LeetMotion
Given a string s and an integer k, return the maximum number of vowel letters in any substring of s with length k.
Vowel letters in English are 'a', 'e', 'i', 'o', and 'u'.
r - k + 1. This means once the window is full, each slide adds one character on the right and removes one on the left simultaneously.count += (s[i] in vowels) adds the new character, and count -= (s[i-k] in vowels) removes the character that fell out of the window. This keeps the count in O(1) per step instead of O(k). The total complexity is O(n).s[0..k-1]. Then start the loop from index k, sliding the window and updating the count and maximum incrementally. This avoids an off-by-one and handles the edge case where the entire string is length k.In computational linguistics and phonetics research, analysts scan large text corpora to find the most "vowel-dense" segment of a fixed syllable count — useful for speech synthesis, poetry generation, and language learning tools. Rather than scoring every possible k-character window from scratch (O(n·k)), a fixed sliding window approach scans the text in a single O(n) pass by maintaining a running vowel count, making it practical for corpora with millions of characters.
Since the window is always exactly k characters wide, we only need one index. We seed the count for the first window, then slide rightward one position at a time: adding the new character's vowel contribution and subtracting the character that just left the window. We track the running maximum throughout.
count = number of vowels in s[0..k-1], set res = count.i from k to len(s)-1:1 if s[i] is a vowel, subtract 1 if s[i-k] is a vowel.res = max(res, count).res.Each character is visited exactly once. The window count update is O(1), giving O(n) total time and O(1) extra space.
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