Stone Game VII

class Solution {
public:
    int calculateMaxScore(vector<int>& stones) {
        int length = stones.size();
        vector<int> cumulativeSum(length + 1);
        vector<vector<int>> dpTable(length, vector<int>(length, 0));
        for (int i = 0; i < length; i++) {
            cumulativeSum[i + 1] = cumulativeSum[i] + stones[i];
        }
        for (int left = length - 2; left >= 0; left--) {
            for (int right = left + 1; right < length; right++) {
                int scoreIfFirstRemoved =
                    cumulativeSum[right + 1] - cumulativeSum[left + 1];
                int scoreIfLastRemoved = cumulativeSum[right] - cumulativeSum[left];
                dpTable[left][right] = max(scoreIfFirstRemoved - dpTable[left + 1][right],
                                           scoreIfLastRemoved - dpTable[left][right - 1]);
            }
        }
        return dpTable[0][length - 1];
    }
};

The provided C++ solution is designed to solve the Stone Game VII problem. This problem involves a game where players alternately remove stones from a row and accumulate points based on the values of the remaining stones. The solution focuses on calculating the maximum score a player can achieve using dynamic programming.

Here's a summary of how the solution operates:

• A function named calculateMaxScore is initialized, which takes a vector containing the arrangement of stones as its parameter.
• It starts by setting up two crucial structures:
  • cumulativeSum, a vector to store prefix sums of stones for quick range sum calculations.
  • dpTable, a 2D vector that will hold the optimal scores depending on which stones are still available in the game.
• The prefix sum array is populated within a loop to facilitate rapid sum queries for any subset of stones.
• The dynamic programming table (dpTable) is filled using a nested loop approach:
  • The outer loop processes stone subsets by starting from the second to last stone and moving leftward.
  • The inner loop then iterates from the current position of the outer loop to the end of the stones array.
  • For each position (denoted by left and right pointers), two potential scores are calculated:
    • scoreIfFirstRemoved: The score if the first stone of the current subset is removed.
    • scoreIfLastRemoved: The score if the last stone of the current subset is removed.
  • The dpTable entry for the current subset is updated to the maximum score achievable by removing either the first or last stone, factoring in the resulting state.

Finally:

• The function returns the value in dpTable[0][length - 1], which represents the maximum score achievable from the full set of stones with optimal play.

This detailed approach ensures an efficient solution by using dynamic programming to eliminate redundant calculations and efficiently compute the optimal game strategy.

class Solution {
    public int calculateMaxDifference(int[] elements) {
        int count = elements.length;
        int[] cumulativeSum = new int[count + 1];
        for (int i = 0; i < count; i++) {
            cumulativeSum[i + 1] = cumulativeSum[i] + elements[i];
        }
        int[][] solution = new int[count][count];
        for (int left = count - 2; left >= 0; left--) {
            for (int right = left + 1; right < count; right++) {
                int scoreIfFirstRemoved = cumulativeSum[right + 1] - cumulativeSum[left + 1];
                int scoreIfLastRemoved = cumulativeSum[right] - cumulativeSum[left];
                solution[left][right] = Math.max(scoreIfFirstRemoved - solution[left + 1][right],
                                                 scoreIfLastRemoved - solution[left][right - 1]);
            }
        }
        return solution[0][count - 1];
    }
}

This solution is tailored for a game involving strategy to calculate the maximum score difference by removing stones from either end of a line of stones consecutively. The approach involves dynamic programming to optimize the process. Here’s how the Java implementation works:

• Start by calculating the cumulative sum of the stone values. This precomputation helps in quickly computing the sum of any subarray of stones, which is essential for determining the score.

• Define a two-dimensional array solution where solution[left][right] will store the optimal score difference possible for the subarray of stones from index left to right.

• Iterate over all possible pairs of left and right indexes in a bottom-up manner. For each pair, calculate:

  • scoreIfFirstRemoved - the score difference if the first stone in the subarray is removed.
  • scoreIfLastRemoved - the score difference if the last stone in the subarray is removed.

• Update the solution[left][right] with the maximum score difference achievable between removing the first or the last stone, using previously computed values in the solution array.

• The entry solution[0][count - 1] will eventually hold the maximum difference in score Alice can achieve over Bob by playing optimally from the full set of stones.

This algorithm leverages dynamic programming to efficiently solve the problem by breaking it down into smaller subproblems, significantly reducing the complexity compared to naive methods. The optimization ensures feasibility even for larger sets of stones.