# 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 non-empty substring from s that contains an odd number of vowels.
• On Bob's turn, he has to remove any non-empty 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 is s = "der".
• Bob plays second, he can delete the underlined substring in s = "der" which contains 0 vowels. The resulting string is s = "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 <= 105
• 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;
}