## The challenge

You are given an `m x n`

binary matrix `grid`

. An island is a group of `1`

‘s (representing land) connected **4-directionally** (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

The **area** of an island is the number of cells with a value `1`

in the island.

Return *the maximum area of an island in *

`grid`

. If there is no island, return ``.**Example 1:**

Input:grid = [[0,0,1,0,0,0,0,1,0,0,0,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,1,1,0,1,0,0,0,0,0,0,0,0],[0,1,0,0,1,1,0,0,1,0,1,0,0],[0,1,0,0,1,1,0,0,1,1,1,0,0],[0,0,0,0,0,0,0,0,0,0,1,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,0,0,0,0,0,0,1,1,0,0,0,0]]Output:6Explanation:The answer is not 11, because the island must be connected 4-directionally.

**Example 2:**

Input:grid = [[0,0,0,0,0,0,0,0]]Output:0

**Constraints:**

`m == grid.length`

`n == grid[i].length`

`1 <= m, n <= 50`

`grid[i][j]`

is either `` or`1`

.

## The solution in Python code

```
class Solution:
def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
def get_neighbor(pos,grid):
y,x = pos
ns = []
if x>=1:
ns.append((y,x-1))
if x<len(grid[0])-1:
ns.append((y,x+1))
if y>=1:
ns.append((y-1,x))
if y<len(grid)-1:
ns.append((y+1,x))
return ns
marked = set()
land = []
for row in range(len(grid)):
for col in range(len(grid[row])):
if grid[row][col] == 1 and (row,col) not in marked:
curr_land_len = 1
marked.add((row,col))
stack = [(row,col)]
while stack:
current = stack.pop()
neighbor = get_neighbor(current,grid)
for n in neighbor:
y,x = n
if grid[y][x] == 1 and (y,x) not in marked:
marked.add((y,x))
curr_land_len += 1
stack.append((y,x))
land.append(curr_land_len)
return (max(land) if len(land)!=0 else 0)
```