Frequency Counter Pattern Using Hash Map in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create an empty frequency counter dictionary.

DSA Python

freq = [1]

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

A{}

B[]

C()

Dset()

Attempts:

3 left

💡 Hint

Common Mistakes

Using square brackets creates a list, not a dictionary.

Using parentheses creates a tuple, not a dictionary.

We use curly braces {} to create an empty dictionary in Python, which is perfect for frequency counting.

2fill in blank

medium

Complete the code to increment the count of character 'char' in the frequency dictionary.

DSA Python

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

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

A//

B+

C*

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-' will decrease the count.

Using '*' or '//' will not correctly increment the count.

We add 1 to the current count to increment the frequency of the character.

3fill in blank

hard

Fix the error in the code to check if two strings have the same frequency of characters.

DSA Python

if freq_s1 == [1]:

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

Afreq_s2

Bfreq

Cfreq_s1

Dfreq_s3

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing the dictionary to itself always returns True.

Using an undefined variable causes errors.

To compare frequencies, we check if the frequency dictionary of the first string equals that of the second string.

4fill in blank

hard

Fill both blanks to create a frequency dictionary for characters in string 'text'.

DSA Python

freq = [1]
for char in [2]:
    freq[char] = freq.get(char, 0) + 1

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

A{}

Btext

Cstring

D[]

Attempts:

3 left

💡 Hint

Common Mistakes

Using a list instead of a dictionary for frequency.

Looping over an undefined variable.

We start with an empty dictionary {} and loop through each character in the string text to count frequencies.

5fill in blank

hard

Fill all three blanks to create a frequency dictionary for words longer than 3 characters in the list 'words'.

DSA Python

freq = [1]
for word in [2]:
    if len(word) [3] 3:
        freq[word] = freq.get(word, 0) + 1

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

A{}

Bwords

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong comparison operator.

Looping over a wrong variable.

Not initializing the dictionary.

We create an empty dictionary, loop through the list words, check if word length is greater than 3, then count frequencies.

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