Maximum Number of Points with Cost

class Solution {
public:
    long long maxScore(vector<vector<int>>& matrix) {
        int nCols = matrix[0].size();
        vector<long long> prevRow(nCols);

        for (auto& row : matrix) {
            long long maxLeft = 0;

            // Traverse from left to right
            for (int j = 0; j < nCols; ++j) {
                maxLeft = max(maxLeft - 1, prevRow[j]);
                prevRow[j] = maxLeft;
            }

            long long maxRight = 0;
            // Traverse from right to left
            for (int j = nCols - 1; j >= 0; --j) {
                maxRight = max(maxRight - 1, prevRow[j]);
                prevRow[j] = max(prevRow[j], maxRight) + row[j];
            }
        }

        // Extract the maximum score
        long long highestScore = 0;
        for (int j = 0; j < nCols; ++j) {
            highestScore = max(highestScore, prevRow[j]);
        }

        return highestScore;
    }
};

Solve the maximum number of points with cost problem by implementing the maxScore function in C++. Begin by determining the number of columns in the matrix using int nCols = matrix[0].size();. Initialize a vector prevRow to store intermediate results.

• Traverse each row in the matrix, updating prevRow with the accumulated maximum scores from the left and right directions to account for the cost constraints when moving to different columns.
• For the left-to-right pass, use the local maximum maxLeft to update the values in prevRow.
• For the right-to-left pass, incorporate another local maximum maxRight to combine the results with those from the leftward pass, adding the current row values for the updated scores.
• After processing all rows, determine the final maximum score by iterating through prevRow and finding the highest value.

Return the highest score as the output of the function to represent the maximum score attainable from the matrix configuration, considering all cost-impact movements between rows and columns. This approach ensures that every cell is evaluated for optimal paths from both directions, effectively accounting for the problem's constraints.

class Solution {

    public long maximumPoints(int[][] grid) {
        int width = grid[0].length;
        long[] priorRow = new long[width];

        for (int[] currentRow : grid) {
            // Store the maximum findable value from previous iterations
            long maxCarry = 0;

            // Process from left to right
            for (int i = 0; i < width; ++i) {
                maxCarry = Math.max(maxCarry - 1, priorRow[i]);
                priorRow[i] = maxCarry;
            }

            maxCarry = 0;
            // Process from right to left
            for (int i = width - 1; i >= 0; --i) {
                maxCarry = Math.max(maxCarry - 1, priorRow[i]);
                priorRow[i] = Math.max(priorRow[i], maxCarry) + currentRow[i];
            }
        }

        // Determine the highest points obtainable in the last computed row
        long totalMaxPoints = 0;
        for (long point : priorRow) {
            totalMaxPoints = Math.max(totalMaxPoints, point);
        }

        return totalMaxPoints;
    }
}

The Java solution provided solves the problem of finding the maximum number of points you can obtain from a grid, where moving horizontally costs you points. The method maximumPoints internally computes the solution utilizing a dynamic programming approach and optimizations for efficient horizontal traversal handling.

• Initialize a priorRow array with the same width as the grid to keep track of the maximum points that can be obtained up to each position.
• Loop through each row of the grid and update the priorRow in two passes:
  1. Left to Right Pass: Update priorRow values by considering the maximum point value that can be carried from the left side minus the cost of moving to the right. This pass ensures that the points from the left are considered.
  2. Right to Left Pass: Similar to the first pass but from the right to the left. This pass makes sure that points from the right are maximized after considering points from the left, storing the best possible value plus the grid value itself.
• Once all rows are processed, the final maximum points which can be obtained are determined by considering the maximum value in the priorRow after processing all rows.

Execute the example provided to integrate this strategy into broader applications where computing optimized paths or scores from 2D matrices is required, especially under constraints that mimic movement penalties.

class Solution:
    def calculateMaxPoints(self, grid: List[List[int]]) -> int:
        column_count = len(grid[0])
        last_row_scores = [0] * column_count

        for grid_row in grid:
            max_from_left = 0

            # Left to right calculation
            for idx in range(column_count):
                max_from_left = max(max_from_left - 1, last_row_scores[idx])
                last_row_scores[idx] = max_from_left

            max_from_right = 0
            # Right to left calculation
            for idx in range(column_count - 1, -1, -1):
                max_from_right = max(max_from_right - 1, last_row_scores[idx])
                last_row_scores[idx] = (
                    max(last_row_scores[idx], max_from_right) + grid_row[idx]
                )

        # Maximum points in the finalized row
        return max(last_row_scores)

This Python solution tackles the problem of finding the maximum points achievable on a m x n grid, each cell containing a point value. At each position, you may choose points based on the cost from moving horizontally across any number of cells. The primary mechanism involves dynamic programming to optimize the results through the rows processing both from left-to-right and right-to-left.

• Initialize a column_count to the number of columns in the grid.
• Prepare an array last_row_scores initialized to zero, with the length equal to column_count, to keep track of the maximum points obtainable until the current row.
• Use a for-loop to traverse each row (grid_row) in the grid:
  • Initialize max_from_left to zero and iterate through indices from left to right. For each column, compare and update max_from_left with the existing maximum reduced by one or the score from last_row_scores at that column. Store the result back into last_row_scores.
  • After the left to right pass, initialize max_from_right to zero and iterate through indices from right to left. Update max_from_right in a similar method and adjust last_row_scores at each index by adding the current grid value at that index. This step combines both directions' calculations to provide the optimal path value for each cell in a row.
• After processing all rows, the final maximum points obtainable is the maximum value from last_row_scores.

Execute the final line of code to return the maximum score from last_row_scores. This implementation ensures each cell's maximum achievable score is computed considering the penalties from horizontal movements both from previous and next columns. By efficiently combining both left and right movements, the solution dynamically updates the potential maximum values row by row, leading to the optimal solution.