Character Frequency Counting in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of character frequency count in a string

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

DSA Java

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

A {h=1, e=1, l=1, o=1}

B {h=1, e=1, l=3, o=1}

C {h=1, e=1, l=2, o=1}

D {h=1, e=2, l=2, o=1}

💡 Hint

Count how many times each letter appears in "hello".

The letter 'l' appears twice, all others once.

🧠 Conceptual

intermediate

Understanding character frequency counting approach

Which data structure is best suited to count the frequency of characters in a string efficiently?

A Array of size 256 indexed by ASCII values

B Linked list storing characters and counts

C Stack to push characters and pop counts

D Queue to enqueue characters and dequeue counts

💡 Hint

Think about direct indexing by character code.

An array indexed by ASCII codes allows O(1) access to counts for each character.

🔧 Debug

advanced

Identify the error in character frequency counting code

What error does the following Java code produce when counting character frequencies? String s = "abc"; Map freq = new HashMap<>(); for (int i = 0; i < s.length(); i++) { char c = s.charAt(i); freq.put(c, freq.get(c) + 1); } System.out.println(freq);

A IndexOutOfBoundsException

B NullPointerException

C ClassCastException

D No error, prints {a=1, b=1, c=1}

💡 Hint

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

freq.get(c) returns null initially, so null + 1 causes NullPointerException.

🚀 Application

advanced

Find the most frequent character in a string

Given a string, which code snippet correctly finds the character with the highest frequency?

DSA Java

String s = "banana";
Map<Character, Integer> freq = new HashMap<>();
for (char c : s.toCharArray()) {
  freq.put(c, freq.getOrDefault(c, 0) + 1);
}
char maxChar = ' ';
int maxCount = 0;
// Which option correctly finds maxChar and maxCount?

A for (char c : freq.keySet()) { maxCount = freq.get(c); maxChar = c; }

B for (char c : s.toCharArray()) { maxCount += freq.get(c); maxChar = c; }

C for (char c : s.toCharArray()) { if (freq.get(c) < maxCount) { maxCount = freq.get(c); maxChar = c; } }

D for (char c : freq.keySet()) { if (freq.get(c) > maxCount) { maxCount = freq.get(c); maxChar = c; } }

💡 Hint

Check which loop and condition correctly track the max frequency.

Option D iterates over unique chars and updates maxCount and maxChar correctly.

❓ Predict Output

expert

Output of character frequency counting with Unicode characters

What is the output of this Java code counting frequencies of Unicode characters in the string "aáa"?

DSA Java

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

A {a=2, á=1}

B {a=3}

C {a=1, á=2}

D {a=1, á=2}

💡 Hint

Remember that 'a' and 'á' are different characters in Unicode.

The string has two 'a' characters and one 'á' character, counted separately.

Practice

1. What does character frequency counting in a string help us find?

easy

A. The first character of the string

B. The length of the string

C. How many times each character appears in the string

D. The position of a character in the string

Solution

1. Step 1: Understand the purpose of frequency counting

   Frequency counting means counting how many times each character appears in a string.

2. Step 2: Identify what frequency counting returns

   It returns a count for each character, not the length or position.

3. Final Answer:

   How many times each character appears in the string -> Option C

4. Quick Check:

   Frequency counting = character counts [OK]

Hint: Frequency counting means counting each character's occurrences [OK]

Common Mistakes:

• Confusing frequency with string length
• Thinking frequency gives character positions
• Assuming frequency returns only unique characters

2. Which Java data structure is best suited to store character frequencies for a string?

easy

A. HashMap

B. ArrayList<Character>

C. LinkedList<Integer>

D. Stack<Character>

Solution

1. Step 1: Identify data structure needs

   We need to map each character to its count, so a key-value pair structure is needed.

2. Step 2: Choose the correct Java collection

   HashMap stores characters as keys and their counts as values.

3. Final Answer:

   HashMap -> Option A

4. Quick Check:

   Map characters to counts = HashMap [OK]

Hint: Use HashMap to map characters to their counts [OK]

Common Mistakes:

• Using ArrayList which doesn't map keys to values
• Choosing LinkedList which is a list, not a map
• Using Stack which is for LIFO, not frequency counting

3. What is the output of the following Java code?

String s = "hello";
Map freq = new HashMap<>();
for (char c : s.toCharArray()) {
  freq.put(c, freq.getOrDefault(c, 0) + 1);
}
System.out.println(freq);

medium

A. {h=1, e=1, l=1, o=1}

B. {h=1, e=1, l=2, o=1}

C. {h=1, e=1, l=3, o=1}

D. Compilation error

Solution

1. Step 1: Count characters in "hello"

   Characters: h(1), e(1), l(2), o(1).

2. Step 2: Check code logic for counting

   Code uses getOrDefault to add 1 to existing count or start at 0, correctly counting frequencies.

3. Final Answer:

   {h=1, e=1, l=2, o=1} -> Option B

4. Quick Check:

   Count 'l' twice, others once = {h=1, e=1, l=2, o=1} [OK]

Hint: Count each character carefully, watch repeated letters [OK]

Common Mistakes:

• Counting 'l' only once
• Confusing getOrDefault usage
• Expecting syntax error incorrectly

4. Find the error in this Java code snippet for counting character frequency:

String s = "test";
Map freq = new HashMap<>();
for (int i = 0; i <= s.length(); i++) {
  char c = s.charAt(i);
  freq.put(c, freq.getOrDefault(c, 0) + 1);
}

medium

A. Using <= in loop causes IndexOutOfBoundsException

B. HashMap cannot store Character keys

C. getOrDefault method is not available in HashMap

D. charAt method does not exist for String

Solution

1. Step 1: Analyze loop condition

   Loop uses i <= s.length(), but valid indices are 0 to s.length()-1.

2. Step 2: Identify error consequence

   Accessing s.charAt(s.length()) causes IndexOutOfBoundsException at runtime.

3. Final Answer:

   Using <= in loop causes IndexOutOfBoundsException -> Option A

4. Quick Check:

   Loop index must be less than length [OK]

Hint: Loop index must be less than string length, not equal [OK]

Common Mistakes:

• Thinking HashMap can't store Character keys
• Assuming getOrDefault is unavailable
• Believing charAt doesn't exist for String

5. Given a string with mixed case letters, how can you count character frequencies ignoring case (treat 'A' and 'a' as same)? Choose the best approach.

hard

A. Ignore characters that are uppercase

B. Count uppercase and lowercase separately and add later

C. Use a TreeMap instead of HashMap

D. Convert string to lowercase before counting frequencies

Solution

1. Step 1: Understand case-insensitive counting

   To treat 'A' and 'a' the same, convert all characters to one case.

2. Step 2: Choose best method

   Converting entire string to lowercase before counting ensures all letters are uniform.

3. Final Answer:

   Convert string to lowercase before counting frequencies -> Option D

4. Quick Check:

   Uniform case simplifies frequency counting [OK]

Hint: Convert string to one case before counting [OK]

Common Mistakes:

• Counting uppercase and lowercase separately
• Using TreeMap which doesn't solve case issue
• Ignoring uppercase letters incorrectly