Character Frequency Counting in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize an empty dictionary for counting characters.

DSA Python

freq = [1]

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

A''

B{}

Cset()

D[]

Attempts:

3 left

💡 Hint

Common Mistakes

Using a list [] instead of a dictionary.

Using a string '' which cannot store counts.

Using a set() which does not store counts.

We use an empty dictionary {} to store character counts.

2fill in blank

medium

Complete the code to loop through each character in the string 'text'.

DSA Python

for [1] in text:
    freq[char] = freq.get(char, 0) + 1

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

Achar

Bletter

Cword

Ditem

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name different from the one inside the loop.

Using 'word' which implies multiple characters.

Using 'item' which is vague.

We use 'char' as the variable name to represent each character in the string.

3fill in blank

hard

Fix the error in the code to correctly update the frequency count.

DSA Python

freq[char] = freq.[1](char, 0) + 1

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

Aget

Bset

Cadd

Dappend

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'add' which is not a dictionary method.

Using 'set' which replaces the whole dictionary.

Using 'append' which is a list method.

The 'get' method returns the current count or 0 if the character is not yet in the dictionary.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that counts characters only if they appear more than once.

DSA Python

freq = {char: text.count(char) for char in [1] if text.count(char) [2] 1}

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

Aset(text)

Blist(text)

Ctext

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using list(text) which includes duplicates.

Using '<' instead of '>' causing wrong filtering.

Using dict(text) which is invalid here.

We use set(text) to avoid counting duplicates multiple times and '>' to filter characters appearing more than once.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that counts characters and includes only those with count at least 2.

DSA Python

freq = [1]: [2] for [3] in set(text) if text.count(char) >= 2

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

A{char

Btext.count(char)

Cchar

Dchar:

Attempts:

3 left

💡 Hint

Common Mistakes

Missing the opening curly brace.

Using wrong variable names.

Incorrect order of key and value.

The dictionary comprehension syntax requires curly brace start '{char', value expression 'text.count(char)', and loop variable 'char'.

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