S2Q2 · Vowel Count of Words¶

⚡ Quick Reference

Type: Full I/O problem

Core idea: split the sentence into words, count vowels in each word, append the count in parentheses, rejoin with spaces.

s = input()
vowels = "aeiouAEIOU"
words = s.split(" ")
result = [f"{w}({sum(1 for c in w if c in vowels)})" for w in words]
print(" ".join(result))

Key rules: - Both lowercase and uppercase vowels count - Split on " " (single space), not split() - to preserve spacing behaviour - Output format: word(count) with no space before ( - Rejoin with a single space between words

Problem Statement¶

Problem (I/O type)

Read a string, count the vowels in each word, and print each word followed immediately by its vowel count in parentheses. Words are separated by spaces; spaces must be retained in the output.

Example:

Input

Computer science is amazing

Output

Computer(3) science(3) is(1) amazing(3)

Understanding the problem¶

Two things to handle per word:

Count vowels - how many characters in the word are in "aeiouAEIOU"
Format output - append the count immediately in parentheses: word(count)

Then join all formatted words back with spaces.

Counting vowels in a word

sum(1 for c in word if c in "aeiouAEIOU")

Walk through each character - if it's a vowel, count it. sum(1 for ...) is the idiomatic counter pattern.

Tracing the example¶

Input: "Computer science is amazing"

Word	Vowels found	Count
`Computer`	`o`, `u`, `e`	3
`science`	`i`, `e`, `e`	3
`is`	`i`	1
`amazing`	`a`, `i`, `a`	3

Output: Computer(3) science(3) is(1) amazing(3)

Solution approaches¶

Pythonic (one-liner)Explanatory (step by step)Pythonic (list comprehension)

s = input()
vowels = "aeiouAEIOU"
print(" ".join(f"{w}({sum(1 for c in w if c in vowels)})" for w in s.split(" ")))

Everything in one print call - split, count, format, and join in a single generator expression.

s = input()
vowels = "aeiouAEIOU"
words = s.split(" ")

result = []
for word in words:
    count = 0
    for c in word:
        if c in vowels:
            count += 1
    result.append(f"{word}({count})")

print(" ".join(result))

Each step named and visible - reading the input, counting vowels per word, building the result list, and printing.

s = input()
vowels = "aeiouAEIOU"
words = s.split(" ")
result = [f"{w}({sum(1 for c in w if c in vowels)})" for w in words]
print(" ".join(result))

Builds the result list with a comprehension, then joins and prints. Cleaner than the loop but still readable.

Why split(" ") and not split()?¶

split() with no argument splits on any whitespace and ignores leading/trailing spaces and multiple consecutive spaces. split(" ") splits on exactly one space and preserves empty strings between consecutive spaces.

For this problem either works since the input is a simple sentence with single spaces. However split(" ") is safer when the problem says "spaces should be retained" - it makes the intent explicit.

Key takeaways¶

01

sum(1 for c in w if c in vowels)

The standard pattern for counting characters matching a condition. Replace vowels with any set of target characters and it generalises to any character-counting problem.

02

f"{w}({count})" for formatting

f-strings handle the word(count) format cleanly. The count is computed inline - no need for a separate variable if using the Pythonic approach.

03

" ".join(list) to reassemble

After transforming each word into a string, " ".join(result) puts them back together with single spaces - exactly matching the expected output format.