Leetcode 200. 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'.

This problem can be solved by many ways. My approach is searching via BFS and modifying the grid to mark as visited to avoid duplicated visit.

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        island = 0
        stack = []
        directions = [(1, 0), (0, 1), (-1, 0), (0, -1)]
        m = len(grid)
        n = len(grid[0])

        for i in range(m):
            for j in range(n):
                if grid[i][j] == '1':
                    stack.append((i, j))

                    while stack:
                        check_i, check_j = stack.pop()

                        grid[check_i][check_j] = 'x'

                        for direction in directions:
                            new_i = check_i + direction[0]
                            new_j = check_j + direction[1]

                            if 0 <= new_i < m and 0 <= new_j < n:
                                if grid[new_i][new_j] == '1':
                                    stack.append((new_i, new_j))
                    
                    island += 1
        
        return island