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1219. Path with Maximum Gold

Description

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 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:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 15
• 0 <= grid[i][j] <= 100
• There are at most 25 cells containing gold.

Solutions

Solution 1: DFS

We can enumerate each cell as the starting point, and then start a depth-first search from the starting point. During the search process, whenever we encounter a non-zero cell, we turn it into zero and continue the search. When we can no longer continue the search, we calculate the total amount of gold in the current path, then turn the current cell back into a non-zero cell, thus performing backtracking.

The time complexity is $O(m \times n \times 3^k)$, where $k$ is the maximum length of each path. Since each cell can only be visited once at most, the time complexity will not exceed $O(m \times n \times 3^k)$. The space complexity is $O(m \times n)$.

• Java
• C++
• Python
• Go
• TypeScript
• Javascript

• class Solution {
      private final int[] dirs = {-1, 0, 1, 0, -1};
      private int[][] grid;
      private int m;
      private int n;
  
      public int getMaximumGold(int[][] grid) {
          m = grid.length;
          n = grid[0].length;
          this.grid = grid;
          int ans = 0;
          for (int i = 0; i < m; ++i) {
              for (int j = 0; j < n; ++j) {
                  ans = Math.max(ans, dfs(i, j));
              }
          }
          return ans;
      }
  
      private int dfs(int i, int j) {
          if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == 0) {
              return 0;
          }
          int v = grid[i][j];
          grid[i][j] = 0;
          int ans = 0;
          for (int k = 0; k < 4; ++k) {
              ans = Math.max(ans, v + dfs(i + dirs[k], j + dirs[k + 1]));
          }
          grid[i][j] = v;
          return ans;
      }
  }
  

• class Solution {
  public:
      int getMaximumGold(vector<vector<int>>& grid) {
          int m = grid.size(), n = grid[0].size();
          function<int(int, int)> dfs = [&](int i, int j) {
              if (i < 0 || i >= m || j < 0 || j >= n || !grid[i][j]) {
                  return 0;
              }
              int v = grid[i][j];
              grid[i][j] = 0;
              int ans = v + max({dfs(i - 1, j), dfs(i + 1, j), dfs(i, j - 1), dfs(i, j + 1)});
              grid[i][j] = v;
              return ans;
          };
          int ans = 0;
          for (int i = 0; i < m; ++i) {
              for (int j = 0; j < n; ++j) {
                  ans = max(ans, dfs(i, j));
              }
          }
          return ans;
      }
  };
  

• class Solution:
      def getMaximumGold(self, grid: List[List[int]]) -> int:
          def dfs(i: int, j: int) -> int:
              if not (0 <= i < m and 0 <= j < n and grid[i][j]):
                  return 0
              v = grid[i][j]
              grid[i][j] = 0
              ans = max(dfs(i + a, j + b) for a, b in pairwise(dirs)) + v
              grid[i][j] = v
              return ans
  
          m, n = len(grid), len(grid[0])
          dirs = (-1, 0, 1, 0, -1)
          return max(dfs(i, j) for i in range(m) for j in range(n))
  
  

• func getMaximumGold(grid [][]int) (ans int) {
  	m, n := len(grid), len(grid[0])
  	var dfs func(i, j int) int
  	dfs = func(i, j int) int {
  		if i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == 0 {
  			return 0
  		}
  		v := grid[i][j]
  		grid[i][j] = 0
  		ans := 0
  		dirs := []int{-1, 0, 1, 0, -1}
  		for k := 0; k < 4; k++ {
  			ans = max(ans, v+dfs(i+dirs[k], j+dirs[k+1]))
  		}
  		grid[i][j] = v
  		return ans
  	}
  	for i := 0; i < m; i++ {
  		for j := 0; j < n; j++ {
  			ans = max(ans, dfs(i, j))
  		}
  	}
  	return
  }
  

• function getMaximumGold(grid: number[][]): number {
      const m = grid.length;
      const n = grid[0].length;
      const dfs = (i: number, j: number): number => {
          if (i < 0 || i >= m || j < 0 || j >= n || !grid[i][j]) {
              return 0;
          }
          const v = grid[i][j];
          grid[i][j] = 0;
          let ans = v + Math.max(dfs(i - 1, j), dfs(i + 1, j), dfs(i, j - 1), dfs(i, j + 1));
          grid[i][j] = v;
          return ans;
      };
      let ans = 0;
      for (let i = 0; i < m; i++) {
          for (let j = 0; j < n; j++) {
              ans = Math.max(ans, dfs(i, j));
          }
      }
      return ans;
  }
  
  

• /**
   * @param {number[][]} grid
   * @return {number}
   */
  var getMaximumGold = function (grid) {
      const m = grid.length;
      const n = grid[0].length;
      const dfs = (i, j) => {
          if (i < 0 || i >= m || j < 0 || j >= n || !grid[i][j]) {
              return 0;
          }
          const v = grid[i][j];
          grid[i][j] = 0;
          let ans = v + Math.max(dfs(i - 1, j), dfs(i + 1, j), dfs(i, j - 1), dfs(i, j + 1));
          grid[i][j] = v;
          return ans;
      };
      let ans = 0;
      for (let i = 0; i < m; i++) {
          for (let j = 0; j < n; j++) {
              ans = Math.max(ans, dfs(i, j));
          }
      }
      return ans;
  };
  
  

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