Character Frequency Counting in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new HashMap for counting characters.

DSA Java

Map<Character, Integer> freqMap = new [1]<>();

Drag options to blanks, or click blank then click option'

AArrayList

BHashMap

CLinkedList

DTreeSet

Attempts:

3 left

💡 Hint

Common Mistakes

Using a list instead of a map for counting frequencies.

Choosing a set which does not store counts.

We use HashMap to store character keys and their frequency counts.

2fill in blank

medium

Complete the code to get the character at index i from the string.

DSA Java

char ch = input.[1](i);

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Asubstring

BindexOf

CcharAt

Dlength

Attempts:

3 left

💡 Hint

Common Mistakes

Using substring which returns a string, not a char.

Using indexOf which finds the position of a character.

charAt(i) returns the character at position i in the string.

3fill in blank

hard

Fix the error in updating the frequency count for the character.

DSA Java

freqMap.put(ch, freqMap.getOrDefault(ch, [1]) + 1);

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A0

B-1

C1

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Using null causes NullPointerException.

Starting count at 1 causes double counting.

getOrDefault returns 0 if the character is not yet in the map, so we can add 1 safely.

4fill in blank

hard

Fill both blanks to print each character and its frequency.

DSA Java

for (Map.Entry<Character, Integer> entry : freqMap.[1]()) {
    System.out.println(entry.[2]() + " -> " + entry.getValue());
}

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AentrySet

BkeySet

CgetKey

DgetEntry

Attempts:

3 left

💡 Hint

Common Mistakes

Using keySet() only gives keys, not entries.

getEntry() is not a valid method.

entrySet() returns all key-value pairs; getKey() gets the character key.

5fill in blank

hard

Fill all three blanks to create a frequency map using streams.

DSA Java

Map<Character, Long> freqMap = input.chars()
    .mapToObj(c -> (char) [1])
    .collect(Collectors.[2](c -> c, Collectors.[3]()));

Drag options to blanks, or click blank then click option'

Ac

BtoMap

CgroupingBy

Dcounting

Attempts:

3 left

💡 Hint

Common Mistakes

Using toMap instead of groupingBy causes errors.

Not casting int to char results in wrong keys.

We cast int to char with (char) c, group by character, and count occurrences.

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