algorithm_practice/number_of_islands.py at main · cris-j-dev/algorithm_practice · GitHub

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"""
200. Number of Islands
Difficulty:Medium

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'.
"""

from typing import List

class Solution:
    def dfs(self, grid: List[List[str]], i: int, j: int):

        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'

        self.dfs(grid, i+1, j)
        self.dfs(grid, i-1, j)
        self.dfs(grid, i, j+1)
        self.dfs(grid, i, j-1)

    def numIslands(self, grid: List[List[str]]) -> int:
        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


if __name__ == "__main__":
    solution = Solution()
    print(solution.numIslands(grid = [
            ["1","1","1","1","0"],
            ["1","1","0","1","0"],
            ["1","1","0","0","0"],
            ["0","0","0","0","0"]
        ])) # 1

    print(solution.numIslands(grid = [
            ["1","1","0","0","0"],
            ["1","1","0","0","0"],
            ["0","0","1","0","0"],
            ["0","0","0","1","1"]
        ])) # 3