1200 - 1299
1219 - Path with Maximum Gold (Medium)
Problem Link
https://leetcode.com/problems/path-with-maximum-gold/
Problem Statement
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 anyposition 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:
- There are at most 25 cells containing gold.
Approach 1: DFS Backtracking
Since only at most cells containing gold, we can try all the possible paths using backtracking.
class Solution {
public:
int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, -1, 0, 1};
int dfs(vector<vector<int>>& grid, int i, int j) {
int m = grid.size(), n = grid[0].size();
int res = 0, orig = grid[i][j];
// mark grid[i][j] as 0 so that we won't visit again in this route
grid[i][j] = 0;
// try all 4 directions
for (int d = 0; d < 4; d++) {
// next (i, j)
int next_i = i + dx[d], next_j = j + dy[d];
// check if next coordinate is still in the grid
if (0 <= next_i && next_i < m && 0 <= next_j && next_j < n && grid[next_i][next_j]) {
// if so, continue with the next position
res = max(res, grid[next_i][next_j] + dfs(grid, next_i, next_j));
}
}
// backtrack
grid[i][j] = orig;
return res;
}
int getMaximumGold(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size(), ans = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j]) {
// start from grid[i][j]
ans = max(ans, grid[i][j] + dfs(grid, i, j));
}
}
}
return ans;
}
};