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

Solution: DFS/Backtracking

For each connected component, if it can reach the boundary then it’s not a closed island.

Time complexity: O(n*m)
Space complexity: O(n*m)

// Author: Huahua
class Solution {
public:
  int closedIsland(vector<vector<int>>& grid) {    
    const int n = grid.size();
    const int m = grid[0].size();    
    function<int(int, int)> dfs = [&](int x, int y) {
      if (x < 0 || y < 0 || x >= m || y >= n) return 0;      
      if (grid[y][x]++) return 1;
      return dfs(x + 1, y) & dfs(x - 1, y) & dfs(x, y + 1) & dfs(x, y - 1);
    };
    
    int ans = 0;
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < m; ++j)
        if (!grid[i][j]) ans += dfs(j, i);      
    return ans;
  }
};

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