[LC] 1219. Path with Maximum Gold

In a gold mine grid of size m * n, each cell in this mine has an integer representing the amount of gold in that cell, 0 if it is empty.

Return the maximum amount of gold you can collect under the conditions:

• Every time you are located in a cell you will collect all the gold in that cell.
• From your position you can walk one step to the left, right, up or down.
• You can't visit the same cell more than once.
• Never visit a cell with 0 gold.
• You can start and stop collecting gold from any position in the grid that has some gold.

Example 1:

Input: grid = [[0,6,0],[5,8,7],[0,9,0]]
Output: 24
Explanation:
[[0,6,0],
 [5,8,7],
 [0,9,0]]
Path to get the maximum gold, 9 -> 8 -> 7.

Example 2:

Input: grid = [[1,0,7],[2,0,6],[3,4,5],[0,3,0],[9,0,20]]
Output: 28
Explanation:
[[1,0,7],
 [2,0,6],
 [3,4,5],
 [0,3,0],
 [9,0,20]]
Path to get the maximum gold, 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7.

class Solution {
    int gRow;
    int gCol;
    int[][] directions = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
    public int getMaximumGold(int[][] grid) {
        gRow = grid.length;
        gCol = grid[0].length;
        int res = 0;
        boolean[][] visited = new boolean[gRow][gCol];
        for (int i = 0; i < gRow; i++) {
            for (int j = 0; j < gCol; j++) {
                res = Math.max(res, dfs(grid, i, j, visited));
            }
        }
        return res;
    }
    
    private int dfs(int[][] grid, int row, int col, boolean[][] visited) {
        if (row < 0 || row >= gRow || col < 0 || col >= gCol || grid[row][col] == 0 || visited[row][col]) {
            return 0;
        }
        int curValue = grid[row][col];
        int nxtMax = 0;
        // mark as visited
        visited[row][col] = true;
        for (int[] direction: directions) {
            nxtMax = Math.max(nxtMax, dfs(grid, row + direction[0], col + direction[1], visited));
        }
        visited[row][col] = false;
        return curValue + nxtMax;
    }
}