Date: 2026-06-06
Time: 16:06
This file solves LeetCode 2586: Count the Number of Vowel Strings in Range. It counts how many strings in a subarray words[left..right] both start and end with a vowel. It's a straightforward array-scan problem classified as Easy.
VOWELS (module-level constant) — A set of the five lowercase vowels. Using a set gives O(1) membership testing, though for a 5-element collection the practical difference from a string in check is negligible.
is_vowel(char: str) -> bool — Thin predicate wrapping the set lookup. Extracted as a named function for readability in the generator expression below.
Solution.vowelStrings(self, words, left, right) -> int — The LeetCode entry point. Iterates indices left through right inclusive, counting words where both words[i][0] and words[i][-1] are vowels. Returns the count as an int.
sum(1 for i in ... if ...) is idiomatic Python for counting matches without materializing a list. It's lazy and memory-efficient.is_vowel lives outside Solution, keeping the class thin and the predicate reusable. This is a common pattern across the repo's solutions.range(left, right + 1)) rather than slicing words[left:right+1], avoiding an intermediate list allocation.Imports: Only typing.List for the type annotation — no external libraries.
Imported by: The testsolution.py in the same directory. The large "Imported By" list in the prompt is misleading — those are other problems' test files that import *their own* solution.py, not this one. The actual consumer is count-the-number-of-vowel-strings-in-range/testsolution.py.
1. Caller provides words, left, right.
2. A generator iterates i from left to right (inclusive).
3. For each i, it checks words[i][0] (first char) and words[i][-1] (last char) against VOWELS via is_vowel.
4. Matches contribute 1 to the running sum.
5. The total count is returned.
Time complexity: O(right - left + 1) — single pass, constant work per word.
Space complexity: O(1) — generator, no auxiliary data structures.
left and right are assumed to be valid indices into words with left <= right. No bounds checking is performed — the LeetCode contract guarantees this.words is assumed non-empty (so words[i][0] and words[i][-1] won't raise IndexError). Again guaranteed by the problem constraints.VOWELS.None. The code trusts the LeetCode runtime to supply valid inputs. An empty word would raise IndexError on words[i][0]; out-of-bounds left/right would silently produce wrong results or raise IndexError. This is appropriate for a competitive-programming context.
count-the-number-of-vowel-strings-in-range/test_solution.py — See which edge cases are tested (empty range, single-char words, all vowels vs. none)count-vowel-substrings-of-a-string/solution.py:vowelStrings — A harder vowel-substring problem that likely uses sliding window instead of a simple scancount-the-number-of-vowel-strings-in-range/review.md — Code review notes may flag alternative approaches (e.g., prefix sums for repeated queries)vowel-check-patterns — Compare how other solutions in the repo handle vowel detection (inline set, string in, regex)vowel-strings-range-linear-scan — vowelStrings performs a single O(n) pass with no precomputation; it does not use prefix sums or cachingis-vowel-lowercase-only — is_vowel matches only lowercase vowels (aeiou); uppercase input would return Falsevowel-set-module-level — The VOWELS set is allocated once at module load, not per method callno-input-validation — Neither is_vowel nor vowelStrings validates inputs; out-of-bounds indices or empty strings will raise unhandled exceptions