695. Max Area of Island

Medium

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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: 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.

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