695. Max Area of Island

Source code notebook Author Update time

Given a non-empty 2D array grid of 0's and 1's, 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.

Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)

Example 1:

[[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]]

Given the above grid, return 6. Note the answer is not 11, because the island must be connected 4-directionally.

Example 2:

[[0,0,0,0,0,0,0,0]]

Given the above grid, return 0.

Note: The length of each dimension in the given grid does not exceed 50.

# @lc code=start
using LeetCode

function max_area_of_island(grid::Matrix{Int})
    function dfs(grid::Matrix{Int}, I)
        grid[I] = 0
        res = 1
        idcs = CartesianIndices(grid)
        for neighb in ((1, 0), (0, 1), (-1, 0), (0, -1))
            new_I = I + CartesianIndex(neighb)
            (new_I ∈ idcs && grid[new_I] == 1) && (res += dfs(grid, new_I))
        end
        return res
    end
    res = 0
    for I in CartesianIndices(grid)
        (grid[I] == 1) && (res = max(res, dfs(grid, I)))
    end
    return res
end
# @lc code=end
max_area_of_island (generic function with 1 method)

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