695. Max Area of Island

class Solution {
public:
    int maxAreaOfIsland(vector<vector<int>>& grid)
    {
        int maxArea = 0;
        for (int i = 0; i < grid.size(); i++)
        {
            for (int j = 0; j < grid[0].size(); j++)
            {
                maxArea = max(maxArea, dfs(grid, i, j));
            }
        }
        return maxArea;
    }

    int dfs(vector<vector<int>>& grid, int i, int j)
    {
        if (i < 0 || j < 0 || i >= grid.size() || j >= grid[0].size() || grid[i][j] == 0)
        {
            return 0;
        }

        int area = 1;
        grid[i][j] = 0;
        vector<vector<int>> directions = {{0, -1}, {0, 1}, {1, 0}, {-1, 0}};
        for (const auto& dir : directions)
        {
            area += dfs(grid, i + dir[0], j + dir[1]);
        }
        return area;
    }
};

• T: $O(M \cdot N)$
• S: $O(M \cdot N)$

class Solution {
public:
    int maxAreaOfIsland(vector<vector<int>>& grid)
    {
        int maxArea = 0;
        vector<vector<int>> directions = {{0, -1}, {0, 1}, {1, 0}, {-1, 0}};
        for (int i = 0; i < grid.size(); i++)
        {
            for (int j = 0; j < grid[0].size(); j++)
            {
                if (grid[i][j] == 0) continue;
                int area = 1;
                queue<vector<int>> q;
                q.push({i, j});
                grid[i][j] = 0;
                while (!q.empty())
                {
                    auto node = q.front(); q.pop();
                    for (const auto& dir : directions)
                    {
                        int new_i = node[0] + dir[0];
                        int new_j = node[1] + dir[1];
                        if (new_i < 0 || new_j < 0 || new_i >= grid.size() || new_j >= grid[0].size() || grid[new_i][new_j] == 0)
                            continue;

                        grid[new_i][new_j] = 0;
                        area++;
                        q.push({new_i, new_j});
                    }
                }
                maxArea = max(maxArea, area);
            }
        }
        return maxArea;
    }
};

• T: $O(M \cdot N)$
• S: $O(M \cdot N)$