Max Area of Island

#leetcode #dsa #theabbie

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.

SOLUTION:

class Solution:
    def DFS(self, grid, i, j, m, n):
        if i >= 0 and j >= 0 and i < m and j < n and grid[i][j] == 1:
            grid[i][j] = 0
            self.ctr += 1
            for x, y in [(i - 1, j), (i + 1, j), (i, j - 1), (i, j + 1)]:
                self.DFS(grid, x, y, m, n)

    def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
        m = len(grid)
        n = len(grid[0])
        mArea = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 1:
                    self.ctr = 0
                    self.DFS(grid, i, j, m, n)
                    mArea = max(mArea, self.ctr)
        return mArea