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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;
}
};

• 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;
}
}
}

############

class Solution {
private int[][] grid;
private int m;
private int n;

public int getMaximumGold(int[][] grid) {
m = grid.length;
n = grid.length;
this.grid = grid;
int ans = 0;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
ans = Math.max(ans, dfs(i, j));
}
}
return ans;
}

private int dfs(int i, int j) {
if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == 0) {
return 0;
}
int t = grid[i][j];
grid[i][j] = 0;
int[] dirs = {-1, 0, 1, 0, -1};
int ans = 0;
for (int k = 0; k < 4; ++k) {
ans = Math.max(ans, t + dfs(i + dirs[k], j + dirs[k + 1]));
}
grid[i][j] = t;
return ans;
}
}

• // 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;
}
};

• class Solution:
def getMaximumGold(self, grid: List[List[int]]) -> int:
def dfs(i, j):
if not (0 <= i < m and 0 <= j < n and grid[i][j]):
return 0
t = grid[i][j]
grid[i][j] = 0
ans = t + max(
dfs(i + a, j + b) for a, b in [[0, 1], [0, -1], [-1, 0], [1, 0]]
)
grid[i][j] = t
return ans

m, n = len(grid), len(grid)
return max(dfs(i, j) for i in range(m) for j in range(n))

############

# 1219. Path with Maximum Gold
# https://leetcode.com/problems/path-with-maximum-gold/

class Solution:
def getMaximumGold(self, grid: List[List[int]]):
rows, cols = len(grid), len(grid)

res = 0

def dfs(x, y):
tmp = grid[x][y]
grid[x][y] = 0

total = 0
for dx,dy in ((x+1, y), (x-1, y), (x, y-1), (x, y+1)):
if 0 <= dx < rows and 0 <= dy < cols and grid[dx][dy] != 0:
total = max(total, grid[dx][dy] + dfs(dx,dy))

grid[x][y] = tmp

for i in range(rows):
for j in range(cols):
if grid[i][j] > 0:
res = max(res, grid[i][j] + dfs(i,j))

return res

• func getMaximumGold(grid [][]int) int {
m, n := len(grid), len(grid)
var dfs func(i, j int) int
dfs = func(i, j int) int {
if i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == 0 {
return 0
}
t := grid[i][j]
grid[i][j] = 0
ans := 0
dirs := []int{-1, 0, 1, 0, -1}
for k := 0; k < 4; k++ {
ans = max(ans, t+dfs(i+dirs[k], j+dirs[k+1]))
}
grid[i][j] = t
return ans
}
ans := 0
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
ans = max(ans, dfs(i, j))
}
}
return ans
}

func max(a, b int) int {
if a > b {
return a
}
return b
}

• /**
* @param {number[][]} grid
* @return {number}
*/
var getMaximumGold = function (grid) {
const m = grid.length;
const n = grid.length;
function dfs(i, j) {
if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == 0) {
return 0;
}
const t = grid[i][j];
grid[i][j] = 0;
let ans = 0;
const dirs = [-1, 0, 1, 0, -1];
for (let k = 0; k < 4; ++k) {
ans = Math.max(ans, t + dfs(i + dirs[k], j + dirs[k + 1]));
}
grid[i][j] = t;
return ans;
}
let ans = 0;
for (let i = 0; i < m; ++i) {
for (let j = 0; j < n; ++j) {
ans = Math.max(ans, dfs(i, j));
}
}
return ans;
};