Frequency Counter Pattern Using Hash Map in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Frequency Counter for Characters

What is the output of the following Java code that counts character frequencies in a string?

DSA Java

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

A {b=1, a=2, n=3}

B {a=2, b=1, n=3}

C {b=1, a=3, n=2}

D {a=3, b=2, n=1}

💡 Hint

Count how many times each character appears in "banana".

The string "banana" has 1 'b', 3 'a's, and 2 'n's. The HashMap stores these counts as key-value pairs.

❓ Predict Output

intermediate

Frequency Counter for Integer Array

What is the output of this Java code that counts frequencies of integers in an array?

DSA Java

import java.util.HashMap;
public class Main {
    public static void main(String[] args) {
        int[] nums = {2, 3, 2, 4, 3, 2};
        HashMap<Integer, Integer> freq = new HashMap<>();
        for (int num : nums) {
            freq.put(num, freq.getOrDefault(num, 0) + 1);
        }
        System.out.println(freq);
    }
}

A {2=2, 3=3, 4=1}

B {2=3, 3=2, 4=1}

C {2=3, 3=1, 4=2}

D {2=1, 3=2, 4=3}

💡 Hint

Count how many times each number appears in the array.

The number 2 appears 3 times, 3 appears 2 times, and 4 appears once.

🔧 Debug

advanced

Identify the Error in Frequency Counter Code

What error will this Java code produce when trying to count character frequencies?

DSA Java

import java.util.HashMap;
public class Main {
    public static void main(String[] args) {
        String input = "apple";
        HashMap<Character, Integer> freq = new HashMap<>();
        for (int i = 0; i < input.length(); i++) {
            char c = input.charAt(i);
            freq.put(c, freq.get(c) + 1);
        }
        System.out.println(freq);
    }
}

A ArrayIndexOutOfBoundsException

B ClassCastException

C No error, prints {a=1, p=2, l=1, e=1}

D NullPointerException

💡 Hint

Check what happens when freq.get(c) is called for a character not yet in the map.

Calling freq.get(c) for a character not yet in the map returns null. Adding 1 to null causes NullPointerException.

🚀 Application

advanced

Find if Two Strings are Anagrams Using Frequency Counter

Which option correctly uses a frequency counter HashMap to check if two strings are anagrams?

A Count characters in first string, decrement counts using second string, then check all counts are zero.

B Sort both strings and compare if they are equal.

C Count characters in first string only and compare length with second string.

D Use a HashMap to count characters in second string only.

💡 Hint

Anagrams have the same character counts in both strings.

Option A uses frequency counters to count characters in the first string, then reduces counts with the second string. If all counts return to zero, strings are anagrams.

🧠 Conceptual

expert

Frequency Counter Pattern Efficiency

What is the time complexity of using a frequency counter HashMap to count characters in a string of length n?

A O(n) because each character is processed once with constant time map operations.

B O(log n) because HashMap operations take logarithmic time.

C O(n^2) because nested loops are needed to count frequencies.

D O(1) because counting is done instantly.

💡 Hint

Consider how many times each character is visited and the cost of HashMap operations.

Each character is processed once, and HashMap get/put operations are average O(1), so total time is O(n).

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