Grid Game

class Solution {
public:
    long long optimalPathSum(vector<vector<int>>& board) {
        long long topRowAccumulate = accumulate(begin(board[0]), end(board[0]), 0LL),
                  bottomRowAccumulate = 0;
        long long smallestMax = LONG_LONG_MAX;
        for (int index = 0; index < board[0].size(); ++index) {
            topRowAccumulate -= board[0][index];
            smallestMax = min(smallestMax, max(topRowAccumulate, bottomRowAccumulate));
            bottomRowAccumulate += board[1][index];
        }
        return smallestMax;
    }
};

The solution provided here addresses the Grid Game problem using C++. The code aims to determine the smallest maximum path sum on a two-row grid, ensuring that the approach is efficient for competitive programming scenarios.

In this code:

• A class Solution is defined with one member function, optimalPathSum, which accepts a 2D vector board containing two rows representing the grid.

• Inside the optimalPathSum function:

  • Two long long variables, topRowAccumulate and bottomRowAccumulate, are initialized. topRowAccumulate uses the accumulate function from the <numeric> header to calculate the sum of the top row, while bottomRowAccumulate starts at zero.
  • Another long long variable, smallestMax, is initialized to the maximum possible value for type long long.

• The function then iterates through each column of the grid using a for-loop. For each column:

  • The current value of the top row at the current index (board[0][index]) is subtracted from topRowAccumulate.
  • The smallestMax value is updated by comparing it to the maximum of topRowAccumulate and bottomRowAccumulate.
  • The value of the bottom row at the current index is added to bottomRowAccumulate.

• After the loop completes, smallestMax, which represents the minimum of the maximum possible accumulated values during the traversal, is returned.

• The solution efficiently determines the optimal path sum by accumulating values as it progresses through the grid, minimizing the maximum possible accumulated sum that a player can be forced to collect.

This approach ensures a time-efficient traversal of the grid, updating accumulated sums and minimal maximum values in a single pass, leveraging dynamic programming principles for optimal performance.

class Solution {

    public long optimalPath(int[][] matrix) {
        // Calculate total sum of first row elements
        long topRowTotal = 0;
        for (int value : matrix[0]) {
            topRowTotal += value;
        }
        long bottomRowTotal = 0, leastMaxSum = Long.MAX_VALUE;
        for (int i = 0; i < matrix[0].length; i++) {
            topRowTotal -= matrix[0][i];
            leastMaxSum = Math.min(
                leastMaxSum,
                Math.max(topRowTotal, bottomRowTotal)
            );
            bottomRowTotal += matrix[1][i];
        }
        return leastMaxSum;
    }
}

The Java solution to the "Grid Game" problem involves calculating the optimal path a player can take in a grid represented by a 2-row matrix, aiming for a strategy that minimizes the maximum possible score an opponent can achieve in the upper row after the player takes their turn in the lower row.

The method optimalPath(int[][] matrix) computes this by:

• Initializing sums of the elements for both the top and bottom rows of the matrix.
• Iteratively updating these sums as if the current column’s top row value is removed (since the player has moved past it), and adding the current column’s bottom row value to the bottom sum (since the player now stands there).
• At each iteration, the method computes the maximum total of remaining values in the two rows (indicating a potential score for an opponent moving only forward).
• Keeping track of the minimal value among these maximum totals, which effectively becomes the least possible score that an opponent could assure if they played optimally given the player's path.

The approach efficiently determines an optimal playing strategy by leveraging simple arithmetic operations and comparisons, ensuring both high performance and clarity. By iterating through the columns just once and using no extra space beyond a few variables, the solution also exhibits an optimal time complexity of O(n) and space complexity of O(1), where (n) is the number of columns in the grid.

• Use arrays to handle matrix operations for a clear and efficient computation.
• Implement simple loop constructs and minimization checks to evaluate the game conditions dynamically.
• Maintain low space usage by avoiding additional data structure beyond necessary variables.

class Solution:
    def gridGame(self, grid: List[List[int]]) -> int:
        upper_row_sum = sum(grid[0])
        lower_row_sum = 0
        min_max_value = float("inf")
        for idx in range(len(grid[0])):
            upper_row_sum -= grid[0][idx]
            min_max_value = min(min_max_value, max(upper_row_sum, lower_row_sum))
            lower_row_sum += grid[1][idx]
        return min_max_value

The given Python solution addresses a "Grid Game" problem, where two dimensional matrix of integers is used as input. The main objective is to determine the minimum possible value by moving through the grid based on specific rules. The solution employs an efficient algorithm to achieve this by leveraging a combination of sum arithmetic and loop-based calculations. Here's an explanation of how the given code achieves this:

• Initiate two accumulative variables (upper_row_sum and lower_row_sum) to keep track of the sum of elements for top and bottom rows, respectively.
• Start by calculating the sum of all elements in the upper row.
• As you iterate over each column through a loop:
  • Subtract the current column's element in the top row from upper_row_sum, effectively updating the sum to represent only the sum of elements to the right of current column.
  • Compute the maximum of the current upper_row_sum and lower_row_sum and then consider it for the minimum max value (min_max_value) encountered so far in the loop, which is updated using the Python min function.
  • Add the current column's element in the bottom row to lower_row_sum, updating it to include all elements from the start to the current column.
• At the end of iterations, min_max_value holds the answer: the minimized maximum value possible from sum of rows based on described rules.

This approach ensures you analyze all relevant scenarios where the robot moves from left to right to induce minimum potential maximum row sum. The method efficiently calculates this with O(n) complexity, where n is the number of columns in the grid, making it suitable for large datasets.