```
class Solution(object):
def maxAreaOfIsland(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
# because
ans = 0
for i in range(len(grid)):
for j in range(len(grid[0])):
if grid[i][j] == 1:
grid[i][j] = 0
ans = max(self.dfs(grid, i, j), ans)
# ans = max(self.bfs(grid, i, j), ans)
return ans
def dfs(self, grid, i, j):
# DFS based on stack
stack = [(i, j)]
area = 0
# Stack for DFS
while stack:
r, c = stack.pop(-1)
area += 1
for nr, nc in ((r - 1, c), (r + 1, c), (r, c - 1), (r, c + 1)):
if (0 <= nr < len(grid) and
0 <= nc < len(grid[0]) and grid[nr][nc]):
stack.append((nr, nc))
grid[nr][nc] = 0
return area
# def bfs(self, grid, x, y):
# # BFS based on queue
# queue = [(x, y)]
# area = 0
# # Stack for DFS
# while queue:
# i, j = queue.pop(0)
# area += 1
# if i - 1 >= 0 and grid[i - 1][j] == 1:
# grid[i - 1][j] = 0
# queue.append((i - 1, j))
# if i + 1 < len(grid) and grid[i + 1][j] == 1:
# grid[i + 1][j] = 0
# queue.append((i + 1, j))
# if j - 1 >= 0 and grid[i][j - 1] == 1:
# grid[i][j - 1] = 0
# queue.append((i, j - 1))
# if j + 1 < len(grid[0]) and grid[i][j + 1] == 1:
# grid[i][j + 1] = 0
# queue.append((i, j + 1))
# return area
# def maxAreaOfIsland(self, grid):
# # Recursive DFS
# seen = set()
# def area(r, c):
# if not (0 <= r < len(grid) and 0 <= c < len(grid[0])
# and (r, c) not in seen and grid[r][c]):
# return 0
# seen.add((r, c))
# return (1 + area(r+1, c) + area(r-1, c) +
# area(r, c-1) + area(r, c+1))
# return max(area(r, c)
# for r in range(len(grid))
# for c in range(len(grid[0])))
```

Download Max Area of Island.pyLeetcode 695 Max Area of Island problem solution in python3 with explanation. This is the best place to expand your knowledge and get prepared for your next interview.

Feedback is the most important part of any website.

If you have any query, suggestion or feedback, Please feel free to contact us.