#+PROPERTY: STUDY_DECK_02
* TODO 2658. Maximum Number of Fish in a Grid :medium:
:PROPERTIES:
:NEETCODE: [[file:../../roadmap.org::*2658. Maximum Number of Fish in a Grid][2658. Maximum Number of Fish in a Grid]]
:END:

You are given a *0-indexed* 2D matrix ~grid~ of size ~m x n~, where ~(r, c)~ represents:

 - A *land* cell if ~grid[r][c] = 0~, or

 - A *water* cell containing ~grid[r][c]~ fish, if ~grid[r][c] > 0~.

A fisher can start at any *water* cell ~(r, c)~ and can do the following operations any number of times:

 - Catch all the fish at cell ~(r, c)~, or

 - Move to any adjacent *water* cell.

Return /the *maximum* number of fish the fisher can catch if he chooses his starting cell optimally, or /~0~ if no water cell exists.

An *adjacent* cell of the cell ~(r, c)~, is one of the cells ~(r, c + 1)~, ~(r, c - 1)~, ~(r + 1, c)~ or ~(r - 1, c)~ if it exists.

*Example 1:*


#+begin_src
Input: grid = [[0,2,1,0],[4,0,0,3],[1,0,0,4],[0,3,2,0]]
Output: 7
Explanation: The fisher can start at cell (1,3) and collect 3 fish, then move to cell (2,3) and collect 4 fish.
#+end_src


*Example 2:*


#+begin_src
Input: grid = [[1,0,0,0],[0,0,0,0],[0,0,0,0],[0,0,0,1]]
Output: 1
Explanation: The fisher can start at cells (0,0) or (3,3) and collect a single fish.
#+end_src


*Constraints:*

 - ~m == grid.length~

 - ~n == grid[i].length~

 - ~1 <= m, n <= 10~

 - ~0 <= grid[i][j] <= 10~

** TODO Approach
Write your approach here.

** TODO Python
#+begin_src python
class Solution:
    def findMaxFish(self, grid: List[List[int]]) -> int:
#+end_src

** TODO C++
#+begin_src cpp
class Solution {
public:
    int findMaxFish(vector<vector<int>>& grid) {
        
    }
};
#+end_src