LeetCode - Number of Islands

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'.

 

ChatGPT's Solution:

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

        def dfs(i, j):
            # If out of bounds or water ('0'), return
            if i < 0 or i >= m or j < 0 or j >= n or grid[i][j] == '0':
                return
            
            # Makr the current cell as visited
            grid[i][j] = '0'

            # Explore all four directions
            dfs(i + 1, j)
            dfs(i - 1, j)
            dfs(i, j + 1)
            dfs(i, j - 1)
        
        num_islands = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] == '1':  # Found an unvisited land cell
                    num_islands += 1
                    dfs(i, j)  # Mark all connected land cells as visited
        
        return num_islands