Stone Game

class Solution {
public:
    bool canWin(vector<int>& stones) {
        int length = stones.size();

        int gameValue[length + 2][length + 2];
        memset(gameValue, 0, sizeof(gameValue));

        for (int size = 1; size <= length; ++size)
            for (int start = 0, end = size - 1; end < length; ++start, ++end) {
                int turn = (end + start + length) % 2;
                if (turn == 1)
                    gameValue[start + 1][end + 1] = max(stones[start] + gameValue[start + 2][end + 1], stones[end] + gameValue[start + 1][end]);
                else
                    gameValue[start + 1][end + 1] = min(-stones[start] + gameValue[start + 2][end + 1], -stones[end] + gameValue[start + 1][end]);
            }

        return gameValue[1][length] > 0;
    }
};

The provided C++ solution addresses the problem of determining if the first player can win a game where players take turns picking stones from either end of an array. The strategy uses a dynamic programming approach.

The function canWin within the Solution class evaluates whether the first player, given optimal play from both sides, can secure a victory. It accomplishes this using a 2D array gameValue to store intermediate results, ensuring each subproblem is only solved once, improving efficiency.

• First, the solution initializes a 2D dynamic programming table gameValue using the memset function to set all values to zero.
• Iterative computations fill in this table:
  • The outer for loop constructs solutions for subarrays of increasing size.
  • Pairs of inner for loops traverse these subarrays, computing the optimal game value for every possible segment of stones from start to end indices.
  • The game strategy alternates turns based on the combined position and size of the current segment, using modulo operation to determine the player's turn (turn).
  • On each player's turn, the strategy compares taking a stone from the start or the end of the current segment:
    • If it's the first player's turn (turn == 1), they aim to maximize their score by choosing the option that gives the higher result.
    • If it's the second player's turn (turn == 0), they aim to minimize the first player's score.
• The final decision, whether the first player can win or not, is determined by the value at gameValue[1][length], which represents the game's outcome when considering the entire array of stones starting with the first player.

By the end of processing, if gameValue[1][length] is positive, the first player can guarantee a victory with optimal play. This decision-making process leverages both game theory and dynamic programming principles to effectively determine the winner.

class Solution {
    public boolean stoneGame(int[] stones) {
        int length = stones.length;

        int[][] gameState = new int[length + 2][length + 2];
        for (int size = 1; size <= length; ++size)
            for (int start = 0; start + size <= length; ++start) {
                int end = start + size - 1;
                int turn = (end + start + length) % 2;
                if (turn == 1)
                    gameState[start + 1][end + 1] = Math.max(stones[start] + gameState[start + 2][end + 1], stones[end] + gameState[start + 1][end]);
                else
                    gameState[start + 1][end + 1] = Math.min(-stones[start] + gameState[start + 2][end + 1], -stones[end] + gameState[start + 1][end]);
            }

        return gameState[1][length] > 0;
    }
}

This solution summary explains the implementation of the "Stone Game" using Java. The algorithm plays out a game where players pick a stone from either end of an array of stones, aiming to maximize their score.

• Start by defining the stoneGame method, which accepts an array of integers stones.
• Determine the length of the array with int length = stones.length.
• Initialize a two-dimensional array gameState to store computed scores for different game states, dimensioned at [length + 2][length + 2] to manage edge cases without additional conditions.
• Use nested loops to iterate through possible game states. The outer loop iterates over different possible subarray sizes (size), while the inner loop chooses the starting point of these subarrays (start).
• For each subarray defined by its start and end, calculate who's turn it is using (end + start + length) % 2. If it's the first player's turn (turn == 1), update the gameState array with the maximum possible score the current player can achieve by picking the beginning or end stone of the current subarray.
• On the opponent's turn, update the gameState by choosing the move which would result in the minimal score gain for the opponent.
• Finally, gameState[1][length] > 0 will return true if the first player can win, otherwise false indicating the second player wins with optimal plays from both sides.

This algorithm efficiently predicts the outcome of the Stone Game using dynamic programming, taking into account the optimal decisions each player can make at each step of the game. The solution builds up answers from smaller sub-problems, ensuring that every possibility is considered to determine the final outcome.

var playStoneGame = function(stones) {
    const totalStones = stones.length;
    const game = Array(totalStones + 2).fill(0).map(() => Array(totalStones + 2).fill(0));

    for (let length = 1; length <= totalStones; ++length)
        for (let start = 0, end = length - 1; end < totalStones; ++start, ++end) {
            let turn = (end + start + totalStones) % 2;
            if (turn == 1)
                game[start+1][end+1] = Math.max(stones[start] + game[start+2][end+1], stones[end] + game[start+1][end]);
            else
                game[start+1][end+1] = Math.min(-stones[start] + game[start+2][end+1], -stones[end] + game[start+1][end]);
        }

    return game[1][totalStones] > 0;
};

In the "Stone Game" solution written in JavaScript, a player tries to determine whether they can win the game by leveraging a dynamic programming approach with a careful strategy to maximize or minimize their score. Initially, the code sets up a two-dimensional array, game, where the dimensions are derived from the size of the stones array plus two. Each cell in this array will store the optimal outcome of scores for a given range of stones.

The solution iterates through all possible subsets of the stones array using nested loops, where the outer loop regulates the size of the subset and the inner loop defines the start and end of the subset. During each iteration:

• Determine the current turn using the modulo operation, which alternatively assigns players to even and odd turns.
• Update the game array based on the player's turn:
  • If it's player one's turn (turn equals 1), focus on maximizing the score by selecting either the first or the last stone of the subset and adding it to the score derived from the remaining subset.
  • If it's player two's turn, the aim is to minimize the score differential, which is done by choosing a stone that - after its removal - leaves the opponent with a smaller edge in the subsequent round.

After processing all possible subsets and combinations, the ultimate strategy decision for player one lies in the top-right cell of the game array, game[1][totalStones]. This value determines whether player one's optimal strategies will lead to a positive score, indicating a win if it's greater than 0. Thus, the function returns a Boolean indicating whether player one has a winning strategy based on the initial configuration of stones. This approach ensures that the algorithm evaluates all scenarios to decide the most favorable outcome efficiently.

from functools import lru_cache

class Solution:
    def stoneGame(self, stones):
        total = len(stones)

        @lru_cache(None)
        def calculate(i, j):
            if i > j: return 0
            player = (j - i - total) % 2
            if player == 1:  # first player
                return max(stones[i] + calculate(i+1,j), stones[j] + calculate(i,j-1))
            else:
                return min(-stones[i] + calculate(i+1,j), -stones[j] + calculate(i,j-1))

        return calculate(0, total - 1) > 0

The given Python code implements a solution for the "Stone Game". This problem involves two players picking stones arranged in a row, where each player aims to collect a maximum stone value. The players take turns, and in each turn, a player can pick a stone either from the start or the end of the row.

Key features of the implementation:

• The solution utilizes a lru_cache decorator from the functools module, which memoizes the results of expensive function calls to optimize the recursive solution.
• A nested function calculate(i, j) determines the optimal play for the subarray of stones from index i to j. The function evaluates the scenario where each player can pick either the first or last stone of the current subarray.
• The player variable alternates between 1 and 0, depending on whose turn it is. This is calculated using modulo arithmetic on the indices difference.
  • For Player 1 (denoted by 1), the function uses a maximizing strategy where the player tries to maximize the score over the next recursive calls.
  • For Player 0, the function uses a minimizing strategy. Here, the calculations adjust scores as negatives to simulate minimizing the opponent's score.
• The stoneGame function starts the recursion from the full range of the stone list (from index 0 to total-1) and returns whether the first player wins by checking if the result is positive (>0).

Usage:

To deploy this solution in practical scenarios:

1. Define your array of stones each with a specific integer value.
2. Instantiate the Solution class and use the stoneGame method by passing the stone list to determine if the first player (assuming optimal play) is guaranteed a win.

The approach ensures that complex scenarios are handled efficiently through memoization, drastically reducing the computation time needed to solve the game even for larger arrays.