1254. Number of Closed Islands

Medium

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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

Constraints:

• 1 <= grid.length, grid[0].length <= 100

• 0 <= grid[i][j] <=1

class Solution:
    def closedIsland(self, grid: List[List[int]]) -> int:
        count = 0
        for row in range(len(grid)):
            for col in range(len(grid[0])):
                if grid[row][col] == 0 and self.dfs(grid, row, col):
                    count += 1
        return count
    
    def dfs(self, grid, row, col):
        if row < 0 or col < 0 or row >= len(grid) or col >= len(grid[0]):
            return False
        # Reached 1's Boundary
        if grid[row][col] == 1:
            return True
        # Prevent Recursion Error by marking it as 1 as visited
        grid[row][col] = 1
        # Previous Line that gave Wrong Answer
        #return self.dfs(grid, row-1, col) and self.dfs(grid, row, col-1) and self.dfs(grid, row+1, col) and self.dfs(grid, row, col+1)
        dfs1 = self.dfs(grid, row-1, col) 
        dfs2 = self.dfs(grid, row, col-1)
        dfs3 = self.dfs(grid, row+1, col)
        dfs4 = self.dfs(grid, row, col+1)
        # Traverese all direction before making decision
        return dfs1 and dfs2 and dfs3 and dfs4