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Alice and Bob play a game with piles of stones. There are an even number of piles arranged in a row, and each pile has a positive integer number of stones piles[i].

The objective of the game is to end with the most stones. The total number of stones is odd, so there is no tie.

Alice and Bob take turns, with Alice starting first. Each player can take a pile of stones from either the beginning or the end of the row.

Return True if Alice wins the game assuming both players play optimally, otherwise return False.

Clarifying Questions

1. What is the maximum length of the piles array?
   • This is important to understand if optimizations are needed for large inputs.
2. Is the sum of the stones always guaranteed to be odd?
   • Yes, as per the problem statement the total number of stones is odd to ensure no tie.
3. How do Alice and Bob take their turns?
   • They take turns alternatively, Alice starts first and they can pick from either end of the row.

Strategy

To solve this problem optimally, we can use dynamic programming. But there’s a simpler realization: Since Alice always starts first and there is an even number of piles, Alice can always ensure she picks either all odd indexed piles or all even indexed piles.

• If piles are indexed 0, 1, 2, 3, ..., n-1, Alice can always select the parity (even-indexed or odd-indexed piles) that gives her the most stones.
• Since both players play optimally and the total number of stones is odd ensuring no tie, Alice can always secure a larger total sum by picking the optimal piles. Hence, Alice always wins.

Code

def stoneGame(piles):
    return True

Time Complexity

• Time Complexity: O(1), because the solution doesn’t require processing the array.
• Space Complexity: O(1), as no additional space is required.

Alice will always win given the conditions of the game and both players playing optimally. This simplification leverages the insight that starting first on an even-length pile array with an odd total sum ensures Alice can always secure the victory.

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