Today I have done leetcode's May LeetCoding Challenge with cpp.

May LeetCoding Challenge 14

Description

Path with Maximum Gold

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.

Solution

class Solution {
  int rows;
  int cols;
  int dfs(vector<vector<int>>& grid, int row, int col) {
    if(row < 0 || col < 0 || row >= rows || col >= cols || !grid[row][col]) return 0;
    int gold = grid[row][col];
    grid[row][col] = 0;

    int result = max({
      dfs(grid, row, col + 1),
      dfs(grid, row, col - 1),
      dfs(grid, row + 1, col),
      dfs(grid, row - 1, col)
    });
    result += gold;
    grid[row][col] = gold;
    return result;
  }
  int allOrNone(vector<vector<int>>& grid) {
    int result = 0;
    for(const auto &row : grid) {
      for(auto n : row) {
        if(!n) return 0;
        result += n;
      }
    }
    return result;
  }
public:
  int getMaximumGold(vector<vector<int>>& grid) {
    int all = allOrNone(grid);
    if(all) return all;

    rows = grid.size();
    cols = grid.front().size();

    int answer = 0;
    for(int i = 0; i < rows; ++i) {
      for(int j = 0; j < cols; ++j) {
        answer = max(answer, dfs(grid, i, j));
      }
    }

    return answer;
  }
};