Implement an algorithm to find the longest ordered subsequence of vowels in a given string.

Why This Pattern Matters: Subsequence DP is foundational for sequence alignment, log parsing (ordered event sequences), and NLP token sequences.

Architectural Logic: We track the longest valid subsequence ending at each vowel (a→e→i→o→u). State: dp[v] = max length ending at vowel v. Transition: when we see vowel v, we can extend from prev vowel. O(n) time, O(1) space—five vowel states.

Scalability: Linear scan, no extra structures—suitable for streaming. For billions of strings (e.g., URL/slug validation): run as Spark map-only job; embarrassingly parallel.

def longest_vowel_subsequence(s):
vowels, dp = 'aeiou', {v: 0 for v in 'aeiou'}
for c in s.lower():
if c in dp:
idx = vowels.index(c)
prev = vowels[idx-1] if idx > 0 else None
dp[c] = max(dp[c], (dp[prev] if prev else 0)) + 1
return dp.get('u', 0)