0974MediumArrayStringPrefix Sum

Vowel-Bounded Labels in Each Stretch

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Trains the technique from

LeetCode 2559Count Vowel Strings in Ranges

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.

Labels are given as words, and each entry of queries is a pair [l, r] naming a stretch of the list, ends included.

A label is vowel-bounded when both its first and its last character are vowels, that is one of 'a', 'e', 'i', 'o' or 'u'. A label of one character counts when that character is a vowel.

Return, for each stretch in turn, how many vowel-bounded labels it holds.

Examples

Example 1

Input: words = ["opal", "brick", "aria", "echo", "damp"], queries = [[0, 4], [1, 3], [2, 2], [4, 4]]
Output: [2, 2, 1, 0]

"aria" and "echo" both begin and end with a vowel; "opal" ends in a consonant, and "brick" and "damp" begin with one. So the whole list holds two, the middle stretch holds two, and the last two stretches hold one and none.

Example 2

Input: words = ["ba"], queries = [[0, 0]]
Output: [0]

The label "ba" begins with a consonant, so it does not count even though it ends with a vowel.

Example 3

Input: words = ["a"], queries = [[0, 0]]
Output: [1]

A label of one character counts when that character is a vowel, so it is both the first and the last.

Constraints

1 <= words.length <= 10^5
1 <= words[i].length <= 40
Every entry of words consists of lowercase English letters only
1 <= queries.length <= 10^5
queries[i].length == 2
0 <= queries[i][0] <= queries[i][1] <= words.length - 1

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 vowel_strings(words: list[str], queries: list[list[int]]) -> list[int]:

Java

public int[] vowelStrings(String[] words, int[][] queries)

The intended approach

Time O(n + q) for n labels and q stretches.Space O(n) for the running totals.

Whether a label is vowel-bounded depends on that label alone, so decide it once for each and forget the labels entirely. That leaves counting marks in a stretch, which running totals answer outright: the total up to a position, less the total up to just before the stretch begins. So build the running totals in one pass over the labels, then answer each stretch with one subtraction. Answering a stretch by walking it would cost the stretch's length every time, which is far too slow when both the labels and the stretches run to a hundred thousand.

Pythonentry point: vowel_strings

def vowel_strings(words, queries):
    vowels = set("aeiou")
    n = len(words)
    running = [0] * (n + 1)
    for i in range(n):
        word = words[i]
        mark = 1 if word[0] in vowels and word[len(word) - 1] in vowels else 0
        running[i + 1] = running[i] + mark
    out = []
    for l, r in queries:
        out.append(running[r + 1] - running[l])
    return out

The obvious version

For each stretch, walk the labels inside it and test each one's two ends. Costs the stretch's length per query and builds nothing in advance.

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_vowel_strings(words, queries):
    vowels = "aeiou"
    out = []
    for l, r in queries:
        count = 0
        for i in range(l, r + 1):
            word = words[i]
            first_ok = False
            last_ok = False
            for v in vowels:
                if word[0] == v:
                    first_ok = True
                if word[len(word) - 1] == v:
                    last_ok = True
            if first_ok and last_ok:
                count += 1
        out.append(count)
    return out

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
3 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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