How do you solve Vowels Game in a String on LeetCode?

Vowels Game in a String (LeetCode #3227) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Alice can win immediately if the total number of vowels is odd.

What is the brute force solution for Vowels Game in a String?

The brute force approach for Vowels Game in a String is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we would try to simulate the game by generating all possible substrings and checking if they contain an odd number of vowels. This method is straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Vowels Game in a String?

Vowels Game in a String uses the following concepts: Math, String, Brainteaser, Game Theory. The recommended patterns to study are: Game Theory, Counting, String Manipulation.

What are the interview tips for Vowels Game in a String?

Always consider edge cases, such as strings with no vowels. Think about the implications of the game's rules on the string's structure. Practice identifying patterns in problems to help streamline your approach.

What are common mistakes when solving Vowels Game in a String?

Not counting vowels correctly or misunderstanding the odd/even requirement. Overcomplicating the problem by trying to simulate every possible game state.

#3227

Vowels Game in a String

Medium

MathStringBrainteaserGame TheoryGame TheoryCountingString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution leverages the count of vowels in the entire string. If Alice starts with an odd number of vowels, she can remove the entire string and win immediately. If there are no vowels, she cannot make a move and loses.

⚙️

Algorithm

4 steps

1Step 1: Count the total number of vowels in the string.
2Step 2: If the count is 0, return false (Alice loses).
3Step 3: If the count is odd, return true (Alice wins).
4Step 4: If the count is even, return false (Alice loses).

solution.py3 lines

1def canAliceWin(s):
2    vowels = sum(1 for c in s if c in 'aeiou')
3    return vowels > 0 and vowels % 2 == 1

ℹ

Complexity note: The time complexity is O(n) because we only need to traverse the string once to count vowels. The space complexity is O(1) since we are using a fixed amount of space regardless of input size.

1Alice can win immediately if the total number of vowels is odd.
2If there are no vowels, Alice cannot make a move and loses.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.