LeetCode #3227 — MEDIUM

Vowels Game in a String

Move from brute-force thinking to an efficient approach using math strategy.

Solve on LeetCode
The Problem

Problem Statement

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.

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

"leetcoder"

Example 2

"bbcd"
Step 02

Core Insight

What unlocks the optimal approach

  • If there are no vowels in the initial string, then Bob wins.
  • If the number of vowels in the initial string is odd, then Alice can remove the whole string on her first turn and win.
  • What if the number of vowels in the initial string is even? What’s the optimal play for Alice’s first turn?
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Upper-end input sizes
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3227: Vowels Game in a String
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;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(log n)
Space
O(1)

Approach Breakdown

ITERATIVE
O(n) time
O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT
O(log n) time
O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.