# 695. Max Area of Island

#### Medium

***

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 `0`.

 

**Example 1:**

![](https://assets.leetcode.com/uploads/2021/05/01/maxarea1-grid.jpg)

```
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: 6
Explanation: 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 `0` or `1`.

```python
class Solution:
    def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
        self.grid = grid
        visited = set()
        max_area = 0
        for row in range(len(grid)):
            for col in range(len(grid[0])):
                if grid[row][col] == 1:
                    max_area = max(self.dfs(row, col, visited), max_area)
        return max_area
    
    def dfs(self, row: int, col: int, visited) -> int:
        if row < 0 or row >= len(self.grid) or col < 0 or col >= len(self.grid[0]) or (row,col) in visited or self.grid[row][col] != 1:
            return 0
        area = 1
        visited.add((row, col))
        for i,j in [(-1,0), (0,1), (1,0), (0,-1)]:
            area += self.dfs(row+i, col+j, visited)
        return area
```