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.

Constraints:

• 1 <= grid.length, grid[i].length <= 15
• 0 <= grid[i][j] <= 100
• There are at most 25 cells containing gold.

class Solution {

public:

    vector<vector<int>> dirs = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

    int getMaximumGold(vector<vector<int>>& grid) {

        int row = grid.size();

        if (row == 0) {

            return 0;

        }

        int ans = 0;

        int col = grid[0].size();

        for (int i = 0; i < row; i++) {

            for (int j = 0; j < col; j++) {

                if (grid[i][j] != 0) {

                    ans = max(ans, dfs(grid, i, j));

                }

            }

        }

        return ans;

    }

    int dfs(vector<vector<int>>& grid, int i, int j) {

        int temp = grid[i][j];

        grid[i][j] = 0;

        int amount = 0;

        for (auto & dir : dirs) {

            int x = i + dir[0];

            int y = j + dir[1];

            if (x < 0 || x >= grid.size() || y < 0 || y >= grid[0].size() ||

                grid[x][y] == 0) {

                continue;

            }

            amount = max(amount, dfs(grid, x, y));

        }

        grid[i][j] = temp;

        return amount + grid[i][j];

    }

};