Stone Game III

class Solution {
public:
    string stoneGameOutcome(vector<int>& values) {
        int length = values.size();
        vector<int> score(4);
        for (int idx = length - 1; idx >= 0; idx--) {
            score[idx % 4] = values[idx] - score[(idx + 1) % 4];
            if (idx + 2 <= length) {
                score[idx % 4] = max(score[idx % 4], values[idx]
                    + values[idx + 1] - score[(idx + 2) % 4]);
            }
            if (idx + 3 <= length) {
                score[idx % 4] = max(score[idx % 4], values[idx] + values[idx + 1]
                    + values[idx + 2] - score[(idx + 3) % 4]);
            }
        }
        if (score[0] > 0) {
            return "Alice";
        }
        if (score[0] < 0) {
            return "Bob";
        }
        return "Tie";
    }
};

The "Stone Game III" problem involves two players, Alice and Bob, selecting stones from piles in a manner that maximizes their score based on the values of the stones. The provided code implements a dynamic programming solution to determine the winning strategy using C++.

Here’s an overview of how the solution is implemented:

• Initialization: A vector<int> score(4) is used to store immediate results, optimized to reduce space complexity by cyclically using the same indices.
• Dynamic Programming Calculation: The solution iterates backwards through the stone values array, dynamically computing the best possible outcome at each stone considering that a player can pick 1, 2, or 3 consecutive stones. The computed values for choices are stored in score, using modulo operation for cycling through indices.
• Comparison for Each Position:
  • For one stone pick: score[idx % 4] = values[idx] - score[(idx + 1) % 4]
  • For two consecutive stones, compare current score against picking two stones: score[idx % 4] = max(score[idx % 4], values[idx] + values[idx+1] - score[(idx + 2) % 4])
  • For three consecutive stones, update the score considering the pick of three stones: score[idx % 4] = max(score[idx % 4], values[idx] + values[idx+1] + values[idx+2] - score[(idx + 3) % 4])
• Final Outcome Decision: Based on the computed score:
  • Return "Alice" if the score indicates a positive value.
  • Return "Bob" if the score indicates a negative value.
  • Return "Tie" if the value is zero.

This effective use of a sliding window of scores within a dynamic programming framework allows for efficiently determining the optimal game strategy to decide the winner or if the game ends in a tie. This strategic use of calculations ensures minimal space usage while maintaining a clear logic flow.

class Solution {
    public String stoneGameIII(int[] stoneValues) {
        int length = stoneValues.length;
        int[] dpTable = new int [4];
        for (int i = length - 1; i >= 0; i--) {
            dpTable[i % 4] = stoneValues[i] - dpTable[(i + 1) % 4];
            if (i + 2 <= length) {
                dpTable[i % 4] = Math.max(dpTable[i % 4], stoneValues[i] + stoneValues[i + 1]
                    - dpTable[(i + 2) % 4]);
            }
            if (i + 3 <= length) {
                dpTable[i % 4] = Math.max(dpTable[i % 4], stoneValues[i] + stoneValues[i + 1]
                    + stoneValues[i + 2] - dpTable[(i + 3) % 4]);
            }
        }
        if (dpTable[0] > 0) {
            return "Alice";
        }
        if (dpTable[0] < 0) {
            return "Bob";
        }
        return "Tie";
    }
}

Implement the "Stone Game III" solution where players Alice and Bob play optimally using an array of stone values. The function stoneGameIII within the Solution class determines the winner using dynamic programming.

• Start by creating a dpTable of size 4 to store temporary results. This table helps in storing the difference in maximum scores possible at each index.
• Use a loop to iterate over the stoneValues array from the last index down to the first. For each stone index i, calculate the optimal score difference Alice can achieve assuming both play optimally from that point.
• For each index i, compute the best score Alice can achieve if she takes 1, 2, or 3 stones.
  • For one stone, the score difference is stoneValues[i] - dpTable[(i + 1) % 4].
  • For two stones, if applicable, the score difference is the maximum of the existing score and stoneValues[i] + stoneValues[i + 1] - dpTable[(i + 2) % 4].
  • For three stones, if applicable, the calculation considers three stones combined.
• Use the modulus operation with 4 on indices to cycle through dpTable, saving memory and maintaining only relevant computations.
• After processing all indices, check if dpTable[0] is greater than, less than, or equal to zero to determine if the winner is Alice, Bob, or if the game results in a Tie, respectively.

The final recommendations of returning the winner or declaring a tie are based on comparisons of accumulated scores in dpTable[0]. This approach efficiently handles the problem using minimal space and provides direct results according to the game's logic and state.

class GameOutcome:
    def calculate_winner(self, stones: List[int]) -> str:
        stone_count = len(stones)
        dp_scores = [0] * 4
        for position in range(stone_count - 1, -1, -1):
            dp_scores[position % 4] = stones[position] - dp_scores[(position + 1) % 4]
            if position + 2 <= stone_count:
                dp_scores[position % 4] = max(dp_scores[position % 4], stones[position]
                                              + stones[position + 1] - dp_scores[(position + 2) % 4])
            if position + 3 <= stone_count:
                dp_scores[position % 4] = max(dp_scores[position % 4], stones[position] 
                                              + stones[position + 1] + stones[position + 2] - dp_scores[(position + 3) % 4])
        if dp_scores[0] > 0:
            return "Alice"
        if dp_scores[0] < 0:
            return "Bob"
        return "Tie"

In the Stone Game III problem, the objective is to determine who wins the game between two players, Alice and Bob, given a series of stones with integer values. The Python code presented utilizes dynamic programming to solve this problem efficiently.

The solution involves defining a class GameOutcome with a method calculate_winner taking a list of integers (stones) representing the stone values. Here's an outline of the steps implemented in the method:

• Initialize stone_count to get the length of the stone list and dp_scores, a list of four elements initialized to zero. The mod operator is used with dp_scores to keep the space complexity constant and manage overlapping subproblems.

• Iteratively calculate the maximum score difference a player can achieve using a reverse loop, starting from the last stone and moving to the first. This considers cases where a player picks 1, 2, or 3 stones consecutively.

• For each stone position:

  • Calculate the score if the current player takes just that stone.
  • Update the score to include taking the current stone and one additional stone if within bounds.
  • Further update the score for taking the current and two additional stones if within bounds.

• After the loop, the first element of dp_scores indicates the result based on the score difference when both players have played optimally. If the difference is positive, Alice wins, if negative, Bob wins, and if zero, it's a tie.

  This dynamic programming approach efficiently decides the game's outcome by considering each possible scenario using previous calculations (memoization via dp_scores). This allows determining the winner without reevaluating all combinations exhaustively.

  With this approach, ensure to properly manage indices and array bounds, particularly when accessing dp_scores modulated by current positions, to avoid out-of-bound errors and correctly reflect the decision space of the problem.