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).

Idea: Start from the four sides of the 0 to the inside floodfill, which becomes 1, so that the rest is the island, and then count connect component, you can also use the fill function to turn 0 into 1;

class Solution {
    public int closedIsland(int[][] grid) {
        if(grid == null || grid.length == 0 || grid[0].length == 0) {
            return 0;
        }
        int n = grid.length;
        int m = grid[0].length;
        
        // from edge into grid, change all 0 to 1;
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                if(i == 0 || i == n - 1 || j == 0 || j == m - 1) {
                    if(grid[i][j] == 0) {
                        fill(grid, i, j);
                    }
                }
            }
        }
        
        // count connected 0 in grid;
        int count = 0;
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                if(grid[i][j] == 0) {
                    count++;
                    fill(grid, i, j);
                }
            }
        }
        return count;
    }
    
    private int[] dx = {0,0,-1,1};
    private int[] dy = {-1,1,0,0};
    
    private void fill(int[][] grid, int x, int y) {
        if(x < 0|| x >= grid.length || y < 0 || y >= grid[0].length || grid[x][y] == 1) {
            return;
        }
        grid[x][y] = 1;
        
        for(int k = 0; k < 4; k++) {
            int nx = x + dx[k];
            int ny = y + dy[k];
            fill(grid, nx, ny);
        }
    }
}