// 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> &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 dx = {1, -1, 0, 0}, dy = {0, 0, 1, -1}; void dfs(int x, int y, vector> &grid, vector> &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> &grid) { int res = 0, n = grid.size(), m = grid[0].size(); vector> vis(n, vector(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 }