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.

Since the count of cells is at maximum 15×15 and we know at maximum there can be 25 cells with gold, we can perform a simple DFS starting with each cell with gold and find the maximum possible gold considering each step as a new start. We can then return the maximum gold that can we aquired of all starting points.

class Solution {
    
    
    public int getMaximumGold(int[][] grid) {
        
        if(grid == null || grid.length == 0 || grid[0].length == 0){
            return 0;
        }
        int m = grid.length;
        int n = grid[0].length;
        int max_gold = 0;
        Set<String> visited = new HashSet<>();
        for(int i = 0 ; i < m ; i++){
            for(int j = 0 ; j < n ; j++){
                max_gold = Math.max(max_gold, DFS(grid, i, j, m, n, visited));
            }
        }
        return max_gold;
    }
    
    private int DFS(int[][] grid, int i, int j, int m, int n, Set<String> visited){
        if(i < 0 || i == m || j < 0 || j == n || grid[i][j] == 0){
            return 0;
        }
        String key = i + "_" + j;
        if(visited.contains(key)){
            return 0;
        }
        int cur_gold = grid[i][j];
        visited.add(key);
        
        int gold_in_top = DFS(grid, i-1, j, m, n, visited);
        int gold_in_bottom = DFS(grid, i+1, j, m, n, visited);
        int gold_in_left = DFS(grid, i, j-1, m, n, visited);
        int gold_in_right = DFS(grid, i, j+1, m, n, visited);
        
        int vertical_max = Math.max(gold_in_top, gold_in_bottom);
        int side_max = Math.max(gold_in_left, gold_in_right);
        
        cur_gold = cur_gold + Math.max(vertical_max,side_max); 
        
        visited.remove(key);
        return cur_gold;
    }
}

Complexity

  • Space: O(n) for the algorithm. O(1) for this question. Since the depth of recursion can go upto max 25 cells.
  • Time: O(4n). Each cell is initiating 4 DFS in 4 directions.

Categories: DFS

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