0392EasyHash TableString

All-Five Vowel Windows

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Trains the technique from

LeetCode 2062Count Vowel Substrings of a String

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A speech-training tool scans a lowercase transcription text for stretches that drill all five vowel sounds.

A vowel window is a non-empty contiguous stretch of text that contains no consonant and in which each of a, e, i, o and u appears at least once. Two windows count separately when they begin or end at different positions, even if they spell the same thing.

Return the number of vowel windows in text.

Examples

Example 1

Input: text = "aeioua"
Output: 3

The windows are "aeiou" starting at 0, "aeioua" starting at 0 and "eioua" starting at 1.

Example 2

Input: text = "oiuae"
Output: 1

The whole transcription is the only window; every shorter stretch is missing a vowel.

Example 3

Input: text = "aeiouzaeiou"
Output: 2

The five characters before the z form one window and the five after it form another; no window can span the z.

Example 4

Input: text = "banana"
Output: 0

No stretch of this transcription contains all five vowels.

Constraints

1 <= text.length <= 100
text consists of lowercase English letters only.

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def count_vowel_substrings(text: str) -> int:

Java

public int countVowelSubstrings(String text)

The intended approach

Time O(n)Space O(1)

A consonant can never sit inside a window, so the transcription splits into runs of vowels and each run is handled on its own; the position just after the last consonant is the earliest a window in the current run may begin. Walk the text keeping the most recent position of each of the five vowels. For the character at index `i`, a window ending there is valid exactly when it starts at or before the earliest of those five positions, so the number of windows ending at `i` is that earliest position minus the run's start, plus one. Clearing the five positions at every consonant restarts the count for the next run.

Pythonentry point: count_vowel_substrings

def count_vowel_substrings(text):
    # Time O(n) -- one pass, constant work per character.
    # Space O(1) -- five last-seen positions.
    vowels = "aeiou"
    last = {"a": -1, "e": -1, "i": -1, "o": -1, "u": -1}
    start = 0
    total = 0
    for i in range(len(text)):
        ch = text[i]
        if ch not in last:
            start = i + 1
            for v in vowels:
                last[v] = -1
            continue
        last[ch] = i
        edge = min(last[v] for v in vowels)
        if edge >= start:
            total += edge - start + 1
    return total

The obvious version

Cut out every contiguous stretch and keep the ones whose character set is exactly the five vowels, which rejects consonants and missing vowels in one test. Cubic in the length of the transcription.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_count_vowel_substrings(text):
    vowels = set("aeiou")
    total = 0
    n = len(text)
    for i in range(n):
        for j in range(i, n):
            piece = text[i:j + 1]
            if set(piece) == vowels:
                total += 1
    return total

How this problem was checked

Reference solution executed against 12 test cases — 12 passed
4 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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