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+// 1568. Minimum Number of Days to Disconnect Island
+#include "leetcode.h"
+
+/*
+ * you are given an 'm * n' binary grid where 1 represents land and 0 represents
+ * water. an island is a maximal 4 directionally horizontal or vertical
+ * connected group of 1's. the grid is said to be connected if we have exactly
+ * one island otherwise it is disconnected. in one day, we are allowed to change
+ * any single land cell 1 into a water cell 0. return the minimum number of days
+ * to disconnect the grid.
+ */
+
+class Solution {
+public:
+ int minDays(vector<vector<int>> &grid) {
+ int islands = cnt_islands(grid), n = grid.size(), m = grid[0].size();
+ if (islands > 1 || !islands)
+ return 0;
+ else
+ for (int i = 0; i < n; i++)
+ for (int j = 0; j < m; j++)
+ if (grid[i][j]) {
+ grid[i][j] = 0;
+ islands = cnt_islands(grid);
+ grid[i][j] = 1;
+ if (islands > 1 || !islands)
+ return 1;
+ }
+ return 2;
+ }
+
+private:
+ vector<int> dx = {1, -1, 0, 0}, dy = {0, 0, 1, -1};
+ void dfs(int x, int y, vector<vector<int>> &grid, vector<vector<int>> &vis) {
+ int n = grid.size(), m = grid[0].size();
+ vis[x][y] = 1;
+ for (int i = 0; i < 4; i++) {
+ int nx = x + dx[i], ny = y + dy[i];
+ if (nx >= 0 and ny >= 0 and nx < n and ny < m and !vis[nx][ny] and
+ grid[nx][ny])
+ dfs(nx, ny, grid, vis);
+ }
+ }
+ int cnt_islands(vector<vector<int>> &grid) {
+ int res = 0, n = grid.size(), m = grid[0].size();
+ vector<vector<int>> vis(n, vector<int>(m, 0));
+ for (int i = 0; i < n; i++)
+ for (int j = 0; j < m; j++)
+ if (!vis[i][j] && grid[i][j]) {
+ dfs(i, j, grid, vis);
+ res++;
+ }
+ return res;
+ }
+};
+
+int main() {
+ Solution obj;
+ vvd g1 = {{0, 1, 1, 0}, {0, 1, 1, 0}, {0, 0, 0, 0}}, g2 = {{1, 1}};
+ printf("%d\n", obj.minDays(g1)); // expect: 2
+ printf("%d\n", obj.minDays(g2)); // expect: 2
+}