---
title: '200. Number of Islands'
description: Given an m x n 2D binary grid grid which represents a map of '1's (land) and '0's (water), return the number of islands
sidebar:
  label: 'Number of Islands'
  badge: 'Medium'
---

Array

### Example 1:
- Input: `grid = [ ["1","1","1","1","0"], ["1","1","0","1","0"], ["1","1","0","0","0"], ["0","0","0","0","0"] ]`
- Output: `1`

### Example 2:
- Input: `grid = [ ["1","1","0","0","0"], ["1","1","0","0","0"], ["0","0","1","0","0"], ["0","0","0","1","1"] ]`
- Output: `3`

### Constraints:

- `m == grid.length`
- `n == grid[i].length`
- `1 <= m, n <= 300`
- grid[i][j] is '0' or '1'.

## Solution

```py
class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        rows, cols = len(grid), len(grid[0])

        def dfs(r, c):
            if r < 0 or r >= rows or c < 0 or c >= cols or grid[r][c] != "1":
                return
            else:
                grid[r][c] = "0"
                dfs(r, c + 1)  # right
                dfs(r + 1, c)  # bottom
                dfs(r, c - 1)  # left
                dfs(r - 1, c)  # top

        num_islands = 0
        for r in range(rows):
            for c in range(cols):
                if grid[r][c] == "1":
                    num_islands += 1
                    dfs(r, c)

        return num_islands
```