Character Frequency Counting in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of character frequency count using dictionary

What is the output of the following Python code that counts character frequency in a string?

DSA Python

text = "hello"
freq = {}
for ch in text:
    freq[ch] = freq.get(ch, 0) + 1
print(freq)

A {"h": 1, "e": 1, "l": 2, "o": 1}

B {"h": 1, "e": 1, "l": 1, "o": 1}

C {"h": 1, "e": 1, "l": 3, "o": 1}

D {"h": 0, "e": 1, "l": 2, "o": 1}

💡 Hint

Count how many times each letter appears in the word 'hello'.

The code loops through each character in the string 'hello'. It adds 1 to the count for each character in the dictionary. The letter 'l' appears twice, so its count is 2. All other letters appear once.

🧠 Conceptual

intermediate

Understanding character frequency counting with default dictionary

Which Python data structure is best suited to count character frequencies without manually checking if a key exists?

A set

B collections.defaultdict(int)

C list

D tuple

💡 Hint

Think about a dictionary that automatically gives zero for missing keys.

defaultdict(int) automatically initializes missing keys with 0, making counting easier without manual checks.

🔧 Debug

advanced

Identify the error in character frequency counting code

What error does the following code produce when counting characters in 'test'?

DSA Python

text = "test"
freq = {}
for ch in text:
    freq[ch] += 1
print(freq)

A KeyError

B SyntaxError

C TypeError

D No error, outputs {'t': 2, 'e': 1, 's': 1}

💡 Hint

Check what happens when you try to add 1 to a key that does not exist yet.

The dictionary freq is empty initially. Trying to do freq[ch] += 1 without initializing freq[ch] causes a KeyError because the key does not exist.

❓ Predict Output

advanced

Output of character frequency count using Counter

What is the output of this Python code using collections.Counter to count characters in 'banana'?

DSA Python

from collections import Counter
text = "banana"
freq = Counter(text)
print(dict(freq))

A {"b": 1, "a": 2, "n": 3}

B {"b": 1, "a": 1, "n": 1}

C {"b": 2, "a": 3, "n": 1}

D {"b": 1, "a": 3, "n": 2}

💡 Hint

Count how many times each letter appears in 'banana'.

Counter counts each character's occurrences. 'a' appears 3 times, 'n' 2 times, and 'b' once.

🚀 Application

expert

Find the character with the highest frequency

Given the string 'abracadabra', which character has the highest frequency and what is its count?

DSA Python

text = "abracadabra"
freq = {}
for ch in text:
    freq[ch] = freq.get(ch, 0) + 1
max_char = max(freq, key=freq.get)
max_count = freq[max_char]
print(max_char, max_count)

A c 1

B b 2

C a 5

D r 2

💡 Hint

Count how many times each letter appears and find the one with the largest count.

The letter 'a' appears 5 times, which is more than any other character in 'abracadabra'.

Practice

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

easy

A. How many times each character appears

B. The length of the string

C. The first character in the string

D. The position of the last character

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: Compare options with the definition

   Only How many times each character appears matches the purpose of character frequency counting.

3. Final Answer:

   How many times each character appears -> Option A

4. Quick Check:

   Frequency counting = count characters [OK]

Hint: Frequency counting means counting characters [OK]

Common Mistakes:

• Confusing frequency with string length
• Thinking it finds character positions
• Mixing frequency with sorting

2. Which of the following Python code snippets correctly counts character frequency in string s?

easy

A. freq = 0; for ch in s: freq += 1

B. freq = []; for ch in s: freq.append(ch)

C. freq = {}; for ch in s: freq[ch] = freq.get(ch, 0) + 1

D. freq = {}; for ch in s: freq[ch] = 1

Solution

1. Step 1: Analyze each code snippet

   freq = {}; for ch in s: freq[ch] = freq.get(ch, 0) + 1 uses a dictionary and increments count for each character correctly. freq = []; for ch in s: freq.append(ch) just collects characters in a list. freq = 0; for ch in s: freq += 1 counts total characters, not frequency per character. freq = {}; for ch in s: freq[ch] = 1 sets frequency to 1 always, ignoring repeats.

2. Step 2: Confirm correct frequency counting logic

   Only freq = {}; for ch in s: freq[ch] = freq.get(ch, 0) + 1 correctly counts how many times each character appears.

3. Final Answer:

   freq = {}; for ch in s: freq[ch] = freq.get(ch, 0) + 1 -> Option C

4. Quick Check:

   Use dict with get() to count [OK]

Hint: Use dict with get() to count frequencies [OK]

Common Mistakes:

• Using list instead of dict for counting
• Overwriting counts instead of incrementing
• Counting total length instead of per character

3. What is the output of this code?

text = 'hello'
freq = {}
for ch in text:
    freq[ch] = freq.get(ch, 0) + 1
print(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. {'h': 1, 'e': 1, 'l': 2}

Solution

1. Step 1: Count characters in 'hello'

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

2. Step 2: Check code logic

   Code correctly counts each character's frequency using dict and get().

3. Final Answer:

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

4. Quick Check:

   Count 'l' twice, others once [OK]

Hint: Count repeated letters carefully [OK]

Common Mistakes:

• Counting 'l' only once
• Missing last character
• Incorrect dictionary format

4. Find the error in this code that counts character frequency:

text = 'apple'
freq = {}
for ch in text:
    freq[ch] += 1
print(freq)

medium

A. freq[ch] is used before initialization

B. Loop variable name is wrong

C. print statement syntax error

D. String 'apple' is invalid

Solution

1. Step 1: Identify how freq[ch] is updated

   freq[ch] += 1 tries to add 1 to freq[ch] without initializing freq[ch] first.

2. Step 2: Understand dictionary key initialization

   Since freq is empty initially, freq[ch] does not exist and causes an error.

3. Final Answer:

   freq[ch] is used before initialization -> Option A

4. Quick Check:

   Initialize dict keys before incrementing [OK]

Hint: Initialize dict keys before incrementing [OK]

Common Mistakes:

• Assuming dict keys auto-initialize
• Ignoring error messages
• Changing loop variable unnecessarily

5. Given a string s = 'banana', which code snippet correctly returns a dictionary with only characters that appear more than once?

hard

A. freq = {ch: 1 for ch in s if s.count(ch) > 1}

B. freq = {ch: s.count(ch) for ch in s}

C. freq = {ch for ch in s if s.count(ch) > 1}

D. freq = {ch: s.count(ch) for ch in set(s) if s.count(ch) > 1}

Solution

1. Step 1: Understand the goal

   We want a dictionary with characters appearing more than once and their counts.

2. Step 2: Analyze each option

   freq = {ch: 1 for ch in s if s.count(ch) > 1} sets count to 1 always, ignoring actual counts. freq = {ch: s.count(ch) for ch in set(s) if s.count(ch) > 1} uses set(s) to avoid duplicates and filters counts > 1 correctly. freq = {ch for ch in s if s.count(ch) > 1} creates a set instead of dictionary with counts. freq = {ch: s.count(ch) for ch in s} includes all characters regardless of count.

3. Final Answer:

   freq = {ch: s.count(ch) for ch in set(s) if s.count(ch) > 1} -> Option D

4. Quick Check:

   Use set to avoid duplicates and filter counts [OK]

Hint: Use set(s) and filter counts > 1 [OK]

Common Mistakes:

• Not using set to avoid duplicate keys
• Assigning wrong counts
• Including characters with single occurrence