1331. Path with Maximum Gold

• Difficulty: Medium

• Topics: Backtracking

• Similar Questions:

Problem:

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.

Solutions:

class Solution {
public:
    int getMaximumGold(vector<vector<int>>& grid) {
        int m = grid.size();
        if (m == 0) return 0;
        int n = grid[0].size();
        if (n == 0) return 0;

        int ret = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] != 0) {
                    ret = max(ret, dfs(grid, m, n, i, j));
                }
            }
        }

        return ret;
    }


private:
    int dfs(vector<vector<int>>& grid, int m, int n, int row, int col) {

        if (row < 0 || row >= m || col < 0 || col >= n || grid[row][col] == 0)   return 0;
        int gold = grid[row][col];
        int sum = gold;

        grid[row][col] = 0;
        sum += max(max(dfs(grid, m, n, row + 1, col), dfs(grid, m, n, row - 1, col)), max(dfs(grid, m, n, row, col - 1), dfs(grid, m, n, row, col + 1)));
        grid[row][col] = gold;

        return sum;
    }

};