The Letter Census

An easy Python interview practice problem on DataDriven. Write the Python and run it against test cases, with instant feedback.

Domain
Python
Difficulty
easy
Seniority
L3
Asked in
Quantiphi

Problem

Given a string, return a dict with keys 'vowels' (count of a, e, i, o, u, case-insensitive) and 'consonants' (count of other alphabetic characters). Non-alphabetic chars are ignored.

Every crowd has its share of talkers and quiet ones.

Example 1

Input:

s = "Hello World"

Output:

{"vowels":3,"consonants":7}

Example 2

Input:

s = "aeiou"

Output:

{"vowels":5,"consonants":0}

Worked solution and explanation

Why this problem exists in real interviews

This tests character classification combined with counting, probing whether a candidate can separate vowels from consonants while ignoring non-alphabetic characters.

Break down the requirements

Step 1: Define the vowel set

Use a set {'a', 'e', 'i', 'o', 'u'} for O(1) lookups.

Step 2: Iterate and classify each character

Check if alphabetic first, then classify as vowel or consonant.

Step 3: Return counts in a dict

Keys are 'vowels' and 'consonants'.

The solution

Character classification with set lookup
def letter_census(s: str) -> dict:
    vowels_set = {'a', 'e', 'i', 'o', 'u'}
    vowel_count = 0
    consonant_count = 0
    for ch in s:
        lower = ch.lower()
        if lower.isalpha():
            if lower in vowels_set:
                vowel_count += 1
            else:
                consonant_count += 1
    result = {'vowels': vowel_count, 'consonants': consonant_count}
    return result

Common follow-up questions