tags: dfs
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
// 16ms, 37%
class Solution {
bool dfsCheck(vector<vector<int>>& grid, int i, int j) {
if (i < 0 || j < 0 || i >= grid.size() || j >= grid[i].size())
return true;
if (grid[i][j] != 0)
return true;
// 0 on the edges
if (i == 0 || j == 0 || i == grid.size() - 1 || j == grid[i].size() - 1)
return false;
// mark as visited
grid[i][j] = -1;
bool res = true;
const int dirs[4][2] = { {0, -1}, {1, 0}, {0, 1}, {-1, 0}};
for (const auto& d : dirs) {
if (!dfsCheck(grid, i + d[0], j + d[1])) {
res = false;
}
}
return res;
}
public:
int closedIsland(vector<vector<int>>& grid) {
int res = 0;
for (int i = 1; i < grid.size() - 1; i++) {
for (int j = 1; j < grid[i].size() - 1; j++) {
if (grid[i][j] == 0) {
res += dfsCheck(grid, i, j);
}
}
}
return res;
}
};