Frequency Counter Pattern Using Hash Map in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of frequency count for characters in a string

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

DSA Python

def char_frequency(s):
    freq = {}
    for char in s:
        freq[char] = freq.get(char, 0) + 1
    return freq

result = char_frequency('banana')
print(result)

A {'b': 2, 'a': 3, 'n': 2}

B {'b': 1, 'a': 2, 'n': 3}

C {'b': 1, 'a': 3, 'n': 3}

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

💡 Hint

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

The function counts each character's occurrences. 'b' appears once, 'a' appears three times, and 'n' appears twice.

❓ Predict Output

intermediate

Frequency count of numbers in a list

What will be printed after running this code that counts frequencies of numbers in a list?

DSA Python

def count_numbers(nums):
    freq = {}
    for num in nums:
        freq[num] = freq.get(num, 0) + 1
    return freq

print(count_numbers([1,2,2,3,3,3,4]))

A {1: 1, 2: 2, 3: 2, 4: 1}

B {1: 1, 2: 2, 3: 3, 4: 1}

C {1: 1, 2: 3, 3: 2, 4: 1}

D {"1": 1, "2": 2, "3": 3, "4": 1}

💡 Hint

Check how many times each number appears in the list.

The list has one 1, two 2s, three 3s, and one 4. The dictionary keys are integers, not strings.

🔧 Debug

advanced

Identify the error in frequency counter code

What error does this code raise when counting frequencies of elements in a list?

DSA Python

def freq_counter(arr):
    freq = {}
    for item in arr:
        freq[item] += 1
    return freq

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

A KeyError

B TypeError

C SyntaxError

D No error, outputs {1:1, 2:2, 3:1}

💡 Hint

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

The code tries to increment freq[item] without initializing it first, causing a KeyError.

🚀 Application

advanced

Find if two strings are anagrams using frequency counter

Given two strings, which code correctly uses frequency counters to check if they are anagrams?

A

def are_anagrams(s1, s2):
    if len(s1) != len(s2):
        return False
    freq1 = {}
    freq2 = {}
    for c in s1:
        freq1[c] = freq1.get(c, 0) + 1
    for c in s2:
        freq2[c] = freq2.get(c, 0) + 1
    return freq1 == freq2

B

def are_anagrams(s1, s2):
    return sorted(s1) == sorted(s2)

C

def are_anagrams(s1, s2):
    freq = {}
    for c in s1:
        freq[c] = freq.get(c, 0) + 1
    for c in s2:
        if c not in freq:
            return False
        freq[c] -= 1
    return all(value == 0 for value in freq.values())

D

def are_anagrams(s1, s2):
    freq = {}
    for c in s1:
        freq[c] = freq.get(c, 0) + 1
    for c in s2:
        freq[c] = freq.get(c, 0) - 1
    return all(value == 0 for value in freq.values())

💡 Hint

Check that frequencies match by incrementing for first string and decrementing for second.

Option C correctly increments counts for s1, decrements for s2, and checks all counts are zero, confirming anagrams.

🧠 Conceptual

expert

Time complexity of frequency counter pattern

What is the time complexity of counting frequencies of elements in a list of length n using a hash map?

A O(n)

B O(n log n)

C O(n^2)

D O(log n)

💡 Hint

Consider how many times each element is processed and the average time for hash map operations.

Each element is processed once, and hash map insertions and lookups are average O(1), so total is O(n).

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