# Max Area of Island

published on 4/29/2022

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 ```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        def dfs(i, j):            if i<0 or j<0 or i>=len(grid) or j>=len(grid) or grid[i][j] != 1:                return 0            maxArea = 1            grid[i][j] = '#'  # this will act as visited set            maxArea += dfs(i+1, j)            maxArea += dfs(i-1, j)            maxArea += dfs(i, j+1)            maxArea += dfs(i, j-1)            return maxArea
for i in range(n):            for j in range(m):                if grid[i][j] == 1:                    maxi = max(dfs(i,j),maxi)        return maxi
```
Created with ❤️ by Avinash Sharma