number_of_islands.py

"""
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.

An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

 

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'.

"""

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        if not grid:
            return 0
            
        m = len(grid)
        n = len(grid[0])

        def dfs(i,j):
            if i < 0 or i >= m or j < 0 or j >= n or grid[i][j] != '1':
                return
            else:
                grid[i][j] = '0' # mark as visited
                dfs(i, j+1) # Go right
                dfs(i+1, j)  # Go down
                dfs(i, j-1)  # Go left
                dfs(i-1, j)  # Go up
        
        num_islands = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] == '1':
                    num_islands = num_islands + 1
                    dfs(i, j)
        
        return num_islands