Frequency Counter Pattern Using Hash Map in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the Frequency Counter Pattern in data structures?

It is a technique to count how many times each value appears in a data set, usually using a hash map to store counts for quick lookup.

beginner

Why use a Hash Map for the Frequency Counter Pattern?

Because a hash map allows fast insertion and lookup of counts by keys, making counting efficient even for large data sets.

beginner

In Java, which class is commonly used to implement a frequency counter?

The java.util.HashMap class is commonly used to map elements to their frequency counts.

beginner

How do you update the count of an element in a frequency counter hash map?

Check if the element exists as a key; if yes, increase its count by 1; if no, add it with count 1.

beginner

What is a real-life example of the Frequency Counter Pattern?

Counting how many times each word appears in a book to find the most common words.

What data structure is best suited for implementing a frequency counter?

A Hash Map

B Array

C Linked List

D Stack

Hash maps provide fast insertion and lookup, ideal for counting frequencies.

In Java, which method increases the count of a key in a HashMap if it exists?

A containsKey(key) then put(key, get(key) + 1)

B get(key) + 1

C remove(key)

D put(key, value + 1)

You check if the key exists, then update its value by adding 1.

What is the initial count assigned to a new element in a frequency counter?

A null

B 0

C -1

D 1

When an element is first seen, its count starts at 1.

Which of these is NOT a use case for the Frequency Counter Pattern?

A Counting word occurrences

B Checking if two arrays have the same elements

C Sorting an array

D Finding duplicates

Sorting is unrelated to frequency counting.

What is the time complexity of building a frequency counter using a hash map for n elements?

A O(n^2)

B O(n)

C O(log n)

D O(1)

Each element is processed once, so time complexity is linear.

Explain how to use a hash map to count the frequency of elements in an array.

Think about how to store and update counts as you see each element.

Describe a real-world problem where the Frequency Counter Pattern is useful and how you would solve it.

Consider counting words, votes, or items sold.

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