1219. Path with Maximum Gold

Problem

In a gold mine grid of size m x 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:

• m == grid.length

• n == grid[i].length

• 1 <= m, n <= 15

• 0 <= grid[i][j] <= 100

• There are at most **25 **cells containing gold.

Solution (Java)

class Solution {
    private int maxGold = 0;

    public int getMaximumGold(int[][] grid) {
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] != 0) {
                    int g = grid[i][j];
                    grid[i][j] = 0;
                    gold(grid, i, j, g);
                    grid[i][j] = g;
                }
            }
        }
        return maxGold;
    }

    private void gold(int[][] grid, int row, int col, int gold) {
        if (gold > maxGold) {
            maxGold = gold;
        }
        if (row > 0 && grid[row - 1][col] != 0) {
            int currGold = grid[row - 1][col];
            grid[row - 1][col] = 0;
            gold(grid, row - 1, col, gold + currGold);
            grid[row - 1][col] = currGold;
        }
        if (col > 0 && grid[row][col - 1] != 0) {
            int currGold = grid[row][col - 1];
            grid[row][col - 1] = 0;
            gold(grid, row, col - 1, gold + currGold);
            grid[row][col - 1] = currGold;
        }
        if (row < grid.length - 1 && grid[row + 1][col] != 0) {
            // flag=false;
            int currGold = grid[row + 1][col];
            grid[row + 1][col] = 0;
            gold(grid, row + 1, col, gold + currGold);
            grid[row + 1][col] = currGold;
        }
        if (col < grid[0].length - 1 && grid[row][col + 1] != 0) {
            int currGold = grid[row][col + 1];
            grid[row][col + 1] = 0;
            gold(grid, row, col + 1, gold + currGold);
            grid[row][col + 1] = currGold;
        }
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).