Number of Islands

Topic: 12. Graphs

Link: LeetCode 200 - Number of Islands

1. Logical Breakdown

• [x] Core Logic: Iterate grid. If '1', increment count and sink island (DFS).
• [x] Sink: Turn connected '1's to '0's.

2. Visualization

graph TD Found["Found '1'"] --> Count["Count++"] Count --> DFS["DFS Neighbors"] DFS --> Sink["Mark as '0'"]

3. Complexity

• Time: O(M * N)
• Space: O(M * N)

4. Code

package com.dsa.graphs;

public class NumberOfIslands {
    public int solve(char[][] grid) {
        if (grid == null || grid.length == 0) return 0;
        int count = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == '1') {
                    count++;
                    dfs(grid, i, j);
                }
            }
        }
        return count;
    }

    private void dfs(char[][] grid, int i, int j) {
        if (i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == '0') {
            return;
        }
        grid[i][j] = '0'; // Mark as visited
        dfs(grid, i + 1, j);
        dfs(grid, i - 1, j);
        dfs(grid, i, j + 1);
        dfs(grid, i, j - 1);
    }
}