## DEV Community

Abhishek Chaudhary

Posted on

# Max Area of Island

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
``````