic types for HashMap

   Keys are characters, so use Character; values are counts, so use Integer.

2. Step 2: Check syntax correctness

   HashMap freq = new HashMap<>(); is valid Java syntax.

3. Final Answer:

   HashMap freq = new HashMap<>(); -> Option B

4. Quick Check:

   Correct generics and syntax = HashMap freq = new HashMap<>(); [OK]

Hint: Use wrapper classes Character and Integer for HashMap keys and values [OK]

Common Mistakes:

• Using primitive types like int in generics
• Missing value type in HashMap declaration
• Incorrect generic syntax

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

import java.util.HashMap;
public class Main {
  public static void main(String[] args) {
    String s = "banana";
    HashMap freq = new HashMap<>();
    for (char c : s.toCharArray()) {
      freq.put(c, freq.getOrDefault(c, 0) + 1);
    }
    System.out.println(freq.get('a'));
  }
}

medium

A. 2

B. null

C. 1

D. 3

Solution

1. Step 1: Count occurrences of 'a' in "banana"

   String "banana" has 'a' at positions 1, 3, and 5, total 3 times.

2. Step 2: Check code logic for frequency counting

   Code uses getOrDefault to count each character's frequency correctly.

3. Final Answer:

   3 -> Option D

4. Quick Check:

   Count of 'a' in "banana" = 3 [OK]

Hint: Count characters manually, then verify with getOrDefault logic [OK]

Common Mistakes:

• Counting 'a' incorrectly as 2 or 1
• Confusing getOrDefault behavior
• Expecting null for existing keys

4. Identify the error in the following Java code snippet that tries to count frequencies of integers:

import java.util.HashMap;
public class Main {
  public static void main(String[] args) {
    int[] nums = {1, 2, 2, 3};
    HashMap freq = new HashMap<>();
    for (int num : nums) {
      freq.put(num, freq.get(num) + 1);
    }
    System.out.println(freq);
  }
}

medium

A. Using freq.get(num) without checking if key exists causes NullPointerException

B. HashMap declaration is incorrect

C. The for loop syntax is invalid

D. Printing freq directly causes syntax error

Solution

1. Step 1: Analyze freq.get(num) usage inside loop

   freq.get(num) returns null if key not present, so null + 1 causes NullPointerException.

2. Step 2: Identify missing default value handling

   Code should use freq.getOrDefault(num, 0) to avoid null on first occurrence.

3. Final Answer:

   Using freq.get(num) without checking if key exists causes NullPointerException -> Option A

4. Quick Check:

   NullPointerException due to missing default value [OK]

Hint: Use getOrDefault to avoid null when key missing [OK]

Common Mistakes:

• Ignoring possibility of null from get()
• Thinking HashMap declaration is wrong
• Assuming printing freq causes error

5. Given two integer arrays arr1 and arr2, which Java code snippet correctly checks if arr2 contains the same frequency of elements as arr1 using the Frequency Counter Pattern?

hard

A. HashMap freq1 = new HashMap<>();
HashMap freq2 = new HashMap<>();
for (int n : arr1) freq1.put(n, freq1.getOrDefault(n, 0) + 1);
for (int n : arr2) freq2.put(n, freq2.getOrDefault(n, 0) + 1);
return freq1.size() == freq2.size();

B. HashMap freq = new HashMap<>();
for (int n : arr1) freq.put(n, freq.getOrDefault(n, 0) + 1);
for (int n : arr2) freq.put(n, freq.getOrDefault(n, 0) - 1);
return freq.size() == 0;

C. HashMap freq = new HashMap<>();
for (int n : arr1) freq.put(n, freq.getOrDefault(n, 0) + 1);
for (int n : arr2) freq.put(n, freq.getOrDefault(n, 0) - 1);
for (int count : freq.values()) if (count != 0) return false;
return true;

D. HashMap freq = new HashMap<>();
for (int n : arr1) freq.put(n, freq.getOrDefault(n, 0) + 1);
for (int n : arr2) freq.put(n, freq.getOrDefault(n, 0) + 1);
return freq.size() == arr1.length;

Solution

1. Step 1: Understand frequency counting for both arrays

   Count frequencies of arr1 elements positively, then subtract frequencies for arr2 elements.

2. Step 2: Verify all counts are zero

   If all counts are zero, frequencies match; otherwise, they differ.

3. Final Answer:

   Code that increments for arr1, decrements for arr2, then checks zero counts -> Option C

4. Quick Check:

   Increment then decrement counts, check zero [OK]

Hint: Add counts for arr1, subtract for arr2, verify all zero [OK]

Common Mistakes:

• Checking sizes of two separate maps instead of contents
• Adding counts for both arrays instead of subtracting
• Checking map size instead of counts