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.

int dfs(vector<vector<int> >& grid, const vector<pair<int, int> >& dirs, int i, int j) {
    if (i < 0 || i >= grid.size() || j < 0 || j >= grid[0].size() || !grid[i][j]) return 0;
    int res = 0, orig = grid[i][j];
    grid[i][j] = 0;
    for (const pair<int, int>& dir : dirs) {
        res = max(res, orig + dfs(grid, dirs, i + dir.first, j + dir.second));
    }
    grid[i][j] = orig;
    return res;
}
int getMaximumGold(vector<vector<int>>& grid) {
    const int m = grid.size(), n = grid[0].size();
    int res = 0;
    const vector<pair<int, int> > dirs({ {-1, 0}, {1, 0}, {0, -1}, {0, 1} });
    for (int i = 0; i < m; ++i) {
        for (int j = 0; j < n; ++j) {
            if (grid[i][j] > 0) res = max(res, dfs(grid, dirs, i, j));
        }
    }
    return res;
}

Last updated

Was this helpful?