# 1254. Number of Closed Islands

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.

Example 1:

Input: grid = [[1,1,1,1,1,1,1,0],[1,0,0,0,0,1,1,0],[1,0,1,0,1,1,1,0],[1,0,0,0,0,1,0,1],[1,1,1,1,1,1,1,0]]
Output: 2
Explanation: 
Islands in gray are closed because they are completely surrounded by water (group of 1s).

Example 2:

Input: grid = [[0,0,1,0,0],[0,1,0,1,0],[0,1,1,1,0]]
Output: 1

Example 3:

Input: grid = [[1,1,1,1,1,1,1],
               [1,0,0,0,0,0,1],
               [1,0,1,1,1,0,1],
               [1,0,1,0,1,0,1],
               [1,0,1,1,1,0,1],
               [1,0,0,0,0,0,1],
               [1,1,1,1,1,1,1]]
Output: 2

# Solution

Approach 1: dfs -- count number of islands, except for islands connected to the borders of the grid.

# Code (Python)

Approach 1:

class Solution:
    def closedIsland(self, grid: List[List[int]]) -> int:
        def dfs(r, c):
            if grid[r][c] != 0:
                return 1
            grid[r][c] = -1
            if r == 0 or c == 0 or r == len(grid) - 1 or c == len(grid[0]) - 1:
                result = 0
            else:
                result = 1
            for delta in ((0, 1), (0, -1), (1, 0), (-1, 0)):
                if not(0 <= r + delta[0] < len(grid) and 0 <= c + delta[1] < len(grid[0])):
                    continue
                if grid[r + delta[0]][c + delta[1]] != 0:
                    continue
                if dfs(r + delta[0], c + delta[1]) == 0:
                    result = 0
            return result
        
        count = 0
        for r in range(len(grid)):
            for c in range(len(grid[0])):
                if grid[r][c] == 0:
                    count += dfs(r, c)
        return count

# Code (C++)

Approach 1:

Approach 2: