284. Max Area of Island

Scan for land cells; from each unvisited one, flood-fill its whole island counting cells, sinking each cell as you visit it so it isn't counted twice.

Walk-through

1. Scan every cell. When you hit a 1, launch a DFS that returns the area of the island it belongs to.
2. The DFS sinks each visited land cell to 0 (marking it visited) and returns 1 + area of the four neighbours.
3. Track the maximum island area across all launches.

dfs(r, c):
    if out of bounds or grid[r][c] == 0: return 0
    grid[r][c] = 0               # sink so we never recount
    return 1 + dfs(r+1,c)+dfs(r-1,c)+dfs(r,c+1)+dfs(r,c-1)

Worked example

[[1,1,0,0],[1,0,0,1],[0,0,1,1]]:

• Island A: (0,0),(0,1),(1,0) → area 3.
• Island B: (1,3),(2,3),(2,2) → area 3.
• Max = 3.

Complexity

• Time: O(m·n) — each cell visited once.
• Space: O(m·n) recursion in the worst case (one giant island).

class Solution {
    private int rows, cols;

    public int maxAreaOfIsland(int[][] grid) {
        if (grid.length == 0 || grid[0].length == 0) return 0;
        rows = grid.length;
        cols = grid[0].length;
        int best = 0;
        for (int r = 0; r < rows; r++) {
            for (int c = 0; c < cols; c++) {
                if (grid[r][c] == 1) {
                    best = Math.max(best, dfs(grid, r, c));
                }
            }
        }
        return best;
    }

    private int dfs(int[][] grid, int r, int c) {
        if (r < 0 || r >= rows || c < 0 || c >= cols || grid[r][c] == 0) return 0;
        grid[r][c] = 0;
        return 1 + dfs(grid, r + 1, c) + dfs(grid, r - 1, c)
                 + dfs(grid, r, c + 1) + dfs(grid, r, c - 1);
    }
}