---
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
icon: dot
topics:
  - { name: "Array", slug: "array" }
  - { name: "Depth-First Search", slug: "depth-first-search" }
  - { name: "Breadth-First Search", slug: "breadth-first-search" }
  - { name: "Union-Find", slug: "union-find" }
  - { name: "Matrix", slug: "matrix" }
issue: "https://github.com/prdlk/leetcode/issues/130"
companies:
  - { name: "Anduril", period: "Thirty Days", frequency: 100 }
  - { name: "Apple", period: "Thirty Days", frequency: 81.7 }
  - { name: "Bloomberg", period: "Thirty Days", frequency: 69.8 }
  - { name: "Amazon", period: "Thirty Days", frequency: 65.8 }
  - { name: "Ola Cabs", period: "Thirty Days", frequency: 65.6 }
  - { name: "Google", period: "Thirty Days", frequency: 50.2 }
  - { name: "Qualcomm", period: "Three Months", frequency: 100 }
  - { name: "TikTok", period: "Three Months", frequency: 89.3 }
  - { name: "Uber", period: "Three Months", frequency: 72.9 }
  - { name: "Infosys", period: "Three Months", frequency: 66 }
  - { name: "Microsoft", period: "Three Months", frequency: 57.2 }
  - { name: "tcs", period: "Three Months", frequency: 47.7 }
  - { name: "Meta", period: "Three Months", frequency: 34.5 }
  - { name: "Autodesk", period: "Six Months", frequency: 100 }
  - { name: "ByteDance", period: "Six Months", frequency: 100 }
  - { name: "Rivian", period: "Six Months", frequency: 100 }
  - { name: "Snap", period: "Six Months", frequency: 100 }
  - { name: "Waymo", period: "Six Months", frequency: 100 }
  - { name: "Walmart Labs", period: "Six Months", frequency: 95.1 }
  - { name: "Visa", period: "Six Months", frequency: 80.9 }
  - { name: "Nvidia", period: "Six Months", frequency: 78.7 }
  - { name: "Oracle", period: "Six Months", frequency: 78.7 }
  - { name: "LinkedIn", period: "Six Months", frequency: 69.8 }
  - { name: "Yandex", period: "Six Months", frequency: 59.6 }
  - { name: "Goldman Sachs", period: "Six Months", frequency: 59 }
sidebar:
  label: 'Number of Islands'
  badge: 'Medium'
---

### 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
```