Formatted question description: https://leetcode.ca/all/1219.html

# 1219. Path with Maximum Gold (Medium)

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.

Related Topics: Backtracking

## Solution 1. DFS

// OJ: https://leetcode.com/problems/path-with-maximum-gold/

// Time: O(MN*4^(MN))
// Space: O(MN) because in the worst case the DFS will visit all cells.
class Solution {
int M, N, ans = 0, dirs = { {0,1},{0,-1},{1,0},{-1,0} };
void dfs(vector<vector<int>> &G, int i, int j, int cnt) {
int g = G[i][j];
G[i][j] = 0;
cnt += g;
ans = max(ans, cnt);
for (auto &[dx, dy] : dirs) {
int x = i + dx, y = j + dy;
if (x < 0 || y < 0 || x >= M || y >= N || G[x][y] == 0) continue;
dfs(G, x, y, cnt);
}
G[i][j] = g;
}
public:
int getMaximumGold(vector<vector<int>>& G) {
M = G.size(), N = G.size();
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (G[i][j] == 0) continue;
dfs(G, i, j, 0);
}
}
return ans;
}
};


Java

class Solution {
int max = 0;

public int getMaximumGold(int[][] grid) {
int rows = grid.length, columns = grid.length;
boolean[][] visited = new boolean[rows][columns];
for (int i = 0; i < rows; i++) {
for (int j = 0; j < columns; j++) {
if (grid[i][j] != 0)
backtrack(grid, i, j, 0, visited);
}
}
return max;
}

public void backtrack(int[][] grid, int row, int column, int amount, boolean[][] visited) {
int rows = grid.length, columns = grid.length;
if (row < 0 || row >= rows || column < 0 || column >= columns || grid[row][column] == 0 || visited[row][column])
max = Math.max(max, amount);
else {
amount += grid[row][column];
visited[row][column] = true;
int[][] directions = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
for (int[] direction : directions)
backtrack(grid, row + direction, column + direction, amount, visited);
amount -= grid[row][column];
visited[row][column] = false;
}
}
}