2658. Maximum Number of Fish in a Grid | leetcode daily challenge | shashcode | java | dsa

[@shashwat_tiwari_st]

like target for this video is 170. Please do like if you have understood the explanation as well as the code😊

[@nikhilmalhotra1986]

Love the way you explain Shahswat. ♥♥

[@NishantPandat-l3q]

nice explain sir

[@sagnikbera4079]

thnx a lot bhaiya, keep teaching like that.

[@workAP-g5g]

Thanks for the solution. Btw why dont u add ur face cam like older videos. it was cool.

[@Mohammad-f8w4k]

♥♥♥♥♥♥

[@DeepakSingh-ew3hi]

Thank you so much 😃

[@PiyushSharma-we8yd]

thankyou for the explanation ❤

[@shashwat_tiwari_st]

@@PiyushSharma-we8yd 🙏🙏

[@abhirajxyz7944]

class Solution:
def findMaxFish(self, grid: List[List[int]]) -> int:
row = len(grid)
col = len(grid[0])
visited = [[False]*col for _ in range(row)]
ans = 0
# first start with the land
for i in range(row):
for j in range(col):
if grid[i][j]>0 and not visited[i][j]:
ans = max(ans,self._dfs(grid,visited,i,j))
return ans
def _dfs(self, grid:List[List[int]],visited:List[List[bool]], i:int, j:int) -> int:
if i < 0 or j < 0 or i >= len(grid) or j >= len(grid[0]) or visited[i][j] or grid[i][j] == 0:
return 0
visited[i][j] = True
current_part = grid[i][j] + self._dfs(grid,visited,i+1,j) + self._dfs(grid,visited,i-1,j) + self._dfs(grid,visited,i,j+1) + self._dfs(grid,visited,i,j-1)
return current_part

[@KedarBhawthankar]

08:13 definately error because of if row or col are out of grid then how can you access then

[@manishchauhan4630]

this problem is exactly the same as 1219. Path with Maximum Gold just without backtracking

[@shashwat_tiwari_st]

@@manishchauhan4630 ya kind of

[@PiyushSharma-we8yd]

first

[@devnarula6733]

error due to not-shortcircuit of grid[r][c] will come