#409 flippedy

Using python to count vowels and flip words; cassidoo’s interview question of the week (2026-01-26).

Notes

The interview question of the week (2026-01-26):

You are given a string consisting of lowercase words, each separated by a single space. Determine how many vowels appear in the first word. Then, reverse each following word that has the same vowel count.

Examples:
flippedy("cat and mice")
> "cat dna mice"

flippedy("banana healthy")
> "banana healthy"

Thinking about the Problem

Seems pretty straight-forward: count the vowels in each word:

vowels: aeiou
don’t forget: case-insensitive

There is a typical requirement ambiguity however: “Determine how many vowels appear in the first word” could mean two different things:

count the number of characters that are vowels i.e “Wheee!” == 3
OR count how many vowels appear in the word i.e “Wheee!” == 1

The examples don’t answer this definitively. I’ll go with the first interpretation

A First Go

Breaking the problem down into 2 simple parts:

count vowels in a word
iterate the words, flipping as required

For a first go, I’ll just used list comprehensions:

def vowel_count(word):
    vowels = 'aeiouAEIOU'
    return sum(1 for char in word if char in vowels)

def flippedy(input):
    words = input.split()
    first_word = words.pop(0)
    target_count = vowel_count(first_word)
    flipped_words = [word[::-1] if vowel_count(word) == target_count else word for word in words]
    return ' '.join([first_word] + flipped_words)

The expression word[::-1] is a neat way of reversing a string in python.

And that works nicely:

$ ./challenge.py "cat and mice"
cat dna mice
$ ./challenge.py "banana healthy"
banana healthy

Refining the Code

The list comprehension is a bit ungainly for counting vowels. How about a regex approach instead:

def vowel_count(word):
    return len(re.findall(r'[aeiou]', word, re.IGNORECASE))

Tests

I’ve setup some validation in test_challenge.py:

$ ./test_challenge.py
....
----------------------------------------------------------------------
Ran 4 tests in 0.000s

OK

Example Code

Final code is in challenge.py:

#! /usr/bin/env python
from sys import argv
import re

def vowel_count(word):
    return len(re.findall(r'[aeiou]', word, re.IGNORECASE))

def flippedy(input):
    words = input.split()
    first_word = words.pop(0)
    target_count = vowel_count(first_word)
    flipped_words = [word[::-1] if vowel_count(word) == target_count else word for word in words]
    flipped_words.insert(0, first_word)
    return ' '.join(flipped_words)

if __name__ == '__main__':
    if len(argv) == 2:
        print(flippedy(argv[1]))
    else:
        print("Usage: challenge.py '<string>'")

Credits and References

cassidoo’s interview question of the week (2026-01-26)

Project Notes