As Far from Land as Possible

class Solution {
public:    
    int calculateMaxDistance(vector<vector<int>>& matrix) {
        int nrows = matrix.size();
        int ncols = matrix[0].size();

        const int MAX_DIST = nrows + ncols + 1;
        
        for (int r = 0; r < nrows; r++) {
            for (int c = 0; c < ncols; c++) {
                if (matrix[r][c]) {
                    matrix[r][c] = 0;
                } else {
                    matrix[r][c] = MAX_DIST;
                    matrix[r][c] = min(matrix[r][c], min(r > 0 ? matrix[r - 1][c] + 1 : MAX_DIST,
                                                         c > 0 ? matrix[r][c - 1] + 1 : MAX_DIST));
                }
            }
        }

        int finalMax = INT_MIN;
        for (int r = nrows - 1; r >= 0; r--) {
            for (int c = ncols - 1; c >= 0; c--) {
                matrix[r][c] = min(matrix[r][c], min(r < nrows - 1 ? matrix[r + 1][c] + 1 : MAX_DIST,
                                                     c < ncols - 1 ? matrix[r][c + 1] + 1 : MAX_DIST));
                finalMax = max(finalMax, matrix[r][c]);
            }
        }
        
        return finalMax == 0 || finalMax == MAX_DIST ? -1 : finalMax;
    }
};

The task "As Far from Land as Possible" involves finding the maximum distance to the nearest land from any cell in a matrix that contains both land (represented by 1s) and water (represented by 0s). This solution is implemented in C++.

The approach taken involves two sweeps of the matrix:

1. Firstly, initialize a new matrix or modify the existing one where all land positions are set to 0 and all water positions to a very high value indicating they are initially inaccessible.
2. In the first sweep (top-left to bottom-right), update every water cell with the minimum distance to land looking upwards (from the above row) and leftwards (from the previous column).
3. In the second sweep (bottom-right to top-left), perform a similar update but this time looking downwards and rightwards.

Throughout the process, keep updating the value of each cell to ensure it holds the minimum value calculated from all possible directions. After completing both sweeps, the maximum value within the matrix, which represents water, is identified as the largest distance from any cell to land.

The function also solves edge cases where updating the matrix might go out of bounds — these are handled with conditionals that check the range of indices in the matrix and a precomputed MAX_DIST value to manage unreachable conditions.

Finally, after both sweeps, the matrix is checked for the maximum distance. If the maximum value remains unchanged from its initialized MAX_DIST or is 0, this would indicate no valid answer (-1), else it returns the maximum distance found.

Thus, the solution ensures:

• Performance is manageable with a two-pass O(n*m) algorithm where n, m are the dimensions of the matrix.
• Edge cases for small matrices or edge placements are correctly handled.

class Solution {
    public int maximumDistance(int[][] matrix) {
        int height = matrix.length;
        int width = matrix[0].length;
        
        final int INFINITY = height + width + 1;

        // Traverse from top-left to bottom-right
        for (int r = 0; r < height; r++) {
            for (int c = 0; c < width; c++) {
                if (matrix[r][c] == 1) {
                    matrix[r][c] = 0;
                } else {
                    matrix[r][c] = INFINITY;
                    matrix[r][c] = Math.min(matrix[r][c], 
                                            Math.min(r > 0 ? matrix[r - 1][c] + 1 : INFINITY,
                                                     c > 0 ? matrix[r][c - 1] + 1 : INFINITY));
                }
            }
        }

        // Traverse from bottom-right to top-left 
        int maxDistance = Integer.MIN_VALUE;
        for (int r = height - 1; r >= 0; r--) {
            for (int c = width - 1; c >= 0; c--) {
                matrix[r][c] = Math.min(matrix[r][c], 
                                        Math.min(r < height - 1 ? matrix[r + 1][c] + 1 : INFINITY,
                                                 c < width - 1 ? matrix[r][c + 1] + 1 : INFINITY));
                
                maxDistance = Math.max(maxDistance, matrix[r][c]);
            }
        }

        return maxDistance == 0 || maxDistance == INFINITY ? -1 : maxDistance;
    }
};

To solve the "As Far from Land as Possible" problem, follow these steps using Java:

1. First, initialize two variables to represent the dimensions (height and width) of the given grid (matrix).

2. Also, declare a constant INFINITY, set to height + width + 1, which effectively acts as a very high distance value used in the solution.

3. Begin with a double loop to iterate over each cell in the matrix:

   • If a cell is 1 (land), reset it to 0 denoting the minimum distance from land.
   • If a cell is 0 (water), initialize it to INFINITY. Update the cell value to be the minimum between its current value and the results of checks in four potential directions (left and top), making sure to add 1 to account for the distance increment.

4. Implement a second double loop to iterate over the matrix, starting from the bottom-right corner and moving to the top-left:

   • For each cell, compare and set the value to be the minimum between its current value and the results from the bottom and right cells, once again, adding 1 to count the distance.
   • Concurrently, update the maxDistance variable to store the maximum distance found from a water cell to land.

5. Finally, check if maxDistance equals 0 or INFINITY. If it does, return -1, as this would mean all cells are land, or water cells are too far from any land respectively. If not, return the maxDistance.

This approach leverages dynamic programming by resolving the shortest possible distance for water cells (0s) from their nearest land cells (1s), considering optimal substructures in matrix traversal both from top-left to bottom-right and bottom-right to top-left. This method ensures that each cell's value is the minimum distance to the nearest land, thereby facilitating the extraction of the maximum distance for any water cell effectively.