Maximum Number of Moves in a Grid

class Solution {
public:
    int optimalMoves(vector<vector<int>>& matrix) {
        int rows = matrix.size(), cols = matrix[0].size();
    
        vector<vector<int>> table(rows, vector<int>(2, 0));
    
        for (int r = 0; r < rows; r++) {
            table[r][0] = 1;
        }
    
        int maxResult = 0;
        for (int col = 1; col < cols; col++) {
            for (int row = 0; row < rows; row++) {
                if (matrix[row][col] > matrix[row][col - 1] && table[row][0] > 0) {
                    table[row][1] = max(table[row][1], table[row][0] + 1);
                }
                if (row - 1 >= 0 && matrix[row][col] > matrix[row - 1][col - 1] &&
                    table[row - 1][0] > 0) {
                    table[row][1] = max(table[row][1], table[row - 1][0] + 1);
                }
                if (row + 1 < rows && matrix[row][col] > matrix[row + 1][col - 1] &&
                    table[row + 1][0] > 0) {
                    table[row][1] = max(table[row][1], table[row + 1][0] + 1);
                }
    
                maxResult = max(maxResult, table[row][1] - 1);
            }
    
            for (int k = 0; k < rows; k++) {
                table[k][0] = table[k][1];
                table[k][1] = 0;
            }
        }
    
        return maxResult;
    }
};

In this CPP (C++) solution, the objective is to determine the maximum number of valid moves in a grid, represented as a matrix where each cell can have varying movement rules. The solution involves populating and analyzing a dynamic programming table for optimal paths.

The optimalMoves function processes the matrix to ascertain the peak sequential move count across a grid. This method employs a memoization table, table, initialized based on the matrix dimensions and used to record possible moves per cell.

The approach uses the following techniques:

• Initializes the first column of the memoization table to signify you can always start moving from any row in the first column.
• Iterates through the matrix using nested loops:
  1. The outer loop iterates through columns.
  2. The inner loop iterates through rows.
• Within the inner loop, evaluates potential moves based on cell comparisons between the current cell and its left, upper-left, and lower-left neighbors, updating the current position's maximum possible count for the sequential moves.
• Updates an accumulator maxResult with the highest value calculated for advance moves throughout the grid configuration.
• At the end of each column processing, transitions the state of the memoization table for the next column's calculations.

Upon completion of the matrix traversal, the function returns maxResult, representing the maximum number of valid consecutive moves attainable within the grid, following the defined movement criteria. This implementation effectively tracks and updates potential moves through dynamic programming, optimizing the calculations for large matrices.

class Solution {
    
    public int calculateMax(int[][] matrix) {
        int rows = matrix.length, cols = matrix[0].length;
    
        int[][] dp = new int[rows][2];
    
        for (int r = 0; r < rows; r++) {
            dp[r][0] = 1;
        }
    
        int maximumMoves = 0;
    
        for (int col = 1; col < cols; col++) {
            for (int row = 0; row < rows; row++) {
                if (matrix[row][col] > matrix[row][col - 1] && dp[row][0] > 0) {
                    dp[row][1] = Math.max(dp[row][1], dp[row][0] + 1);
                }
                if (row - 1 >= 0 && matrix[row][col] > matrix[row - 1][col - 1] && dp[row - 1][0] > 0) {
                    dp[row][1] = Math.max(dp[row][1], dp[row - 1][0] + 1);
                }
                if (row + 1 < rows && matrix[row][col] > matrix[row + 1][col - 1] && dp[row + 1][0] > 0) {
                    dp[row][1] = Math.max(dp[row][1], dp[row + 1][0] + 1);
                }
    
                maximumMoves = Math.max(maximumMoves, dp[row][1] - 1);
            }
    
            for (int r = 0; r < rows; r++) {
                dp[r][0] = dp[r][1];
                dp[r][1] = 0;
            }
        }
    
        return maximumMoves;
    }
}

In the "Maximum Number of Moves in a Grid" problem, the implementation involves calculating the path in a matrix that allows the maximum number of increasing moves in any direction (right or diagonally) from a starting point. This task is managed by employing a dynamic programming approach with Java.

• Define a method calculateMax that computes the maximum number of allowable sequential moves in an integer matrix.
• Utilize a 2D integer array (dp) with the same number of rows as the matrix and two columns to track the interim and final results for the maximum number of moves from each cell.
• Initialize the first column of this dp array to 1 for all rows, reflecting the minimum possible moves from such a position.
• Iterate through each column (starting from the second column) and explore three potential moves for each cell: directly to the right, diagonally right-up, and diagonally right-down. Use conditions to ensure these moves follow an increasing order and are within the matrix boundaries.
• Update the dp array's current values based on the maximum moves calculated from the above conditions.
• After processing each column, transfer the computed possible moves to be the previous column values for the next iteration, and reset the current values.
• Finally, track and update a variable maximumMoves through each iteration, which holds the global maximum of moves encountered.

The method correctly concludes by returning the calculated maximum moves, which takes into account the nature of transitions between cells and correctly manages boundaries and maximum calculations. The solution leverages dynamic programming effectively to optimize the retrieval of results, making the process efficient and the implementation compact.

class GameSolver:
    def bestPath(self, maze: List[List[int]]) -> int:
        rows, cols = len(maze), len(maze[0])
    
        # Initialize DP table for storing intermediate results
        dp_table = [[0] * 2 for _ in range(rows)]
    
        # Mark first column as accessible
        for row in range(rows):
            dp_table[row][0] = 1
    
        max_paths = 0
    
        # Processing all columns except the first
        for col in range(1, cols):
            for row in range(rows):
                # Continue path from left in the same row
                if maze[row][col] > maze[row][col - 1] and dp_table[row][0] > 0:
                    dp_table[row][1] = max(dp_table[row][1], dp_table[row][0] + 1)
    
                # Continue path from upper left diagonal cell
                if (
                    row - 1 >= 0
                    and maze[row][col] > maze[row - 1][col - 1]
                    and dp_table[row - 1][0] > 0
                ):
                    dp_table[row][1] = max(dp_table[row][1], dp_table[row - 1][0] + 1)
    
                # Continue path from lower left diagonal cell
                if (
                    row + 1 < rows
                    and maze[row][col] > maze[row + 1][col - 1]
                    and dp_table[row + 1][0] > 0
                ):
                    dp_table[row][1] = max(dp_table[row][1], dp_table[row + 1][0] + 1)
    
                # Record the highest number of paths from all possibilities
                max_paths = max(max_paths, dp_table[row][1] - 1)
    
            # Update for the next column
            for r in range(rows):
                dp_table[r][0] = dp_table[r][1]
                dp_table[r][1] = 0
    
        return max_paths

The Python code defines a class GameSolver which contains a method bestPath to find the maximum number of moves possible through a grid (or maze), based on strictly increasing values from one cell to an adjacent cell (right, upper right diagonal, lower right diagonal). Here's a brief overview of the solution approach implemented in this code:

• Initialize a 2D list dp_table to store the intermediate results of the possible paths up to each cell. This table leverages dynamic programming by using two columns to represent the current and previous results, thereby minimizing space usage.

• The first column of the grid is set as accessible, implying that at least one move is possible starting from any cell in the first column.

• Iterate over all columns of the grid except the first one, and for each cell:

  • Consider continuation from the left cell if it's a valid increment.
  • Consider diagonally from the upper left cell if it's a valid increment.
  • Consider diagonally from the lower left cell if it's a valid increment.
  • Keep track of the maximum paths possible by comparing and storing the highest values.

• After processing a column, transfer the newly computed values in the second column of dp_table back to the first for use in the next iteration, resetting the second column afterwards.

• The function returns max_paths, reflecting the maximum number of moves from the starting column to the last, following the rules of movement defined.

This code effectively finds the path with the highest number of moves satisfying the conditions given by the maze matrix, ensuring an optimal solution through dynamic programming techniques and careful updating of state between iterations.