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).