0200: Number of Islands

Problem Statement

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'.

Code Solution

from collections import deque

class Solution:
    def numIslands(self, grid):
        nR = len(grid)
        nC = len(grid[0])
        def bfs(rx, cx, vset):
            dq = deque([(rx, cx)])
            vset.add((rx, cx))
            while dq:
                rr, cc = dq.popleft()
                for dr, dc in [(0, 1), (1, 0), (-1, 0), (0, -1)]:
                    nr = rr + dr
                    nc = cc + dc
                    if 0 <= nr < nR and 0 <= nc < nC:
                        if grid[nr][nc] == "1" and (nr, nc) not in vset:
                            vset.add((nr, nc))
                            dq.append((nr, nc))
        count = 0
        cache = set()
        for ir in range(nR):
            for ic in range(nC):
                if grid[ir][ic] == "1" and (ir, ic) not in cache:
                    bfs(ir, ic, cache)
                    count = count + 1
        return count