Q19: How do you solve Number of Islands using Graph Search (DFS / BFS)?

Iterate through the 2D grid matrix. When encountering '1' (land), increment island count and trigger DFS/BFS to sink connected land cells by mutating grid cells to '0' (or marking them visited). Repeat for all unvisited land cells.

DFS Number of Islands Solution

public class NumberOfIslands {
    public static int numIslands(char[][] grid) {
        if (grid == null || grid.length == 0) return 0;
        int count = 0;
        for (int r = 0; r < grid.length; r++) {
            for (int c = 0; c < grid[0].length; c++) {
                if (grid[r][c] == '1') {
                    count++;
                    dfs(grid, r, c);
                }
            }
        }
        return count;
    }
    private static void dfs(char[][] grid, int r, int c) {
        if (r < 0 || c < 0 || r >= grid.length || c >= grid[0].length || grid[r][c] == '0') {
            return;
        }
        grid[r][c] = '0'; // Sink land
        dfs(grid, r + 1, c);
        dfs(grid, r - 1, c);
        dfs(grid, r, c + 1);
        dfs(grid, r, c - 1);
    }
}

💡 Traverses and sinks adjacent land cells recursively when a new island is detected.

Key Takeaways

2D Grid Graph Traversal using DFS or BFS.
In-place land sinking ('1' $ o$ '0') eliminates $O(m imes n)$ auxiliary visited array memory.

Knowledge Check

What is the time complexity of the Number of Islands graph traversal on an M x N grid?

Q19: How do you solve Number of Islands using Graph Search (DFS / BFS)?

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