Number of Islands

Problem Statement

You’re an explorer charting a 2D grid map of '1's (land) and '0's (water). Count the number of distinct islands—connected land masses (horizontally/vertically). This medium-level DFS/BFS challenge is a nautical adventure—sink those islands!

Example

Input: grid = [["1","1","1","1","0"],["1","1","0","1","0"],["1","1","0","0","0"],["0","0","0","0","0"]]

Output: 1 (One large island)

Input: grid = [["1","1","0","0","0"],["1","1","0","0","0"],["0","0","1","0","0"],["0","0","0","1","1"]]

Output: 3 (Three separate islands)

Input: grid = [["0"]]

Output: 0

Code

class Solution {
    public int numIslands(char[][] grid) {
        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') {
                    dfs(grid, i, j);
                    count++;
                }
            }
        }
        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] != '1') 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);
    }
}
            

class Solution:
    def numIslands(self, grid):
        if not grid:
            return 0
        count = 0
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j] == '1':
                    self.dfs(grid, i, j)
                    count += 1
        return count
    
    def dfs(self, grid, i, j):
        if i < 0 or i >= len(grid) or j < 0 or j >= len(grid[0]) or grid[i][j] != '1':
            return
        grid[i][j] = '0'  # Sink the land
        self.dfs(grid, i+1, j)
        self.dfs(grid, i-1, j)
        self.dfs(grid, i, j+1)
        self.dfs(grid, i, j-1)
            

function numIslands(grid) {
    if (!grid.length) return 0;
    let count = 0;
    for (let i = 0; i < grid.length; i++) {
        for (let j = 0; j < grid[0].length; j++) {
            if (grid[i][j] === '1') {
                dfs(grid, i, j);
                count++;
            }
        }
    }
    return count;
    
    function dfs(grid, i, j) {
        if (i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] !== '1') return;
        grid[i][j] = '0';
        dfs(grid, i+1, j);
        dfs(grid, i-1, j);
        dfs(grid, i, j+1);
        dfs(grid, i, j-1);
    }
}
            

Explanation

• DFS Insight: Flood-fill each island, marking visited land to count distinct ones.
• Flow: Scan grid; on '1', DFS to sink connected land, increment count.
• Example Walkthrough: [["1","1"],["0","1"]] → DFS sinks top-left island, then bottom-right → 2.
• BFS Alt: Queue-based flood-fill—same result, different traversal.
• Optimization: In-place marking saves extra space.

Note