1219. Path with Maximum Gold

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.

1219. Path with Maximum Gold
struct Solution;

impl Solution {
    fn get_maximum_gold(mut grid: Vec<Vec<i32>>) -> i32 {
        let n = grid.len();
        let m = grid[0].len();
        let mut sum = 0;
        let mut res = 0;
        for i in 0..n {
            for j in 0..m {
                Self::dfs(i, j, &mut sum, &mut res, &mut grid, n, m);
            }
        }
        res
    }

    fn dfs(
        i: usize,
        j: usize,
        sum: &mut i32,
        max: &mut i32,
        grid: &mut Vec<Vec<i32>>,
        n: usize,
        m: usize,
    ) {
        if grid[i][j] == 0 {
            return;
        }
        let val = grid[i][j];
        *sum += val;
        *max = (*max).max(*sum);
        grid[i][j] = 0;
        if i > 0 {
            Self::dfs(i - 1, j, sum, max, grid, n, m);
        }
        if j > 0 {
            Self::dfs(i, j - 1, sum, max, grid, n, m);
        }
        if i + 1 < n {
            Self::dfs(i + 1, j, sum, max, grid, n, m);
        }
        if j + 1 < m {
            Self::dfs(i, j + 1, sum, max, grid, n, m);
        }
        grid[i][j] = val;
        *sum -= grid[i][j];
    }
}

#[test]
fn test() {
    let grid = vec_vec_i32![[0, 6, 0], [5, 8, 7], [0, 9, 0]];
    let res = 24;
    assert_eq!(Solution::get_maximum_gold(grid), res);
    let grid = vec_vec_i32![[1, 0, 7], [2, 0, 6], [3, 4, 5], [0, 3, 0], [9, 0, 20]];
    let res = 28;
    assert_eq!(Solution::get_maximum_gold(grid), res);
}