LeetCode-in-Java

1219. Path with Maximum Gold

Medium

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:

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:

Solution

public class Solution {
    private int maxGold = 0;

    public int getMaximumGold(int[][] grid) {
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] != 0) {
                    int g = grid[i][j];
                    grid[i][j] = 0;
                    gold(grid, i, j, g);
                    grid[i][j] = g;
                }
            }
        }
        return maxGold;
    }

    private void gold(int[][] grid, int row, int col, int gold) {
        if (gold > maxGold) {
            maxGold = gold;
        }
        if (row > 0 && grid[row - 1][col] != 0) {
            int currGold = grid[row - 1][col];
            grid[row - 1][col] = 0;
            gold(grid, row - 1, col, gold + currGold);
            grid[row - 1][col] = currGold;
        }
        if (col > 0 && grid[row][col - 1] != 0) {
            int currGold = grid[row][col - 1];
            grid[row][col - 1] = 0;
            gold(grid, row, col - 1, gold + currGold);
            grid[row][col - 1] = currGold;
        }
        if (row < grid.length - 1 && grid[row + 1][col] != 0) {
            // flag=false;
            int currGold = grid[row + 1][col];
            grid[row + 1][col] = 0;
            gold(grid, row + 1, col, gold + currGold);
            grid[row + 1][col] = currGold;
        }
        if (col < grid[0].length - 1 && grid[row][col + 1] != 0) {
            int currGold = grid[row][col + 1];
            grid[row][col + 1] = 0;
            gold(grid, row, col + 1, gold + currGold);
            grid[row][col + 1] = currGold;
        }
    }
}