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 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[0])
return max(dfs(i, j) for i in range(m) for j in range(n))class Solution {
private int[][] grid;
private int m;
private int n;
public int getMaximumGold(int[][] grid) {
m = grid.length;
n = grid[0].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;
}
}class Solution {
public:
vector<int> dirs = {-1, 0, 1, 0, -1};
int getMaximumGold(vector<vector<int>>& grid) {
int ans = 0;
for (int i = 0; i < grid.size(); ++i)
for (int j = 0; j < grid[0].size(); ++j)
ans = max(ans, dfs(i, j, grid));
return ans;
}
int dfs(int i, int j, vector<vector<int>>& grid) {
if (i < 0 || i >= grid.size() || j < 0 || j >= grid[0].size() || grid[i][j] == 0) return 0;
int t = grid[i][j];
grid[i][j] = 0;
int ans = 0;
for (int k = 0; k < 4; ++k) ans = max(ans, t + dfs(i + dirs[k], j + dirs[k + 1], grid));
grid[i][j] = t;
return ans;
}
};func getMaximumGold(grid [][]int) int {
m, n := len(grid), len(grid[0])
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[0].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;
};