Max Area of Island

Problem

Given a binary grid where 1 = land, 0 = water, return the area of the largest 4-connected island (0 if none).

Constraints

• 1 ≤ rows, cols ≤ 50

Approach — Graphs

This is a Graphs problem. The idea: explore the graph with BFS or DFS, marking nodes visited so you never process one twice. Work through the reference code below line by line, then re-derive it yourself in the editor — that's how the pattern sticks.

Complexity: O(V + E) time.

Solution code

Python

class Solution:
    def maxAreaOfIsland(self, grid):
        rows, cols = len(grid), len(grid[0])
        def dfs(r, c):
            if r < 0 or c < 0 or r >= rows or c >= cols or grid[r][c] != 1:
                return 0
            grid[r][c] = 0
            return 1 + dfs(r + 1, c) + dfs(r - 1, c) + dfs(r, c + 1) + dfs(r, c - 1)
        best = 0
        for r in range(rows):
            for c in range(cols):
                if grid[r][c] == 1:
                    best = max(best, dfs(r, c))
        return best

Java

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

    private int dfs(int[][] grid, int r, int c) {
        if (r < 0 || c < 0 || r >= grid.length || c >= grid[0].length) {
            return 0;
        }
        if (grid[r][c] != 1) {
            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);
    }
}

Practice it

Reading a solution isn't the same as being able to write it under pressure. Open this problem in the in-browser editor, hide the solution, and solve it from scratch — your code runs against real test cases instantly.