LeetCode [1219] Path with Maximum Gold

 1219. Path with Maximum Gold

Medium

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.

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class Solution { int maxSum = 0; int[][] grid; int m, n; int[][] dir = new int[][]{{-1,0},{1,0},{0,-1},{0,1}}; public int getMaximumGold(int[][] grid) { m = grid.length; if(m==0) return 0; n = grid[0].length; if(n==0) return 0; this.grid = grid; for(int i=0; i<m; ++i){ for(int j=0; j<n; ++j){ if(grid[i][j]>0){ helper(i, j, 0); } } } return maxSum; } void helper(int i, int j, int sum){ sum += grid[i][j]; maxSum = Math.max(sum, maxSum); int t = grid[i][j]; grid[i][j] = 0; for(int[] d: dir){ int ii = i+d[0], jj = j+d[1]; if(ii>=0 && ii<m && jj>=0 && jj<n && grid[ii][jj]>0){ helper(ii, jj, sum); } } grid[i][j] = t; } }