877. Stone Game

Problem Description

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 across all the piles is odd, so there are no ties.

Alice and Bob take turns, with Alice starting first. Each turn, a player takes the entire pile of stones either from the beginning or from the end of the row. This continues until there are no more piles left.

Assuming Alice and Bob play optimally, return true if Alice wins the game, or false if Bob wins.

Examples:

Input: piles = [5,3,4,5]
Output: true
Explanation: 
Alice starts first, and can only take the first 5 or the last 5.
Say she takes the first 5, leaving piles = [3,4,5].
If Bob takes 3, then the board is [4,5], and Alice takes 5 to win with 10 points.
If Bob takes the last 5, then the board is [3,4], and Alice takes 4 to win with 9 points.
This demonstrated that taking the first 5 was a winning move for Alice, so we return true.

Input: piles = [3,7,2,3]
Output: true

Constraints:

• 2 ≤ piles.length ≤ 500
• piles.length is even
• 1 ≤ piles[i] ≤ 500
• sum(piles[i]) is odd

Python Solution

class Solution:
    def stoneGame(self, piles: List[int]) -> bool:
        # Alice always wins!
        # Since the number of piles is even and sum is odd,
        # Alice can always choose all even-indexed or all odd-indexed piles
        # One of these choices will have more stones than the other
        return True

Implementation Notes:

• Mathematical proof shows Alice always wins
• Time complexity: O(1)
• Space complexity: O(1)

Java Solution

class Solution {
    public boolean stoneGame(int[] piles) {
        // Alice always wins!
        // Since the number of piles is even and sum is odd,
        // Alice can always choose all even-indexed or all odd-indexed piles
        // One of these choices will have more stones than the other
        return true;
    }
}

C++ Solution

class Solution {
public:
    bool stoneGame(vector& piles) {
        // Alice always wins!
        // Since the number of piles is even and sum is odd,
        // Alice can always choose all even-indexed or all odd-indexed piles
        // One of these choices will have more stones than the other
        return true;
    }
};

JavaScript Solution

/**
 * @param {number[]} piles
 * @return {boolean}
 */
var stoneGame = function(piles) {
    // Alice always wins!
    // Since the number of piles is even and sum is odd,
    // Alice can always choose all even-indexed or all odd-indexed piles
    // One of these choices will have more stones than the other
    return true;
};

C# Solution

public class Solution {
    public bool StoneGame(int[] piles) {
        // Alice always wins!
        // Since the number of piles is even and sum is odd,
        // Alice can always choose all even-indexed or all odd-indexed piles
        // One of these choices will have more stones than the other
        return true;
    }
}

Implementation Notes:

• Alice can always win by choosing the correct strategy
• She can choose either all even-indexed or all odd-indexed piles
• Since total sum is odd, one choice will always be larger