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3227. Vowels Game in a String
Description
Alice and Bob are playing a game on a string.
You are given a string s
, Alice and Bob will take turns playing the following game where Alice starts first:
 On Alice's turn, she has to remove any nonempty substring from
s
that contains an odd number of vowels.  On Bob's turn, he has to remove any nonempty substring from
s
that contains an even number of vowels.
The first player who cannot make a move on their turn loses the game. We assume that both Alice and Bob play optimally.
Return true
if Alice wins the game, and false
otherwise.
The English vowels are: a
, e
, i
, o
, and u
.
Example 1:
Input: s = "leetcoder"
Output: true
Explanation:
Alice can win the game as follows:
 Alice plays first, she can delete the underlined substring in
s = "leetcoder"
which contains 3 vowels. The resulting string iss = "der"
.  Bob plays second, he can delete the underlined substring in
s = "der"
which contains 0 vowels. The resulting string iss = "er"
.  Alice plays third, she can delete the whole string
s = "er"
which contains 1 vowel.  Bob plays fourth, since the string is empty, there is no valid play for Bob. So Alice wins the game.
Example 2:
Input: s = "bbcd"
Output: false
Explanation:
There is no valid play for Alice in her first turn, so Alice loses the game.
Constraints:
1 <= s.length <= 10^{5}
s
consists only of lowercase English letters.
Solutions
Solution 1: Brain Teaser
Let’s denote the number of vowels in the string as $k$.
If $k = 0$, meaning there are no vowels in the string, then Little Red cannot remove any substring, and Little Ming wins directly.
If $k$ is odd, then Little Red can remove the entire string, resulting in a direct win for Little Red.
If $k$ is even, then Little Red can remove $k  1$ vowels, leaving one vowel in the string. In this case, Little Ming cannot remove any substring, leading to a direct win for Little Red.
In conclusion, if the string contains vowels, then Little Red wins; otherwise, Little Ming wins.
The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.

class Solution { public boolean doesAliceWin(String s) { for (int i = 0; i < s.length(); ++i) { if ("aeiou".indexOf(s.charAt(i)) != 1) { return true; } } return false; } }

class Solution { public: bool doesAliceWin(string s) { string vowels = "aeiou"; for (char c : s) { if (vowels.find(c) != string::npos) { return true; } } return false; } };

class Solution: def doesAliceWin(self, s: str) > bool: vowels = set("aeiou") return any(c in vowels for c in s)

func doesAliceWin(s string) bool { vowels := "aeiou" for _, c := range s { if strings.ContainsRune(vowels, c) { return true } } return false }

function doesAliceWin(s: string): boolean { const vowels = 'aeiou'; for (const c of s) { if (vowels.includes(c)) { return true; } } return false; }