Frequency Counter Pattern Using Hash Map in DSA Python - Build from Scratch

Frequency Counter Pattern Using Hash Map

📖 Scenario: You are working in a library to keep track of how many times each book title has been borrowed. This helps the library know which books are popular.

🎯 Goal: Build a program that counts how many times each book title appears in a list of borrowed books using a hash map (dictionary).

📋 What You'll Learn

Create a list called borrowed_books with exact book titles given.

Create an empty dictionary called frequency_counter to store counts.

Use a for loop to count how many times each book appears in borrowed_books.

Print the frequency_counter dictionary to show the counts.

💡 Why This Matters

🌍 Real World

Libraries, stores, and websites often count how many times items are used or bought to understand popularity.

💼 Career

Knowing how to count frequencies with hash maps (dictionaries) is a basic skill for data analysis, software development, and many programming jobs.

1

Create the list of borrowed books

Create a list called borrowed_books with these exact book titles in order: 'Harry Potter', 'The Hobbit', 'Harry Potter', '1984', 'The Hobbit', 'The Hobbit'.

DSA Python

# Create the list borrowed_books with the exact titles
# Your code here

Hint

Use square brackets [] to create a list and put the book titles as strings inside, separated by commas.

2

Create the frequency counter dictionary

Create an empty dictionary called frequency_counter to store the count of each book.

DSA Python

borrowed_books = ['Harry Potter', 'The Hobbit', 'Harry Potter', '1984', 'The Hobbit', 'The Hobbit']
# Create an empty dictionary called frequency_counter
# Your code here

Hint

Use curly braces {} to create an empty dictionary.

3

Count the frequency of each book

Use a for loop with variable book to go through borrowed_books. Inside the loop, update frequency_counter to count how many times each book appears.

DSA Python

borrowed_books = ['Harry Potter', 'The Hobbit', 'Harry Potter', '1984', 'The Hobbit', 'The Hobbit']
frequency_counter = {}
# Use a for loop to count each book's frequency
# Your code here

Hint

Check if the book is already a key in the dictionary. If yes, add 1 to its count. If no, set its count to 1.

4

Print the frequency counter dictionary

Print the frequency_counter dictionary to show the count of each book.

DSA Python

borrowed_books = ['Harry Potter', 'The Hobbit', 'Harry Potter', '1984', 'The Hobbit', 'The Hobbit']
frequency_counter = {}
for book in borrowed_books:
    if book in frequency_counter:
        frequency_counter[book] += 1
    else:
        frequency_counter[book] = 1
# Print the frequency_counter dictionary
# Your code here

Hint

Use print(frequency_counter) to display the dictionary.

Practice

1. What does the frequency counter pattern using a hash map help you do in a list of items?

easy

A. Count how many times each item appears

B. Sort the list in ascending order

C. Remove duplicate items from the list

D. Find the largest item in the list

Solution

1. Step 1: Understand frequency counting

   The frequency counter pattern counts occurrences of each item in a list using a hash map (dictionary).

2. Step 2: Compare options

   Only Count how many times each item appears describes counting occurrences. Sorting, removing duplicates, or finding largest are different tasks.

3. Final Answer:

   Count how many times each item appears -> Option A

4. Quick Check:

   Frequency counter = count occurrences [OK]

Hint: Frequency counter counts items, not sorts or removes [OK]

Common Mistakes:

• Confusing frequency counting with sorting
• Thinking it removes duplicates automatically
• Assuming it finds max/min values

2. Which of the following is the correct way to create an empty frequency counter using a hash map in Python?

easy

A. freq = []

B. freq = {}

C. freq = ()

D. freq = set()

Solution

1. Step 1: Identify data structure for frequency counter

   A frequency counter uses a hash map, which in Python is a dictionary, created with curly braces {}.

2. Step 2: Check options

   freq = {} uses {}, which creates an empty dictionary. Others create list, tuple, or set, which are incorrect here.

3. Final Answer:

   freq = {} -> Option B

4. Quick Check:

   Empty dict = {} for frequency counter [OK]

Hint: Use {} to create empty dictionary for frequency counter [OK]

Common Mistakes:

• Using [] which creates a list, not a dictionary
• Using () which creates a tuple
• Using set() which creates a set, not a dict

3. What will be the output of the following code?

def freq_counter(lst):
    freq = {}
    for item in lst:
        freq[item] = freq.get(item, 0) + 1
    return freq

print(freq_counter(['a', 'b', 'a', 'c', 'b', 'a']))

medium

A. {'a': 3, 'b': 2, 'c': 1}

B. {'a': 1, 'b': 1, 'c': 1}

C. {'a': 2, 'b': 2, 'c': 2}

D. Error: KeyError

Solution

1. Step 1: Count each item in the list

   Loop goes through ['a', 'b', 'a', 'c', 'b', 'a']. 'a' appears 3 times, 'b' 2 times, 'c' 1 time.

2. Step 2: Use freq.get(item, 0) + 1 to update counts

   freq.get(item, 0) returns current count or 0 if not found, then adds 1.

3. Final Answer:

   {'a': 3, 'b': 2, 'c': 1} -> Option A

4. Quick Check:

   Count 'a'=3, 'b'=2, 'c'=1 [OK]

Hint: Use dict.get(key,0)+1 to count frequencies safely [OK]

Common Mistakes:

• Forgetting to add 1 to count
• Using freq[item] without get causing KeyError
• Miscounting occurrences

4. Find the error in this frequency counter code and choose the fix:

def freq_counter(lst):
    freq = {}
    for item in lst:
        if item in freq:
            freq[item] += 1
        else:
            freq[item] = 0
    return freq

print(freq_counter(['x', 'y', 'x']))

medium

A. Initialize freq as a list instead of dict

B. Remove the if-else and use freq[item] += 1

C. No error, code works fine

D. Change freq[item] = 0 to freq[item] = 1

Solution

1. Step 1: Identify initialization error

   When item is new, freq[item] is set to 0, but it should start at 1 because we found it once.

2. Step 2: Fix initialization

   Change freq[item] = 0 to freq[item] = 1 to count first occurrence correctly.

3. Final Answer:

   Change freq[item] = 0 to freq[item] = 1 -> Option D

4. Quick Check:

   First occurrence count must be 1 [OK]

Hint: First time count must be 1, not 0 [OK]

Common Mistakes:

• Setting initial count to 0 instead of 1
• Using list instead of dict for frequency
• Ignoring the else block

5. Given two lists of integers, write a function using frequency counter pattern to check if the second list contains the same frequency of elements as the first list.
What will be the output of this code?

def same_frequency(lst1, lst2):
    if len(lst1) != len(lst2):
        return False
    freq1 = {}
    freq2 = {}
    for num in lst1:
        freq1[num] = freq1.get(num, 0) + 1
    for num in lst2:
        freq2[num] = freq2.get(num, 0) + 1
    return freq1 == freq2

print(same_frequency([1,2,2,3], [2,1,3,2]))
print(same_frequency([1,2,2,3], [2,1,3,3]))

hard

A. False\nFalse

B. False\nTrue

C. True\nFalse

D. True\nTrue

Solution

1. Step 1: Build frequency dictionaries for both lists

   freq1 counts {1:1, 2:2, 3:1}, freq2 for first call counts {2:2, 1:1, 3:1} which matches freq1.

2. Step 2: Compare frequency dictionaries

   For second call, freq2 counts {2:1, 1:1, 3:2} which differs from freq1, so returns False.

3. Final Answer:

   True False -> Option C

4. Quick Check:

   Same frequency = True, different frequency = False [OK]

Hint: Compare frequency dicts to check equal element counts [OK]

Common Mistakes:

• Only comparing lengths, not frequencies
• Not building frequency for both lists
• Returning True when frequencies differ