# Max Area of Island

published on 4/29/2022You 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

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 0 or 1.

### Solution explanation:

- First, we run dfs on the grid to find all the islands.
- Then,for each island, we find the maximum area of each island.
- And thne update the maxi with the maximum area for the island.
- Finally, we return the maxi.

class Solution:def maxAreaOfIsland(self, grid: List[List[int]]) -> int:n,m,maxi = len(grid),len(grid[0]),0def dfs(i, j):if i<0 or j<0 or i>=len(grid) or j>=len(grid[0]) or grid[i][j] != 1:return 0maxArea = 1grid[i][j] = '#' # this will act as visited setmaxArea += dfs(i+1, j)maxArea += dfs(i-1, j)maxArea += dfs(i, j+1)maxArea += dfs(i, j-1)return maxAreafor i in range(n):for j in range(m):if grid[i][j] == 1:maxi = max(dfs(i,j),maxi)return maxi