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
0gold. - 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.lengthn == grid[i].length1 <= m, n <= 150 <= grid[i][j] <= 100- There are at most 25 cells containing gold.
DFS.
class Solution:
def getMaximumGold(self, grid: List[List[int]]) -> int:
def dfs(i, j):
if i < 0 or i >= len(grid) or j < 0 or j >= len(grid[0]) or grid[i][j] == 0 or vis[i][j]:
return 0
vis[i][j] = True
t = 0
for x, y in [[0, -1], [0, 1], [1, 0], [-1, 0]]:
t = max(t, dfs(i + x, j + y))
vis[i][j] = False
return t + grid[i][j]
m, n = len(grid), len(grid[0])
ans = 0
vis = [[False] * n for _ in range(m)]
for i in range(m):
for j in range(n):
ans = max(ans, dfs(i, j))
return ansclass Solution {
private int[][] grid;
private boolean[][] vis;
public int getMaximumGold(int[][] grid) {
int m = grid.length, n = grid[0].length;
this.grid = grid;
this.vis = new boolean[m][n];
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 >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == 0 || vis[i][j]) {
return 0;
}
vis[i][j] = true;
int t = 0;
int[][] dirs = new int[][]{{0, -1}, {0, 1}, {1, 0}, {-1, 0}};
for (int[] dir : dirs) {
t = Math.max(t, dfs(i + dir[0], j + dir[1]));
}
vis[i][j] = false;
return t + grid[i][j];
}
}class Solution {
public:
vector<vector<int>> grid;
vector<vector<int>> dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
int getMaximumGold(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size();
this->grid = grid;
vector<vector<bool>> vis(m, vector<bool>(n, false));
int ans = 0;
for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j)
ans = max(ans, dfs(i, j, vis));
return ans;
}
int dfs(int i, int j, vector<vector<bool>>& vis) {
if (i < 0 || i >= grid.size() || j < 0 || j >= grid[0].size() || grid[i][j] == 0 || vis[i][j]) return 0;
vis[i][j] = true;
int t = 0;
for (auto& dir : dirs)
t = max(t, dfs(i + dir[0], j + dir[1], vis));
vis[i][j] = false;
return t + grid[i][j];
}
};func getMaximumGold(grid [][]int) int {
m, n := len(grid), len(grid[0])
vis := make([][]bool, m)
for i := range vis {
vis[i] = make([]bool, n)
}
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 || vis[i][j] {
return 0
}
vis[i][j] = true
dirs := [4][2]int{{0, -1}, {0, 1}, {1, 0}, {-1, 0}}
t := 0
for _, dir := range dirs {
t = max(t, dfs(i+dir[0], j+dir[1]))
}
vis[i][j] = false
return t + grid[i][j]
}
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
}