1254. Number of Closed Islands

[chengchengxu15]

leetcode:

Given a 2D grid consists of 0s (land) and 1s (water). An island is a maximal 4-directionally connected group of 0s and a closed island is an island totally (all left, top, right, bottom) surrounded by 1s.

Return the number of closed islands.

[chengchengxu15]

same as 200, just need to delete the lands in sides. https://github.com/chengchengxu15/leet_code/issues/46 3 solutions.

my solution:

class Solution:
    def closedIsland(self, grid: List[List[int]]) -> int:
        m = len(grid)
        if m == 0:return 0
        n = len(grid[0])
        res = 0
        def delete_one_land(cell,grid,m,n):
            list_cell_for_this_land = [cell]
            while list_cell_for_this_land:
                cell = list_cell_for_this_land.pop()
                i,j = cell
                if grid[i][j] == 1:
                    continue
                grid[i][j] = 1

                if i > 0 and  grid[i-1][j] == 0:list_cell_for_this_land.append((i-1,j))
                if j > 0 and  grid[i][j-1] == 0:list_cell_for_this_land.append((i,j-1))
                if i < m-1 and  grid[i+1][j] == 0:list_cell_for_this_land.append((i+1,j))
                if j < n-1 and  grid[i][j+1] == 0:list_cell_for_this_land.append((i,j+1))
        for i in (0,m-1):
            for j in range(n):
                if grid[i][j] == 0:
                    cell = (i,j)
                    delete_one_land(cell,grid,m,n)

        for j in (0,n-1):
            for i in range(1,m-1):                
                if grid[i][j] == 0:
                    cell = (i,j)
                    delete_one_land(cell,grid,m,n)

        for i in range(1,m-1):
            for j in range(1,n-1):
                if grid[i][j] == 0:
                    res +=1
                    cell = (i,j)
                    delete_one_land(cell,grid,m,n)
        return res