Max area of island

class Solution:
    def __init__(self):
        self.directions = [(1, 0), (-1, 0), (0, -1), (0, 1)]
        self.visited = set()
        self.res = 0

    def bfs(self, grid, i, j):
        q = [(i, j)]
        curr = 0
        while q:
            for i in range(len(q)):
                x, y = q.pop()
                curr += 1
                for dx, dy in self.directions:
                    newx, newy = x + dx, y + dy
                    if 0 <= newx < len(grid) and 0 <= newy < len(grid[0]):
                        if (newx, newy) not in self.visited and grid[newx][newy] == 1:
                            self.visited.add((newx, newy))
                            q.append((newx, newy))

        self.res = max(self.res, curr)

    def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if (i, j) not in self.visited and grid[i][j] == 1:
                    self.visited.add((i, j))
                    self.bfs(grid, i, j)

        return self.res

Max Area of Island

Difficulty: Medium

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.